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Continuous practice using S Chand Class 10 Maths Solutions ICSE Chapter 12 Similar Triangles Ex 12(b) can lead to a stronger grasp of mathematical concepts.

S Chand Class 10 ICSE Maths Solutions Chapter 12 Similar Triangles Ex 12(b)

Question 1.
In each question, of the three triangles, name the one that is different, i.e., not similar to the other two.

Solution:
In ∆ (a), third angle = 180° – (75° + 55°) = 180° – 130° = 50°
∴ Three angles are 75°, 55°, 50°
In ∆ (b), third angle is = 180° – (55° + 80°) = 180° – 135° = 45°
Three angles are 55°, 80°, 45°
In ∆ (c), third angle is = 180° – (45° + 80°) = 180°- 125° = 55°
Three angles are 45°, 80°, 55°
We see that triangle (a) is different

(ii) In ∆ (a), third angle is = 180° – (100°+ 45°) = 180° – 145° = 35°
∴ Three angles will be 100°, 45°, 35°
In ∆ (b), third angle is = 180° – (100° + 35°) = 180° + 35° = 45°
∴ Three angles will be = 100°, 35°, 45°
In ∆ (c), third angle is = 180° – (25° + 45°) = 180° – 70° = 110°
∴ Three angles will be 110°, 25°, 45°
We see that triangle (c) is different

(iii) In ∆ (a), two sides are equal
In ∆ (b), no two sides are equal
In ∆ (c), two sides are equal
∴ ∆ (b) is different

Question 2.
Write down the ratio of the corresponding sides for each pair of triangles and check that it is the same.

Solution:
(i) In the first pair of triangle
The ratio in corresponding sides are
\(\frac{A B}{Y Z}=\frac{B C}{X Y}=\frac{A C}{X Z}\)
⇒ \(\frac{6}{9}=\frac{4.5}{6.75}=\frac{7}{10.5}\)
⇒ \(\frac{2}{3}=\frac{450}{675}=\frac{70}{105}\)
⇒ \(\frac{2}{3}=\frac{2}{3}=\frac{2}{3}\)
Yes, it is the same ratio
\(\frac{A B}{Y Z}=\frac{B C}{X Y}=\frac{A C}{X Z}=\frac{2}{3}\)

(ii) In the second pair of triangles
The ratio in the corresponding sides are
\(\frac{\mathrm{LM}}{\mathrm{RP}}=\frac{\mathrm{MN}}{\mathrm{RQ}}=\frac{\mathrm{LN}}{\mathrm{PQ}}\)
⇒ \(\frac{16}{2.4}=\frac{8}{1.2}=\frac{14}{2.1}\)
⇒ \(\frac{16 \times 10}{24}=\frac{8 \times 10}{12}=\frac{14 \times 10}{21}\)
⇒ \(\frac{20}{3}=\frac{20}{3}=\frac{20}{3}=\frac{2}{3}\) (Dividing by 10)
Yes, it is the same ratio
∴ \(\frac{\mathrm{LM}}{\mathrm{RP}}=\frac{\mathrm{MN}}{\mathrm{RQ}}=\frac{\mathrm{LN}}{\mathrm{PQ}}=\frac{2}{3}\)

Question 3.
In ∆ABCD and E are points on the sides AB and AC respectively such that DE || BC.
(i) If \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{4}{5}\) and EC = 2.5 cm, find AE.
(ii) If AD = 4, AE = 8, DB = x – 4, and EC = 3x – 19, find x.
(in) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the vlaue of x.
(iv) If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC.
(v) If \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{2}{3}\) and AC = 18 cm, find AE.
(vi) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.
(vii) If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.
Solution:
In ∆ABC, D and E are points on AB and AC respectively and DE || BC
(i) \(\frac { AD }{ BD }\) = \(\frac { 4 }{ 5 }\), EC = 2.5 cm

In ∆ABC,
∵ DE || BC
∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}} \Rightarrow \frac{4}{5}=\frac{\mathrm{AE}}{2.5}\)
⇒ AE = \(\frac{4 \times 2.5}{5}\) = 2 cm
∴ AE = 2 cm

(ii) In the figure, in ∆ABC, DE || BC

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
AD = 4, AE = 8, DB = x – 4 and EC = 3x – 19
⇒ \(\frac{4}{x-4}=\frac{8}{3 x-19}\)
⇒ 12x – 76 = 8x – 32 (By cross multiplication)
⇒ 12x – 8x = – 32 + 76
⇒ 4x = 44 ⇒ x = \(\frac { 44 }{ 4 }\) = 11
∴ x = 11

(iii) In ∆ABC, DE || BC

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
AD = x, DB = x – 2, AE = x + 2 and EC = x – 1
∴ \(\frac{x}{x-2}=\frac{x+2}{x-1}\)
⇒ x (x – 1) = (x + 2) (x – 2) (By cross multiplication)
⇒ x² – x = x² – 4
⇒ – x = – 4
⇒ x = 4
∴ x = 4

(iv) In ∆ABC, DE || BC

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
AD = 6 cm, DB = 9 cm, AE = 8 cm
∴ \(\frac{6}{9}=\frac{8}{\mathrm{EC}} \Rightarrow \mathrm{EC}=\frac{9 \times 8}{6}\)
⇒ EC = 12 cm
∴ AC = AE + EC = 8 + 12 = 20 cm

(v) In ∆ABC, DE || BC

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
\(\frac { AD }{ DB }\) = \(\frac { 2 }{ 3 }\), AC = 18 cm
Let AE = x cm
∴ EC = AC – AE = (18 -x) cm
∴ \(\frac { 2 }{ 3 }\) = \(\frac { x }{ 18-x }\) ⇒ 3x = 36 – 2x (By cross multiplication)
⇒ 3x + 2x = 36 ⇒ 5x = 36
⇒ x = \(\frac { 2 }{ 3 }\) = 7.2
∴ AE = 7.2

(vi) In ∆ABC, DE || BC

∴ \(\frac { AD }{ DB }\) = \(\frac { AE }{ EC }\)
AD = 8 cm, AB = 12 cm, AE = 12 cm DB = AB – AD = 12 – 8 = 4 cm
\(\frac { 8 }{ 4 }\) = \(\frac { 12 }{ CE }\) ⇒ CE = \(\frac { 4×12 }{ 8 }\) = 6
∴ CE = 6 cm

(vii) In ∆ABC, DE || BC

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
AD = 2 cm, AB = 6 cm, AC = 9 cm
DB = AB – AD = 6 – 2 = 4 cm
Let AE = x, then EC = AC – AE = 9 – x
∴ \(\frac{2}{4}=\frac{x}{9-x}\) ⇒ 4x = 18 – 2x
⇒ 4x + 2x = 18
⇒ 6x = 18
⇒ x = \(\frac { 18 }{ 6 }\) = 3
∴ AE = 3 cm

Question 4.
In a ∆ABC, D and E are points on the sides AB and AC respectively, for each of the following cases, show that DE || BC.
(i) AD = 3, BD = 4.5, AE = 4, CE = 6.
(ii) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.
(iii) AB = 5.6 cm, AD = 1.4 cm, AC 7.2 cm and AE = 1.8 cm.
Solution:
(i) In ∆ABC, D and E are the points on the sides AB and AC respectively such that AD = 3, BD = 4.5, AE = 4 and CE = 6

(ii) In ∆ABC, D and E are the points on the sides AB and AC respectively such that AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm

∴ DB = AB – AD = 12 – 8 = 12 – 8 = 4 cm
EC = AC – AE = 18 – 12 = 6 cm
Now \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{8}{4}=\frac{2}{1}\)
and \(\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{12}{6}=\frac{2}{1}\)
∵ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
∴ DE || BC

(iii) In ∆ABC, D and E are the points on the sides AB and AC respectively such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

∴ DB = AB – AD = 5.6 – 1.4 = 4.2 cm
EC = AC – AE = 7.2 – 1.8 = 5.4 cm
Now \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{1.4}{4.2}=\frac{1}{3}\)
and \(\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{1.8}{5.4}=\frac{1}{3}\)
∵ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
∴ DE || BC

Question 5.
In the figure, ABC and DEF and similar triangles, find the values of x and y, if the sides are as marked.

Solution:
In the figure, ∆ABC and ∆DEF are similar
AB = x, AC = 7, BC = y
DE = 5, DF = 3 and EF = 8
∵ ∆s are similar
Corresponding sides are proportional
∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)
\(\frac{x}{5}=\frac{7}{3}=\frac{y}{8}\)
\(\frac{x}{5}=\frac{7}{3} \Rightarrow x=\frac{5 \times 7}{3}=\frac{35}{3}=11 \frac{2}{3}\)
and \(\frac{y}{8}=\frac{7}{3} \Rightarrow y=\frac{8 \times 7}{3}=\frac{56}{3}=18 \frac{2}{3}\)

Question 6.
In the figure, ∆ABO ~ ADCO. If CD = 2 cm, AB = 3 cm, OC = 3.2 cm, OD = 2.4 cm, determine OA and OB.

Solution:
∵ ∆ABO ~ ADCO
∴ Their corresponding sides are proportional
∴ \(\frac{\mathrm{AB}}{\mathrm{CD}}=\frac{\mathrm{AO}}{\mathrm{OD}}=\frac{\mathrm{BO}}{\mathrm{OC}}\)
But AB = 3 cm, CD = 2 cm, OC = 3.2 cm and OD = 2.4 cm

Hence OA = 3.6 cm and OB = 4.8 cm

Question 7.
In the figure, ∆ABC ~ ∆ADE. If AD : DB = 2:3 and DE = 5 cm. (i) find BC (ii) if x be the length of the perpendicular from A to DE, find the length of the perpendicular from A to BC in terms of x.

Solution:
In the figure,

Question 8.
In the figure, ABCD is a trapezium with AB parallel to DC. Given that AB = 4 cm, BC = 3 cm and CD = 6 cm.
(i) Name two triangles in the figure which are similar.
(ii) Calculate the length of EB.

Solution:
In trapezium ABCD, AB || DC
AB = 4 cm, BC = 3 cm and CD = 6 cm

(i) In the figure,
∆EDC and ∆EAB
∠E = ∠E (common)
∠EDC = ∠EAB (corresponding angles)
∠ECD = ∠EBA (corresponding angles)
∴ ∆EDC ~ ∆EAB

(ii) Let EB = x, then EC = x + 3
∵ ∆EAB ~ ∆EDC
∴ \(\frac{E B}{E C}=\frac{A B}{D C}\)
⇒ \(\frac{x}{x+3}=\frac{4}{6}\) ⇒ 6x = 4x + 12
⇒ 6x – 4x = 12 ⇒ 2x = 12 12
⇒ x = \(\frac { 12 }{ 2 }\) = 6
∴ EB = 6 cm

Question 9.
A man of height 1.8 m is standing 5 m away from a lamp post and observes that the length of his shadow is 1.5 m. Find the height of the lamp post.
Solution:
∵ The mean and the lamp post are perpendicular on the ground
Height of man PQ = 1.8 m
and his shadow QN = 1.5 m

Let LM be the lamp post and LM = x m
Now in ∆LMN
∵ PQ || LM
(both perpendiculars on the same line)
∴ ∆PNQ ~ ∆LNM
∵ \(\frac{\mathrm{PQ}}{\mathrm{LM}}=\frac{\mathrm{NQ}}{\mathrm{NM}}\)
⇒ \(\frac{1.8}{x}=\frac{1.5}{1.5+5} \Rightarrow \frac{1.8}{x}=\frac{1.5}{6.5}\)
⇒ x = \(\frac{1.8 \times 6.5}{1.5}\)
∴ Height of the lamp post = 7.8 m

Question 10.
In figure, ACE and BCD are two straight lines.

∠A = 40° and ∠B = 85°. Using the measurements given in the figure, complete the following true statements
(i) Triangles ABC and CDE are similar because ……… and ………
(ii) The size of ∠D is ………
(iii) If AB = x cm, then ED = ………
Solution:
Two lines ACE and BCD intersect each other at C
AC = 4 cm, CE = 6 cm
BC = 3 cm and CD = 8 cm
(i) In ∆ABC and ∆CDE
\(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4} \text { and } \frac{\mathrm{CE}}{\mathrm{CD}}=\frac{6}{8}=\frac{3}{4}\)
∠ACB = ∠DCE (vertically opposite angles)
∴ ∆ABC ~ ACDE because \(\frac{B C}{A C}=\frac{C E}{C D}=\frac{3}{4}\) and ∠ACB = ∠DCE

(ii) ∠D = ∠A = 40° (corresponding angles)

(iii) AB = x cm
and \(\frac{\mathrm{AC}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{4}{8}=\frac{1}{2}\)
⇒ \(\frac{x}{\mathrm{DE}}=\frac{1}{2}\)
⇒ DE = 2x
∴ DE = 2x

Question 11.
The triangle ABC is right-angled at C. From P, a point on the hypotenuse, PQ is drawn parallel to AC cutting BC at Q. If AC 2.5 cm, BC = 6 cm and PQ = 1 cm, find
(i) BQ
(ii) BP
Solution:
In right angle ∆ABC, ∠C = 90
P is any point on AB

From P, PQ is drawn parallel to AC which meets BC at Q
AC = 2.5 cm, BC = 6 cm, PQ = 1 cm
In ∆ABC
∵ PQ || AC

(i) ∴ ∆ABC ~ APBQ
\(\frac{\mathrm{AC}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{BQ}} \Rightarrow \frac{2.5}{1}=\frac{6}{\mathrm{BQ}}\)
⇒ BQ = \(\frac{6 \times 1}{2.5}=\frac{6}{2.5}\) = 2.4 cm

(ii) In right ∆PQB (∵ ∠Q = ∠C = 90°)
BP² = PQ² + BQ²
= (1)² + (2.4)²
= 1 + 5.76
= 6.75
= (2.6)²
BP = 2.6 cm

Question 12.
State whether the following statement is true or false, briefly giving the reasons. If two angles of one triangle are 72° and 80° respectively and that of another triangle are 28° and 72° respectively, then the triangles are similar.
Solution:
Let in one triangle ABC,
∠A = 72°, ∠B = 80°
Then ∠C =180°- (72° + 80°) = 180° – 152° = 28°
and in the other triangle PQB
∠P = 28°, ∠Q = 72°
∴ ZR = 180° – (28° + 72°) = 180° – 100° = 80°
∵ ∠A = ∠Q, ∠B = ZR and ∠C = ∠P
∴ ∆ABC ~ APQR (AAA axiom)

Question 13.
In figure, ∠ADE = ∠B, show that ∆ABC ~ ∆ADE. If AD = 2.7 cm, AE = 2.5 cm, BE = 1.1 cm and BC = 5.2 cm, find DE.

Solution:
In the figure, AD = 2.7 cm, AE = 2.5 cm, BE = 1.1 cm and BC = 5.2 cm
(i) In ∆ABC and ∆ADE
∠A = ∠A (common)
∠ABC=∠ADE (given)
∴ ∆ABC ~ ∆ADE (AA axiom)

(ii) ∴ \(\frac{B C}{D E}=\frac{A B}{A D}=\frac{A C}{A E}\)
(corresponding sides are proportional)
⇒ \(\frac{5.2}{\mathrm{DE}}=\frac{(1.1+2.5)}{2.7}\)
⇒ \(\frac{5.2}{\mathrm{DE}}=\frac{3.6}{2.7} \Rightarrow \mathrm{DE}=\frac{5.2 \times 2.7}{3.6}\) = 3.9
Hence DE = 3.9 cm

Question 14.
In figure, AB, EF and CD are parallel lines. Given that AB = 15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate EF and AC.

Solution:
In the figure, ABC is a triangle AB, EF and CD are parallel line
AB = 15 cm, EG = 5 cm, BC = 10 cm and DC =18 cm
We have to find EF and AC
(i) In ∆EFG and ∆CGD
∠EGF = ∠CGD
(vertically opposite angles)
∠EFG = ∠GDC (alternate angles)
∴ ∆EFG ~ ∆CGD (∆A axiom)
∴ \(\frac{E G}{C G}=\frac{E F}{C D}\)
(corresponding sides are proportional)
⇒ \(\frac{5}{10}=\frac{\mathrm{EF}}{18} \Rightarrow \mathrm{EF}=\frac{5 \times 18}{10}\) = 9
∴ EF = 9 cm

(ii) Similarly in ∆ABC and AECF
∠C = ∠C (common)
∠CAB = ∠CEF (corresponding angles)
∆ABC ~ AECF (AA axiom)
∴ \(\frac{A C}{E C}=\frac{A B}{E F} \Rightarrow \frac{A C}{10+5}=\frac{15}{9} \Rightarrow \frac{A C}{15}=\frac{15}{9}\)
⇒ AC = \(\frac{15 \times 15}{9}=\frac{225}{9}\) = 25
∴ AC = 25 cm

Question 15.
Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4, 7 and 8 cm.
Solution:
Let ∆ABC ~ ∆PQR and let
in ∆PQR, PQ = 4 cm, QR = 7 cm and RP = 8 cm
and let in ∆ABC, AB = 6 cm, BC = x and CA = y
Then \(\frac{P Q}{A B}=\frac{Q R}{B C}=\frac{R P}{C A}\) (∵ As are similar)
⇒ \(\frac{4}{6}=\frac{7}{x}=\frac{8}{y}\)
Now \(\frac{7}{x}=\frac{4}{6}\) ⇒ x = \(\frac{7 \times 6}{4}=\frac{21}{2}\) = 10.5
and \(\frac{8}{y}=\frac{4}{6} \Rightarrow y=\frac{6 \times 8}{4}\) = 12
∴ Other two corresponding sides are 10.5 cm and 12 cm

Question 16.
In ∆ABC, DE || BC
(a) If AD = 3 cm, DB = 4 cm, EC = 12 cm, findAE.
(b) If AE = 2.7 cm, EC = 4.5 cm, AD = 2.4 cm, find BD.
(c) If AD = 2.6 cm, DB = 6.5 cm and AE = 3 cm, find EC.
(d) If AB = 6 cm, AD = 2 cm and AC = 9 cm, calculate the length of CE.

Solution:
(a) In ∆ABC, DE || CB
∴ ∆ABC ~ ∆ADE
Now AD = 3 cm, DB = 4 cm, EC = 12 cm,
Let AE = x cm
∵ In ∆ABC, DE || BC
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \Rightarrow \frac{3}{4}=\frac{\mathrm{AE}}{12}\)
⇒ \(\frac{3}{4}=\frac{x}{12} \Rightarrow x=\frac{3 \times 12}{4}\) = 9
∴ AE = 9 cm

(b) In ∆ABC, DE || BC
∴ \(\frac{A D}{D B}=\frac{A E}{E C}\)
⇒ \(\frac{2.4}{\mathrm{BD}}=\frac{2.7}{4.5}\)
⇒ BD = \(\frac{2.4 \times 4.5}{2.7}\)
∴ BD = 4 cm

(c) In ∆ABC, DE || BC
∴ \(\frac{A D}{D B}=\frac{A E}{E C}\)
⇒ \(\frac{2.6}{6.5}=\frac{3}{\mathrm{EC}} \Rightarrow \mathrm{EC}=\frac{6.5 \times 3}{2.6}\)
⇒ EC = \(\frac { 15 }{ 2 }\) = 7.5
∴ EC = 7.5 cm

(d) In ∆ABC, DE || BC
∴ ∆ABC ~ ∆ADE
∴ \(\frac{A D}{D B}=\frac{A E}{A C}\)
⇒ \(\frac{2}{6}=\frac{A E}{9}\)
⇒ AE = \(\frac { 2×9 }{ 6 }\) = 3
∴ CE = AC – AE = 9 – 3 = 6 cm

Effective S Chand Class 10 Maths Solutions ICSE Chapter 12 Similar Triangles Ex 12(a) can help bridge the gap between theory and application.

S Chand Class 10 ICSE Maths Solutions Chapter 12 Similar Triangles Ex 12(a)

Question 1.
Draw rectangular axes. Scale them from 0 to 18. Plot the points:
A (3, 4), B (9, 4), C (9, 6), D (3, 6). Join the points in alphabetical order to form a rectangle. Transform each of the point (x, y) on the points (x₁, y₁) by the transformation (x, y) → (2x, 2y)
Lable image A₁B₁C₁D₁ to correspond with ABCD.
(i) What do you notice about the length of each side of the image when you compare it with the corresponding side of the original rectangle?
(ii) Are the rectangles similar in shape?
(iii) Write the ratio \(\frac{\text { Area of image } A_1 B_1 C_1 D_1}{\text { Area of original } A B C D}\) in the lowest terms.
Solution:

Plot the points A (3, 4), B (9, 4), C (9, 6), D (3, 6) on the graph as shown
Join them to form rectangle ABCD
The ratio scale of transformation is (x, y) → (2x, 2y) i.e. 1 : 2
Now plot the points A’ (6, 8), B’ (18, 8), C’ (18, 12) and D’ (6, 12) on the same graph and join them to form another rectangle A’B’C’D’

• Length of the sides of the image of rectangle ABCD is 1 : 2 i.e. double of the original rectangle
• Yes the rectangles are similar and ratio is 1 : 2
• Ratio in area = \(\frac{\text { Area of image } A_1 B_1 C_1 D_1}{\text { Area of original } A B C D}=\frac{12 \times 4}{6 \times 2}=\frac{48}{12}=\frac{4}{1}\)
  i.e. 4 : 1

Question 2.
(a) What would be the effect (i) on the length of sides, (ii) on the area of the transformation (x, y) → (3x, 3y) ?
Label the image A₁B₁C₁D₁ to correspond with ABCD.
(b) Which transformation would make the area of the image sixteen times the area of original rectangle?
(c) Which transformation would made the length of each side of the image half the length of the corresponding side of the original rectangle ?
Solution:
(i) ∵ (x, y) → (3x, 3y)
∴ The image A₁B₁C₁D₁ is three times in length of the figure ABCD

(ii) ∵ The area of the image is 16 times the area of the original rectangle
∴ (x, y) → (4x, 4y) (∵ \(\sqrt{16}\) = 4)

(iii) In the given figure,
The image is half the length of the corres-ponding sides of the original figure
∴ (x, y) → (\(\frac { 1 }{ 2 }\)x, \(\frac { 1 }{ 2 }\)y)

Question 3.
In the figure, O is the centre of dilatation and the dilatation factor is 3.
(i) Complete

Preimage	Image
1. AB = 5 units	A’B’ = ?
2. BC = ?	B’C’ = 18 units
3. AC = 2 units	A’C’ = ?
4. OA = 7 units	OA’ = ?
5. OC = ?	OC’ = 39 units
6. OB = m units m ∈ R	OB’ = ?

(ii) Determine:
(a) \(\frac{\mathrm{OA}^{\prime}}{\mathrm{OA}}\)
(b) \(\frac{\mathrm{OB}^{\prime}}{\mathrm{OB}}\)
(c) \(\frac{\mathrm{OC}^{\prime}}{\mathrm{OC}}\)
(d) \(\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}\)
(e) \(\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}\)
(f) \(\frac{C^{\prime} A^{\prime}}{C A}\)
Solution:
O is the centre of dilatation and the dilatation factor is 3
i.e. (x, y) → (3x, 3y)

Preimage	Image
1. AB = 5 units	A’B’ = 3 x 5 = 15 units
2. BC = \(\frac { 1 }{ 3 }\) x 18 = 6 units	B’C’ = 18 units
3. AC = 2 units	A’C’ = 3 x 2 = 6 units
4. OA = 7 units	OA’ = 3 x 7 = 21 units
5. OC = \(\frac { 1 }{ 3 }\) x 39 = 13 units	OC’ = 39 units
6. OB = m units m ∈ R	OB’ = m x 3 = 3 m units m ∈ R

(ii) ∵ (x, y) → (3x, 3y)
(a) \(\frac{\mathrm{OA}^{\prime}}{\mathrm{OA}}\) = \(\frac { 3 }{ 1 }\)
(b) \(\frac{\mathrm{OB}^{\prime}}{\mathrm{OB}}\) = \(\frac { 3 }{ 1 }\)
(c) \(\frac{\mathrm{OC}^{\prime}}{\mathrm{OC}}\) = \(\frac { 3 }{ 1 }\)
(d) \(\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}\) = \(\frac { 3 }{ 1 }\)
(e) \(\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}\) = \(\frac { 3 }{ 1 }\)
(f) \(\frac{C^{\prime} A^{\prime}}{C A}\) = \(\frac { 3 }{ 1 }\)

Question 4.
A triangle ABC has its vertices at (1, 1), (1, 3) and (3, 1) respectively. Enlarge this triangle by using a scale factor of 2 with the origin as the centre of enlargement.
Solution:
The coordinates of the vertices of a triangle are A (1, 1), B (1, 3) and C (3, 1) respectively
Enlargement of scale factor is 2 i.e. (x, y) → (2x, 2y)

∴ Coordinates of enlargement triangle will be
A’ (2, 2), B’ (2, 6) and C’ (6, 2) respectively. O is the origin (centre) of the enlargement triangle whose vertices are A’ (2, 2), B’ (2, 6) and C’ (6, 2)

Question 5.
The coordinates of vertices being given, enlarge the given figure by using the given scale factor. For each question you may use a grid.

Coordinates of vertices	Scale factor	Scale factor
(a) P (1, 2), Q (2, 2), R (2, 1)	4	4
(b) A (0, 0), B (3, 2), C (4, 0)	2	2
(c) A (5, 5), B (5, 7), C (7, 7), D (7, 5)	4	4
(d) A (5, 4) X (5, 8), Y (7, 8), Z (7, 4)	2	2

Solution:
(a) Coordinates of vertices are P (1, 2), Q (2, 2), R (2, 1)
Scale factor 1 : 4
and centre of enlargement is origin (0, 0)
∴ (x, y) → (4x, 4y)
∴ Coordinates of the enlarged figure will be P’ (4, 8), Q’ (8, 8) and R’ (8, 4)

(b) Coordinates of vertices are A (0, 0), B (3, 2), C (4, 0)
Scale factor is 1 : 2
(x, y) → (2x, 2y)
∴ Coordinates of enlarged figure will be A’ (0, 0), B’ (6, 4) and C’ (8, 0)

(c) Coordinates of figures are A (5, 5), B (5, 7), C (7, 7) and D (7, 5)
and enlargement factor is 1 : 4
∴ (x, y) → (4x, 4y)
and centre of enlargement is the point (6, 6)
∴ Co-ordinates of the vertices of enlarged figure will be A’ (2,2), B’ (10, 2), C’ (10, 10) and D’ (2,10)

(d) Coordinates of figure are A (5, 4), X (5, 8), Y (7, 8) and Z (7, 4)
and enlargement factor is 1 : 2 and centre of enlargement is the point (6, 6)
The vertices of the enlarged figure A’X’Y’Z’ will be A’ (4, 2), X’ (4, 10), Y’ (8, 10) and Z’ (8, 2)

Question 6.
Draw the image of the shape KLMN after an enlargement by scale factor \(\frac { 1 }{ 2 }\) with centre C. Label the image K’L’M’N’.
Solution:
Take CK = 6 cm and take its mid point K’ such that CK’ = 3 cm.
Join CK, LM, MC and NC.

Draw K’L’ || KL, K’N’ || KN, N’M’ || NM and L’M’ || LM.
Then figure, K’L’M’N’ is the image of figure KLMN by the scale factor \(\frac { 1 }{ 2 }\) with centre C.

Question 7.
Figure shows the enlargement transformation of shape ABCD to PQRS by a scale factor of 2. Find the centre of enlargement and state its coordinates.

Solution:
Join PA, QB, RC and SD and produce them. Which meet all of them at a point L. This is the required centre of enlargement and its co-ordiantes are (-2, 3).

Question 8.
A square with side 3 cm is drawn any where in a plane which is then enlarged about any point in the plane with a scale factor 2. What is the area of the image of the square?
Solution:
Side of a square = 3 cm
When it is enlarged, the scale factor = 2
i.e. k = 2
Area of the square = (side)² = (3)² = 9 cm²
∴ Area of the image of square = k² (Area of square)
= (2)² x 9 = 4 x 9 = 36 cm²

Question 9.
Construct a square ABCD of side 2 cm. Enlarge ABCD with enlargement factor m = 2 and label its image A’B’C’D’. What is the relation between the area ABCD and its image A’B’C’D’.
Solution:
Draw a square ABCD with each side 2 cm. Join AC and BD and produced them both sides. .
Take A’B’ = 4 cm (as m = 2) and complete the required square A’B’C’D’ with each side 4 cm.
Now area of square ABCD = 2 x 2 = 4 cm²
and area of square A’B’C’D’ with each side 2 x 2 = 4 cm
= 4 x 4 = 16 cm²
Hence, area of the image of square ABCD = 16 cm²

Question 10.
A triangle, whose area is 12 cm², is transformed under enlargement about a point in space. If the area of its image is 108 cm², find the scale factor of the enlargement.
Solution:
Area of the triangle = 12 cm²
Area of enlarged triangle = 108 cm² .
Let scale factor = k and then scale factor of area = k²
Then k² x 12 = 108 ⇒ k² = \(\frac { 108 }{ 12 }\) = 9
∴ k = \(\sqrt{9}\) = 3

Question 11.
The parallelogram ABCD has vertices (6, 3), (9, 3), (12, 9), (9, 9) respectively. Copy the grid and the parallelogram. An enlargement scale factor \(\frac { 1 }{ 3 }\) and centre (0, 0) transforms parallelogram ABCD onto parallelogram A’B’C’D’.

(a) (i) Draw the parallelogram A’B’C’D’.
(ii) Calculate the area of parallelogram A’B’C’D’.

(b) The side AB has length 3 cm. The original shape ABCD is now enlarged with a scale factor of \(\frac { 2 }{ 5 }\) to give A”B”C”D”. Calculate the length of the side A”B”.
Solution:
A parallelogram ABCD is given whose vertices are A (6, 3), B (9, 3), C (12, 3), D (9, 9) scale factor is \(\frac { 1 }{ 3 }\) and centre O (0, 0) transforms parallelogram ABCD on to ||gm A’B’C’D’.

(a) (i) Join OA, OB, OC and OD.
Take a point A’ on OA such that OA’ = \(\frac { 1 }{ 3 }\) OA
Similarly take points B’, C’ and D’
Join A’B’, B’C’, C’D’ and D’A’
A’B’C’D’ is the required || gm

(ii) Area of ||gm ABCD = A’B’ x Perpendicular between A’B’ and C’D’ =1 x 2 = 2 cm²

(b) Let AB = 3 cm
Now ||gm ABCD is enlarged with a scale factor

\(\frac { 2 }{ 5 }\) given new ||gm A”B”C”D”.
Length of side A”B” = 1.2 cm
Join OA, OB, OC and OD
Take a point A” (6 x \(\frac { 2 }{ 5 }\), 3 x \(\frac { 2 }{ 5 }\)) or (2.4, 1.2)
Similarly point B” (2.4, 1.2), C” (3.6, 3.6) and D” (4.8,3.6) and join A”B”, B”C”, C”D”, D”A” respectively.
A”B”C”D” is the required parallelogram.
On measuring length of A”B” = 1.2 cm.

Question 12.
A model of a van is constructed using a scale of 1 : 40. If the length of the real van is 8 metres, what is the length of the model?
Solution:
Scale in model of a van is the real van = 1 : 40
Length of the real van = 8 m
∴ Length of the model = \(\frac { 8×1 }{ 40 }\) m
= \(\frac { 8×100 }{ 40 }\) = 20cm

Question 13.
Akhil is drawing a plan of his classroom using a scale of 1 : 50. If the length and width of the classroom are 8 metres and 6 metres, how long will they each be on his drawing?
Solution:
Scale of the plan of a classroom = 1 : 50
Length of classroom = 8 m
and breadth = 6 m
∴ Length of the drawing of the room = \(\frac { 8×1 }{ 50 }\) x 100 cm = 16 cm
and breadth = \(\frac { 6×1 }{ 50 }\) x 100 = 12 cm

Question 14.
Nikhil is drawing a plan of the gymnasium at his school using a scale of 1 : 75. If the dimensions of the gymnasium on his plan are 28 cm and 16 cm, what are the dimensions of the real gymnasium?
Solution:
Scale of the plan of the gymnasium of the school = 1 : 75
In the plan,
Length = 28 cm and breadth = 16 cm
∴ Length of actual gymnasium = \(\frac { 28×75 }{ 1 }\) cm
= \(\frac { 28×75 }{ 100 }\) m = 21 m
and breadth = \(\frac { 16×75 }{ 100 }\) = 12 m
∴ Dimension of gymnasium = 21 m by 12 m

Question 15.
Krishna has a model of the Eiffel Tower which is 16 cm high. If the height of the real tower is 320 m, what is the scale of the model?
Solution:
Model of a Eiffel Tower = 16 cm high
But real height = 320 m
∴ Scale = 16 : 320 x 100
= 1 : 20 x 100
= 1 : 2000

Question 16.
The dimensions of the model of a multi-storey building are 1.2 m x 75 cm x 2 m. If the scale factor is 1 : 30, find the actual dimensions of the building.
Solution:
Dimension of the model of a multi-storey.
building are 1.2 m x 75 cm x 2 m
= 120 cm x 75 cm x 200 cm
Scale = 1 : 30, then
Actual length = \(\frac { 12×30 }{ 1 }\) cm
= \(\frac { 12×30 }{ 100 }\) = 36 m
Breadth = \(\frac { 375×30 }{ 100 }\) = \(\frac { 45 }{ 2 }\) = 22.5 m
Height = \(\frac { 20×30 }{ 100 }\) = 60 m
Hence, dimensions are 36 m x 22.5 m x 60 m

Question 17.
On a map drawn to a scale of 1 : 2500, a triangular plot of land has the following measurements: AB = 3 cm, BC = 4 cm and ∠ABC = 90°.
(i) Calculate the actual lengths of AB and AC in km;
(ii) The actual area of the plot in km².
Solution:
Scale of map = 1 : 2500
Dimension of a triangular plot are AB = 3 cm, BC = 4 cm and ∠ABC = 90°

(i) ∴ Actual length of AB
= \(\frac{3 \times 2500}{1}\)cm
= \(\frac{3 \times 2500}{100}\) = 75 m
and length of BC = \(\frac{4 \times 2500}{100}\) = 100 m

(ii) Actual area = \(\frac { 1 }{ 2 }\) x BC x AB
= \(\frac { 1 }{ 2 }\) x 75 x 100 m²
= 3750 m²
= \(\frac{3750}{1000 \times 1000}\) = 0003750 km²
= 0003750 km²

Question 18.
The scale of a map is 1 : 200000. A plot of land of area 20 km² is to be represented on the map; find :
(i) the number of kilometres on the ground which is represented by 1 centimetre on the map;
(ii) the area in km² that can be represented by 1 cm²;
(iii) the area on the map that represents the plot of land.
Solution:
It means that 1 cm on the map will represent 200000 cm 2 on the ground
(i) ∴ 1 cm = 20000 cm
= \(\frac{200000}{100 \times 1000}\) km = 2 km

(ii) ∵ 1 cm on the map = 2 km on the ground and 1 cm² = 2 x 2 = 4 km² on the ground

(iii) Area of a plot on the ground = 20 km²
Area of plot to be represented on the map will be = \(\frac { 20 }{ 4 }\) = 5 cm²

Question 19.
A model of a ship is made to a scale of 1 : 200.
(i) The length of the model is 4 m. Calculate the length of the ship.
(ii) The area of the deck of the ship is 160000 m². Find the area of the deck of the-model.
(iii) The volume of the model is 200 litres. Calculate the volume of the ship in m³.
Solution:
Scale of the model of a ship = 1 : 200
(i) Length of model = 4 m
∴ Actual length = \(\frac { 4×200 }{ 1 }\) = 800 m

(ii) Area of deck = 160000 m2
∴ Area on the model = \(\frac { 160000 }{ 200× }\) = 4 m²

(iii) Volume of model = 200 litres
= \(\frac { 200 }{ 1000 }\) = 0.2 m²
∴ Volume of actual ship = 1 m³ = 1000 l = 0.2 x (200)³ m³
= \(\frac { 2 }{ 10 }\) x 200 x 200 x 200 = 1600000 m³

Access to comprehensive S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(f) encourages independent learning.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(f)

Question 1.
In figure, PT is a tangent to a circle. If ∠BTA = 45° and ∠PTB = 75°, find ∠ABT.

Solution:
In the figure, PQ is the tangent to the circle at T
∠BTP = 75° and ∠ATB = 45°
But ∠QTA + ∠ATB + ∠BTP = 180° (Angles of a line)
∠QTA + 45° + 75° = 180° ⇒ ∠QTA + 120° = 180°
⇒ ∠QTA = 180° – 120° = 60°
Now QTP is the tangent and TA is the chord
∴ ∠QTA = ∠ABT (Angle in the alternate segment)
⇒ ∠ABT = 60°

Question 2.
In figure, TAS is a tangent to the circle, with centre O, at the point A. If ∠OBA = 32°, find the values of x and y.

Solution:
In the figure, a circle with centre O
TAS is a tangent to the circle at A
AB, AC and BC are chords of the circle
∠OBA = 32°
In △OAB,
OA = OB (radii of the same circle)
∴ ∠OAB = ∠OBA = 32°
In △AOB,
∠AOB + ∠OAB + ∠OBA = 180° (Sum of angles of a triangle)
⇒ ∠AOB + 32° + 32° = 180°
⇒ ∠AOB + 64° = 180°
⇒ ∠AOB = 180° – 64° = 116°
Now arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴ ∠AOB = 2 ∠ACB
⇒ 116° = 2y ⇒ y = \(\frac{116^{\circ}}{2}\) = 58°
But ST is the tangent and AB is the chord
∴ ∠BAS = ∠ACB
⇒ x = y ⇒ x = y = 58°

Question 3.
In figure, KLMN is a cyclic quadrilateral and PQ is a tangent to the circle at K. If LN is a diameter of the circle, ∠KLN = 30° and ∠MNL = 60°, determine
(i) ∠QKN,
(ii) ∠PKL,
(iii) ∠MLK.

Solution:
In the circle with centre O,
LN is the diameter
PQ is a tangent to the circle at K
∠KLN = 30° and ∠MNL = 60°
In △LKN,
∠LKN = 90° (Angle in a semicircle)
∠KLN = 30°
∴ ∠KNL = 90° – 30° = 60°

(i) Now PQ is tangent and KN is chord of the circle at K
∴ ∠QKN = ∠KLN = 30° (Angle in the alternate segment)

(ii) Again PQ is tangent and KL is the chord
∴ ∠PKL = ∠KNL = 60°

(iii) In cyclic quad LMNK,
∠KNM + ∠KLM = 180° (sum of opposite angles)
⇒ ∠KNL + ∠LNM + ∠MLK = 180°
⇒ 60° + 60° + ∠MLK = 180°
⇒ ∠MLK + 120° = 180°
⇒ ∠MLK = 180°- 120°
⇒ ∠MLK = 60°

Question 4.
In figure, AT is a tangent to the circle. If ∠ABC = 50° and AC = BC, find ∠BAT.

Solution:
In the figure, AT is the tangent to the circle and ZABC = 50°
AC = BC
∴ ∠BAC = ∠ABC = 50° (opposite of equal sides)
But ∠ACB + ∠BAC + ∠ABC = 180° (sum of angles of a triangle)
ZACB+ 50°+ 50°= 180°
⇒ ∠ACB + 100°= 180°
⇒ ∠ACB = 180° – 100° = 80°
∵ AT is the tangent and AB is chord
∠BAT = ∠ACB = 80° (Angles in the alternate segment)

Question 5.
In figure, O is the centre of the circumcircle of △XYZ. Tangents at X and Y intersects at T. Given, ∠XTY = 80° and ∠XOZ = 140°. Calculate ∠ZXY.

Solution:
In the figure, O is the centre of circumcircle of △XYZ.
Tangents at X and Y are drawn to meet at T
∠XTY = 80° and ∠XOZ = 140°
Join OY
XT and YT are tangents to the circle
∴ ∠XTY + ∠XOY = 180° (Angles are supplementary)

⇒ 80° + ∠XOY = 180°
⇒ ∠XOY =180° – 80° = 100°
But ∠XOY + ∠YOZ + ∠ZOX = 360° (Angles at a point)
⇒ 100° +∠YOZ + 140° = 360°
⇒ ∠YOZ + 240° = 360°
∠YOZ = 360° – 240° = 120°
Now arc YZ subtends ∠YOZ at the centre and ∠ZXY at the remaining part of the circle
∴ ∠YOZ = 2 ∠ZXY
⇒ ∠ZXY = \(\frac { 1 }{ 2 }\)∠YOZ = \(\frac { 1 }{ 2 }\) × 120° = 60°

Question 6.
In figure, O is the centre of the circle and AB is a chord of the circle. Line QBS is a tangent to the circle at B. If ∠AOB = 110°, find ∠APB and ∠ABQ.

Solution:
O is the centre of the circle, AB is the chord and QBS is the tangent to the circle at B
∠AOB = 110°
Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle
∴ ∠AOB = 2 ∠APB ⇒ ∠APB = \(\frac { 1 }{ 2 }\)∠AOB
⇒ ∠APB = \(\frac { 1 }{ 2 }\) × 110° = 55°
Now QBS is the tangent and AB is the chord
∴ ∠ABQ = ∠APB = 55°
Hence ∠APB = ∠ABQ = 55°

Question 7.
In figure, AB is a diameter and AC is a chord of a circle such that ∠BAC = 30°. The tangent at C intersects AB produced at D. Prove that BC = BD.

Solution:
Given : In a circle with centre O and
AB is its diameter
At C, a tangent is drawn to the circle which
meet AB on producing at D
∠BAC = 30°
To prove : BC = BD
Construction : Join BC
Proof: CD is tangent and CB is chord
∴ ∠DCB = ∠BAC = 30° (Angles in the alternate segment)
In △ABC,
∠ACB = 90° (Angle in a semicircle)
∴ ∠BAC + ∠CBA = 90°
⇒ 30° + ∠CBA = 90°
⇒ ∠CBA = 90°- 30° = 60°
Now in △BCD
Ext. ∠CBA = ∠BCD + ∠BDC
⇒ 60° = 30° + ∠BDC ⇒ ∠BDC = 60° – 30° = 30°
∵ ∠BCD = ∠BDC = 30°
BC = BD (Sides opposite to equal angles)
Hence proved.

Question 8.
In figure, DE is a tangent to the circumcircle of △ABC at the vertex A such that DE || BC. Show that AB = AC.

Solution:
In a circle, DE is a tangent at A to the circumcircle of △ABC in which DE || BC
To prove : AB = AC
Proof : ∵ DE is the tangent and AB is the chord of the circle
∴ ∠DAB = ∠ACB ….(i) (Angles in the alternate segment)
But DE || BC
∴ ∠DAB = ∠ABC ….(ii) (Alternate angles)
From (i) and (ii)
∠ACB = ∠ABC
∴ AB=AC (Sides opposite equal angles) Hence proved.

Question 9.
In figure, two circles intersect in B and C. Lines ABD and ACE are drawn to meet the second circle at points D and E, AF is a tangent at A. Prove that AF || DE.
[Hint Join B to C]

Solution:
Given : Two circles intersect each other at B and C. Lines ABD and ACE are drawn to meet the smaller circle at D and E respectively. AF is the tangent to the first circle at A
To prove : AF || DE
Construction : Join BC
Proof: AF is the tangent and AB is the chord of the first circle
∴ ∠BAF = ∠ACB ….(i) (Angles in the alternate segment)
In cyclic quadrilateral BCED,
Ext. ∠ACB = Int. opp. ∠BDE ….(ii)
From (i) and (ii)
∠BAF = ∠BDE
But these are alternate angles
∴ AF || DE
Hence proved.

Question 10.
In the figure, CD is the tangent line at C to the circumcircle of △ABC intersecting AB produced in D. Show that △DBC ~ △DCA.

Solution:
A circumcircle of △ABC, a tangent DC is drawn at C and AB is produced to meet the tangent at D
To prove : △DBC ~ △DCA
Proof: CD is tangent and BC is the chord of the circle
∴ ∠BCD = ∠BAC (Angles in the alternate segment)
Now in △DBC and △DCA,
∠D = ∠D (common)
∠BCD = ∠BAC or ∠DAC (proved)
∴ △DBC ~ △DCA
Hence proved.

Students can cross-reference their work with S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(e) to ensure accuracy.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(e)

Question 1.
In fig., if AB and CD are two chords of a circle intersecting at a point P inside the circle such that
(i) AP = 8 cm, CP = 6 cm and PD = 4 cm, find PB.
(ii) AB = 12 cm, AP = 2 cm and PD = 4 cm, find CP.
(iii) AP = 6 cm, PB = 5 cm and CD = 13 cm, find CP.

Solution:
∵ AB and CD are two chords which intersect
at P inside the circle
∴ AP × PB = CP × PD

(i) Now AP = 8 cm, CP = 6 cm and PD = 4 cm
∴ 8 × PB = 6 × 4 ⇒ PB = \(\frac{6 \times 4}{8}\) cm
∴ PB = 3 cm

(ii) AB = 12 cm, AP = 2 cm and PD = 4 cm
∴ PB = AB – AP = 12 – 2 = 10 cm
Now AP × PB = CP × PD
⇒ 2 × 10 = CP × 4 ⇒ CP = \(\frac{2 \times 10}{4}\) = 5 cm
∴ CP = 5 cm

(iii) AP = 6 cm, PB = 5 cm and CD = 13 cm
Let CP = x, then PD = CD – CP
= (13 – x) cm
Now AP × PB = CP × PD
⇒ 6 × 5 = x(13 – x) ⇒ 30 = 13 x – x²
⇒ x² – 13x + 30 = 0
⇒ x² – 10x – 3x + 30 = 0

⇒ x(x – 10) – 3 (x – 10) = 0
⇒ (x- 10)(x – 3) = 0
Either x – 10 = 0, then x = 10
or x – 3 = 0, then x = 3
CP = 10 cm or 3 cm

Question 2.
In figure, if AB and CD are two chords of a circle which when produced meet at a point P such that
(i) PA = 10 cm, PB = 4 cm and PC = 8 cm, find PD.
(ii) PC = 15 cm, CD = 7 cm and PA = 12 cm, find AB.
(iii) PA = 16 cm, PC = 10 cm and PD = 8 cm, find AB.

Solution:
∵ Chords AB and CD of a circle intersect each other at P outside the circle
∴ AP × PB = CP × PD

(i) PA = 10 cm, PB = 4 cm and PC = 8 cm
Now AP × PB = CP × PD
⇒ 10 × 4 = 8 × PD
⇒ PD= \(\frac{10 \times 4}{8}\) = 5
∴ PD = 5 cm

(ii) PC = 15 cm, CD = 7 cm and PA = 12 cm
PD = CP – CD = 15 – 7 = 8 cm
Now PA × PB = PC × PD
12 × PB = 15 × 8 ⇒ PB = \(\frac{15 \times 8}{12}\) = 10 cm
∴ AB = PA – PB = 12 – 10 = 2 cm

(iii) PA = 16 cm, PC = 10 cm and PD = 8 cm
Now PA × PB = PC × PD
⇒ 16 × PB = 10 × 8
⇒ PB = \(\frac{10 \times 8}{16}\) = 5
∴ AB = PA – PB = 16 – 5 = 11 cm

Question 3.
(i) In figure, if PT is a tangent to the circle, PB = 4 cm and AB = 12 cm, then PT = …….cm.

(ii) From an external point P, the tangent PT and a secant PAB are drawn. If PA = 9.6 cm and PB = 2.4 cm, determine PT.
Solution:
(i) In the figure, PT is tangent and PBA is a secant to the circle PB = 4 cm, AB = 12 cm

∴ PA = PB + AB = 4 + 12 = 16 cm
∴ PT² = PA × PB
= 16 × 4 = 64 = (8)²
∴ PT = 8 cm

(ii) From an external point P, PT is the tangent to the circle and PAB is a secant

∴ and PA = 9.6 cm, PB = 2.4 cm
Now PT² = PA × PB = 9.6 × 2.4
= 23.04 = (4.8)²
∴ PT = 4.8 cm

Question 4.
The angle A of the triangle ABC is a right angle. The circle on AC as diameter cuts BC at D. If BD = 9, and DC = 7, calculate the length of AB.

Solution:
In △ABC, ∠A = 90°
A circle is drawn on AC as diameter which intersects BC at D
AD is joined
BD = 9 and DC = 7

We see that in the circle AB is the tangent at A and BDC is the secant of the circle
∴ AB² = BD × BC
= 9 × 16 {∵ BC = BD + DC = 9 + 7 = 16}
= 144 = (12)²
∴ AB = 12

Question 5.
In △ABC, AB = 9, AC = 12, F is the mid-point of AC; the circle BFC intersects AB at E; find BE.
Solution:
Steps of construction :
(i) Draw a line segment AB = 9
At A draw ∠BAC = 90° and cut off AC = 12
(ii) Join BC. Take F as midpoint of AC.
(iii) Now draw perpendicular bisector of FC and BC intersecting each other at O.
(iv) With centre O, and radius OC draw a circle with passes through B, F and C and intersects AB at E on measuring BE = 1.

Question 6.
In △ABC, ∠BAC = 90°, AB = 4, AC = 3, AD is an altitude, find BD.
Solution:
ABC ABC is a right angled triangle in which ∠A = 90°,
AD ⊥ BC
AB = 4, AC = 3

∴ BC² = AB2+AC2 (Pythagoras theorem)
= (4)² + (3)² = 16 + 9 = 25 = (5)²
∴ BC = 5
Let BD = x, then DC = 5 – x
Now AB² = BD × BC
(4)² = x × 5 ⇒16 = 5x ⇒ x = \(\frac { 16 }{ 5 }\) = 3\(\frac { 1 }{ 5 }\)
∴ BD = 3\(\frac { 1 }{ 5 }\)

Question 7.
From the external point P, PA is a tangent to the circle at A. PBC is a secant intersecting the circle at B and C. What is the power of P with respect to the circle if PA = 7 ? What is the value of PB.PC ?
Solution:
From a point P,
Outside the circle PA is tangent and PBC is the secant PA = 7

Power of P with respect to the circle
= PA² = (7)² = 49
∵ PA is tangent and PBC is the secant
∴ PA² = PB × PC => 49 = PB × PC
Hence PB.PC = 49

Question 8.
In figure, ABC is a triangle inscribed in a circle. AB =AC = 10 cm, BC = 16 cm. The chord AE is at right angles to the chord BC at D. Calculate DE and the radius of the circle.

Solution:
△ABC in which AB = AC = 10 cm and BC = 16 cm, is inscribed in a circle AE is a chord which is at right angle to BC at D
In △ABC,
AB = AC and AD ⊥ BC
∴ BD = DC = \(\frac { 16 }{ 2 }\) = 8 cm
Now in right △ABD,
AB² = BD² + AD² ⇒ (10)² = (8)² + AD²
⇒ 100 = 64 + AD²
⇒ AD² = 100 – 64 = 36 = (6)²
∴ AD = 6
Now two chords AE and BC intersect each other at D
BD × DC = AD × DE
⇒ 8 × 8 = 6 × DE ⇒ DE = \(\frac{8 \times 8}{6}\) = \(\frac{64}{6}\) = \(\frac{32}{3}\)
∴ DE = \(\frac{32}{3}\) = 10\(\frac{2}{3}\) cm
Now AE = AD + DE = 6 + 10\(\frac{2}{3}\) = 16\(\frac{2}{3}\) cm
∴Radius = \(\frac{1}{3}\) AE (∵ AE is diameter)
= \(\frac{1}{3}\) × 16\(\frac{2}{3}\) = 8\(\frac{1}{3}\) cm

Question 9.
In figure, XY is a tangent to the circle with centre O. XCD is a secant. Calculate r, the radius of the circle.

Solution:
In the figure, XY is a tangent to the circle with centre O. XCD is a secant
ON ⊥ DX
Join OC and join OB
CX = 4 and XB = 6
∵XB is tangent and XCD is the secant
Let CD = x
∴ XB² = XC × XD ⇒ (6)² = 4 (4 + x)
= 4(4 + x) = 36 ⇒ 4 + x = \(\frac{36}{4}\) = 9
x = 9 – 4 = 5
∴ CD = 5
∵ ON ⊥ CD
∴ CN = ND = \(\frac{5}{2}\)
∴ Radius OB = NX = CN + XC
= \(\frac{5}{2}\) + 4 = 6\(\frac{1}{2}\) cm

Question 10.
In figure, O is the centre of the circle. If
(i) AX = 5 cm, XD = 7 cm, CX = 10 cm; find BX.
(ii) OA = 6 cm, BX = 5 cm, OX = 4 cm; find XC.
(iii) CD = 2 cm, DP = 6 cm, BP = 3 cm; find AB.

Solution:
In the figure, PBA, PDC are the secants PT is the tangents to the circle with centre O AD and BC are joined which intersect each other at X
(i) AX = 5 cm, XD = 7 cm, CX = 10 cm
∵ Two chords AD and BC intersect each other at X
∴ AX.XD = CX.XB
⇒ 5 × 7 = 10 × XB
⇒ XB = \(\frac{5 \times 7}{10}\) = \(\frac{7}{2}\) = 3.5 cm

(ii) OA}=6 cm, BX = 5 cm, OX = 4 cm
Join OA, OX and produce it to both sides meeting the circle at P and Q

OA = OP = OQ (radii)
XP = OP – OX = OA – OX = 6 – 4 = 2 cm
XQ = OX + OQ = OX + OA = 6 + 4 = 10 cm
∵ Chord BC and PQ intersect each other at X
∴ XC × 5 = 2 × 10 ⇒ XC = \(\frac{2 \times 10}{5}\) = 4
∴ XC = 4 cm

(iii) CD = 2 cm, DP = 6 cm, BP = 3 cm
Let AB = x
∵ Chords AB and CD intersect each other at P outside the circle

∴ PA × PB = PC × PD
(x + 3) × 3 = (2 + 6) × 6
(x + 3) × 3 = 8 × 6 ⇒ x + 3 = \(\frac{8 \times 6}{3}\) = 16
x = 16 – 3 = 13
∴ AB = 13 cm

Question 11.
In figure, two circles intersect each other at the points P and Q. If AB and AC are the tangents to the two circles from a point A on QP produced, show that AB = AC.

Solution:
Given : Two circles intersect each other at P and Q
AB and AC are the tangents on QP produced from A
To prove: AB = AC
Proof : ∵AB is the tangent and APQ is a secant
∴ AB² = AP × AQ …(i)
Similarly AC is the tangent and APQ is the secant
∴ AC² =AP × AQ …(ii)
From (i) and (ii)
AB² = AC²
⇒ AB = AC
Hence proved.

Question 12.
In figure, AB is any chord of a circle with centre O and P is any point on this chord. A perpendicular drawn through P on OP cuts the circumference in X. Prove that AP.PB = PX².

Solution:
Given : In a circle AB is a chord of a circle with centre O
P is any point on AB and PX ⊥ OP
To prove : AP.PB = PX²
Construction : Produce XP to meet the circle at Y

Proof : ∵ XY is a chord and OP ⊥ XY
∴ P is the midpoint of XY
∴ PX = PY
Now chord AB and XY intersect each other at P
∴ AP × PB = XP × PY
⇒ AP.PB = XP × XP (∵ XP = PY)
⇒ AP.PB = XP² or AP.PB = PX²
Hence proved.

Question 13.
In figure, the two circles intersects at S and T, and STP, BSC, BAP and CDP are st. lines. Prove that
(i) the quad. PATD is cyclic;
(ii) PA.PB = PD.PC.

Solution:
Given : Two circles intersect each other at T and S. STP, BSC, BAP and CDP are straight lines drawn
To prove :
(i) Quad. PATD is a cyclic
(ii) PA.PB = PD.PC
Construction : Join AT and TD
Proof:
(i) ∵ ABST is a cyclic quad.
∴ Ext. ∠ATP = Int. opp. ∠B (i)
Similarly in cyclic quad. CDTS,
Ext. ∠DTP = ∠C ….(ii)
Adding (i) and (ii)
∠ATP + ∠DTP = ∠B + ∠C ….(iii)
But in △PBC,
∠BPC + ∠B + ∠C= 180° (Sum of angles of a triangle)
⇒ ∠B + ∠C = 180° – ∠BPC From (iii)
∠ATP + ∠DTP =180°- ∠BPC
⇒ ∠ATD + ∠BPC = 180°
But these are sum of opposite angles of quad. PATD
∴ Quad. PATD is a cyclic quad.,

(ii) ∵ Chords BA and ST intersect at P outside the circle
∴ PA × PB = PT × PS ….(i)
Similarly CD and ST chords meet at P
∴ PD × PC = PT × PS ….(ii)
From (i) and (ii)
PA.PB = PD.PC Hence proved.

Question 14.
Two circles intersect at A. Chords PAQ and RAS are drawn through A, each passing through the centre of one of the two circles and terminated by the circum-ference. Prove PA.AQ = RA.AS.
Solution:
Given : Two circles intersect each other at A
Chords PAQ and RAS are drawn such that PAQ passes through O’ and RAS, passes through O, the centres of circles respectively
To prove : PA.AQ = RA.AS
Construction : Join PR and QS

Proof: In △APR and △ASQ,
∠P = ∠S (each 90° being in a semicircle)
∠PAR = ∠QAS (Vertically opposite angles)
∴ △APR ~ △ASQ (AA axiom)
∴ \(\frac{\text { PA }}{\text { AS }}\) = \(\frac{\text { RA }}{\text { AQ }}\)
∴ PA.AQ = RA.AS (By cross multiplication)
Hence proved.

Question 15.
Two circles intersect at points A and B. From a point P on the common chord BA produced, secants PCD and PEH are drawn one to each circle. Prove that the points C, D, H, E are concyclic.
Solution:
Given : Two circles intersect each other at A and B.
P is a point on BA produced through P, PCD and PEH are secant drawn to each circle.

To prove : C, D, H, E are concyclic
Proof : ∵ DC and BA are chords which intersect at P outside the first circle
∴ PC × PD = PA × PB ….(i)
Similarly, chords BA and HE intersect each other at P outside the second circle
∴ PA × PB = PE × PH ….(ii)
From (i) and (ii)
PC × PD = PE × PH
But there are two chords DC and HE which meet at P outside the circle
∴ C, D, H and E are concyclic Hence proved.

Question 16.
In figure, PB = BT and PT is tangent to the circle, then prove that
(i) △PTC is isosceles
(ii) PB.PC = TC²

Solution:
Given : In the figure, PB = BT PT is the tangent to the circle
PB is produced to meet the circle at C TC is joined
To prove :
(i) △PTC is an isosceles
(ii) PB.PC = TC²
Proof:
(i) In △PBT,
PB = BT (given)
∴ ∠1 = ∠2 ….(i)
But PT is tangent and BT is the chord of the circle
∴ ∠2 = ∠3 ….(ii)
(Angle in the alternate segment) From (0 and (ii)
∠1 = ∠3
∴ In ATPC,
TP = TC (Sides opposite to equal angles)
∴ △PTC is an isosceles triangle

(ii) ∵ PT is tangent and PBC is secant of the circle
∴ PT² = PB.PC
⇒ PB.PC = PT² = TC²
[∵ PT = TC proved in (i)]
⇒ PB.PC = TC²
Hence proved.

Students can cross-reference their work with OP Malhotra Class 10 ICSE Solutions Chapter 11 Coordinate Geometry Ex 11(c) to ensure accuracy.

S Chand Class 10 ICSE Maths Solutions Chapter 11 Coordinate Geometry Ex 11(c)

Question 1.
State the slope of the line
(i) 3x + 4y = 9
(ii) 3x + 2y – 4
Solution:
(i) 3x + 4y = 9 ⇒ 4y = – 3x + 9
⇒ y = \(\frac { -3 }{ 4 }\)x + \(\frac { 9 }{ 4 }\)
Which is in the form of y = mx + c
∴ Slope (m) = \(\frac { -3 }{ 4 }\)

(ii) 3x + 2y = 4 ⇒ 2y = – 3x + 4
⇒ y = \(\frac { -3 }{ 2 }\)x + \(\frac { 4 }{ 2 }\) ⇒ y = \(\frac { -3 }{ 2 }\)x + 2
∴ Slope (m) = \(\frac { -3 }{ 2 }\)

Question 2.
Given 3x + 2y + 4 = 0
(i) Express the equation in the form y = mx + c.
(ii) Find the slope and the y-intercept of the line 3x + 2y + 4 = 0.
(iii) Use your answer to (ii) above and on a graph paper draw the graph of the straight line 3x + 2y + 4 = 0.
Solution:
(i) 3x + 2y + 4 = 0 ⇒ 2y – -3x – 4
⇒ y = \(\frac { -3 }{ 2 }\)x – 2 which is in the form of y – mx + c

(ii) Slope (m) = \(\frac { -3 }{ 2 }\) and y-intercept (c) = – 2

(iii) The line has been drawn on the graph paper as given here.

Question 3.
State the equation of the line which has the y-intercept
(a) 2 and a slope 7 ;
(b) – 3 and a slope – 4 ;
(c) – 1 and is parallel to y = 5x – 7 ;
(d) 2 and is inclined at 45° to the x-axis ;
(e) – 5 and is equally inclined to the axes.
Solution:
We know that the equation of a line is y = mx + c
Where m is the slope and c is the y-intercept.
Therefore
(a) y-intercept (c) = 2
slope (m) = 7
∴ Equation of the line will be y = 7x + 2⇒7x-y-t-2 = 0

(b) Slope (m) = – 4
and y-intercept (c) = – 3
∴ Equation of the line will be y = – 4x – 3 or 4x + y + 3 = 0

(c) y-intercept (c) = – 1
∵ The line is parallel to the line y = 5x – 7
Slope (m) = 5
∴ The required line will be
y = 5x – 1
(∵ parallel line have equal slopes)

(d) y-intercept (c) = 2
∵ The equation inclined at 45° to the x-axis
∴ Slope (m) = tan θ = tan 45° = 1
∴ Equation of the required line will be
y = 1 x + 2 ⇒ y = x + 2
x – y + 2 = 0

(e) y-intercept (c) = – 5
∵ The line is equally inclined to the axes
∴ Slope (m) = tan θ = + 1
∴ Equation of the line will be
y – 1x – 5 ⇒ y = x – 5
x – y = 5
or y = – 1x – 5 ⇒ y = – x – 5 or x + y + 5 = 0

Question 4.
What will be the value of m and c if the straight line y = mx + c passes through the points (3, – 4) and (- 1, 2) ?
Solution:
The equation of line is y = mx + c … (i)
∵ It passes through the points (3, – 4) and (-1, 2)
Substituting the values of x and y in (i)
– 4 = 3 m + c … (ii)
and 2 = – 1m + c ⇒ 2 = – m + c … (iii)
Subtracting we get,
– 6 = 4 m ⇒ m = \(\frac { -6 }{ 4 }\) = \(\frac { -3 }{ 2 }\)
From (ii)
– 4 = 3 × (\(\frac { -3 }{ 2 }\)) + c ⇒ – 4\(\frac { -9 }{ 2 }\) + c
⇒ c = – 4 + \(\frac { -9 }{ 2 }\) = \(\frac { -8+9 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
∴ m = \(\frac { -3 }{ 2 }\) and c = \(\frac { 1 }{ 2 }\)

Question 5.
The graph of the equation y = mx + c passes through the points (1, 4) and (-2, -5). Determine the values of m and c.
Solution:
∵ The line y = mx + c passes through the points (1, 4) and (-2, -5)
∴ Substituting the values of x and y, we get
4 = 1m + c ⇒ 4 = m + c
– 5 = – 2m + c ⇒ – 5 = – 2m + c … (ii)
Subtracting, we get
9 = 3m ⇒ m = \(\frac { 9 }{ 3 }\) = 3
From (i)
4 = 3 + c ⇒ c = 4 – 3 = 1
Hence m = 3 and c = 1

Question 6.
Find the equation of the straight line through the given point P and having the given slope m if
(a) P (- 4, 7); m = – \(\sqrt{3}\);
(b) P (- 1, – 5), m = \(\frac { -6 }{ 11 }\).
Solution:
(a) The equation of the line will be y – y₁ = m (x – x₁)
⇒ Here it passes through P (- 4, 7) and m = – \(\sqrt{3}\)
∴ y – 7 = \(\sqrt{3}\) [x – (- 4)]
⇒ y – 7 = – \(\sqrt{3}\) x – 4\(\sqrt{3}\)
⇒ \(\sqrt{3}\)x + y = 7 – 4\(\sqrt{3}\)

(b) ∵ The line passes through the point P (- 1, – 5)
and has slope = \(\frac { -6 }{ 11 }\)
∴ Equation of the line will be
y – y₁ = m (x – x₁).
⇒ y + 5 = \(\frac { -6 }{ 11 }\) (x + 1)
⇒ 11 y + 55 = – 6x – 6
⇒ 6x + 11y + 55 + 6 = 0
⇒ 6x + 11y + 61 = 0

Question 7.
Find the equation to the straight line passing through :
(a) the origin and perpendicular to x + 2v = 4;
(b) the point (4, 3) and parallel to 3x + 4y = 12;
(c) the point (4, 5) and (i) parallel to, (ii) perpendicular to 3x – 2y + 5 = 0.
Solution:
(a) Slope of the line x + 2y = 4 ⇒ 2y = – x + 4
⇒ y = \(\frac { -1 }{ 2 }\)x + 2
m₁ = \(\frac { -1 }{ 2 }\) and slope of the line perpendicular
∵ The required equation passes through the origin (0, 0)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – 0 = 2(x – 0) ⇒ y – 2x
⇒ 2x – y = 0

(b) Slope of the given line 3x + 4y = 12
⇒ 4y = – 3x + 12 ⇒ y = \(\frac { -3 }{ 4 }\)x + 3
m = \(\frac { -3 }{ 4 }\)
∴ Slope of the line parallel to the given line = \(\frac { -3 }{ 4 }\)
∵ It passes through the point (4, 3)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac { -3 }{ 4 }\)(x – 4)
⇒ 4y – 12 = – 3x + 12
⇒ 3x + 4y = 12 + 12
⇒ 3x + 4y = 24
⇒ 3x + 4y – 24 = 0

(c) Slope of the given line 3x – 2y + 5 = 0
⇒ 2y = 3x + 5 ⇒ y = \(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 2 }\)
m = \(\frac { 3 }{ 2 }\)
(i) Slope of the required line which is parallel to the given line = m = \(\frac { 3 }{ 2 }\)
∵ It passes through the point (4, 5)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – 5 = \(\frac { 3 }{ 2 }\) (x – 4) ⇒ 2y – 10 = 3x – 12
⇒ 3x – 2y + 10 – 12 = 0
⇒ 3x – 2y – 2 = 0

(ii) Slope of the line perpendicular to the given line = \(\frac { -1 }{ m }\)
m₁ = \(\frac { -2 }{ 3 }\)
∵ It passes through the point (4, 5)
∴ Equation of the line will be
y – y₁ = m (x – x₁) ⇒ y – 5 = \(\frac { -2 }{ 3 }\) (x – 4)
⇒ 3y – 15 = – 2x + 8 ⇒ 2x + 3y – 15 – 8 = 0
⇒ 2x + 3y – 23 = 0 ⇒ 2x + 3y = 23

Question 8.
Find the equations of the line joining the points
(a) A (1, 1) and B (2, 3);
(b) P (3, 3) and Q (7, 6);
(c) L (a, b) and M (b, a).
Solution:
The equation of a line which passes through two points is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
(a) ∵ The required line passes through the points
A (1, 1) and B (2, 3)
∴ Equation of the line will be
y – 1 = \(\frac{3-1}{2-1}\) (x – 1) ⇒ y – 1 = \(\frac { 2 }{ 1 }\) (x – 1)
⇒ y – 1 = 2x – 2 ⇒ 2x – y – 2 + 1 = 0
⇒ 2x – y – 1 = 0

(b) ∵ The required line passes through the points P (3, 3) and Q (7, 6)
∴ Equation of the line will be
⇒ y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
⇒ y – 3 = \(\frac{6-3}{7-3}\) (x – 3)
⇒ y – 3 = \(\frac { 3 }{ 4 }\) (x – 3) ⇒ 4y – 12 = 3x – 9
3x – 4y + 12 – 9 = 0 ⇒ 3x – 4y + 3 = 0

(c) ∵ The required line passes through L (a, b) and M (b, a)
∴ Equation of the line will be
y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
⇒ y – 3 = \(\frac{a-b}{b-a}\) (x – a)
⇒ y – b = \(\frac{(a-b)}{-(a-b)}\) (x – a)
⇒ y – b = – 1 (x – a) ⇒ y – b = – x + a
⇒ x + y = a + b

Question 9.
The lines represented by 3x + 4y = 8 and px + 2y = 7 are parallel. Find the value of p.
Solution:
Slope of line 3x + 4y = 8 ⇒ 4y = – 3x + 8 – 3
⇒ y = \(\frac { -3 }{ 4 }\)x + 2
∴ (m₁) = \(\frac { -3 }{ 4 }\)
and slope of line px + 2y = 7 ⇒ 2y = – px + 7
⇒ y = \(\frac{-p}{2} x+\frac{7}{2}\)
m₂ = \(\frac { -p }{ 2 }\)
∵ The lines are parallel
∴ Their slopes are equal
∴ m₁ = m₂ ⇒ y = \(\frac{-3}{4}=\frac{-p}{2}\)
⇒ p = \(\frac{3 \times 2}{4}=\frac{3}{2}\)

Question 10.
The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find :
(i) the gradient of EF ;
(ii) the equation of EF ;
(iii) the coordinates of the point where the line EF intersects the x-axis.
Solution:
Co-ordinates of two points are E (0, 4), F (3, 7)
(i) ∴ Slope (W) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{7-4}{3-0}=\frac{3}{3}\) = 1
∴ Gradient of EF = 1

(ii) Now equation of EF will be
⇒ y – y₁ = m(x – x₁)
⇒ y – 7 = 1 (x – 3) ⇒ y – 7 = x – 3
⇒ x – y + 7 – 3 = 0 ⇒ x – y + 4 = 0

(iii) Let the line EF intersect the x-axis at P then
y-co-ordinates of P will be 0
Let the co-ordinates of P be (x₁, 0)
∵ P lies on EF
∴ It will satsify the equation of EF
x – 0 + 4 = 0 ⇒ x + 4 = 0 ⇒ x = – 4
∴ Co-ordinates of P will be (- 4, 0)

Question 11.
P, Q, R have coordinates (-2, 1), (2, 2) and (6, -2) respectively. Write down
(i) the gradient of QR ;
(ii) the equation of the line through P perpendicular to QR.
Solution:
Co-ordinates of points are P (- 2, 1), Q (2, 2) and R (6, – 2)
(i) Gradient of QR (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-2-2}{6-2}\) = \(\frac { -4 }{ 4 }\) = 1

(ii) Slope of line perpendicular to QR = \(\frac { -1 }{ m }\)
= -(-1) = 1
∵ It is passes through P
∴ Equation of the line will be
⇒ y – y₁ = m(x – x₁)
⇒ y – 1 = 1(x + 2) ⇒ y – 1 = x + 2
x – y + 2 + 1 = 0 ⇒ x – y + 3 = 0

Question 12.
A line 3x – 4y + 12 = 0 meets the x-axis at the point P. Find the equation of the line through P, perpendicular to the line 3x + 5y – 15 = 0
Solution:
The line 3x – 4y + 12 = 0 meets x-axis at P
∴ y-coordinate of P will be 0
Let the co-ordinates of P be (x, 0)
∵ It lies on the line 3x – 4y+ 12 = 0
∴ It will satisfy the equation
∴ 3x – 4 x 0 + 12 = 0 ⇒ 3x + 12 = 0
⇒ 3x = – 12 ⇒ x = \(\frac { -12 }{ 3 }\) = – 4
∴ Co-ordinates of P will be (- 4, 0)
Now slope of the line 3x + 5y – 15 = 0 ⇒ 5y = – 3x + 15
⇒ y = \(\frac { -3 }{ 5 }\)x + 3
(m) = \(\frac { -3 }{ 5 }\)
∴ Slope of the line perpendicular to this line
= \(\frac{-1}{m}=-\left(\frac{-5}{3}\right)=\frac{5}{3}\)
∴ Equation of the perpendicular line through P will be
y – y₁ = m (x – x₁) ⇒ y – 0 = \(\frac { 5 }{ 3 }\)(x + 4)
⇒ y = \(\frac { 5 }{ 3 }\) (x + 4) ⇒ 3y = 5x + 20
⇒ 5x – 3y + 20 = 0

Question 13.
The line segment joining P (5, – 2) and Q (9, 6) is divided in the ratio 3 : 1 by a point A on it. Find the equation of a line through the point A perpendicular to the line x – 3y + 4 = 0.
Solution:
Point A divides the line joining the point P (5, -2) and Q (9, 6) in the ratio 3 : 1
∴ m₁ = m₂ = 3 : 1
Now co-ordinates of A will be
\(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)
= \(\left(\frac{3 \times 9+1 \times 5}{3+1}, \frac{3 \times 6+1 \times(-2)}{3+1}\right)\)
= \(\left(\frac{27+5}{4}, \frac{18-2}{4}\right) \text { or }\left(\frac{32}{4}, \frac{16}{4}\right)\) or (8, 4)
Slope of the line x – 3y + 4 = 0 ⇒ 3y = x + 4
⇒ y = \(\frac { 1 }{ 3 }\)x + \(\frac { 4 }{ 3 }\) will be (m) = \(\frac { 1 }{ 3 }\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\)
= – (\(\frac { 3 }{ 1 }\)) = – 3
Equation of the line through A will be
y – y₁ = m(x – x₁)
⇒ y – 4 = – 3(x – 8) ⇒ y – 4 = – 3x + 24
⇒ 3x + y = 24 + 4
⇒ 3x + y = 28
⇒ y + 3x – 28 = 0

Question 14.
Write down the equation of the line parallel to x – 2y + 8 = 0 passing through the point (1, 2).
Solution:
The slope of line x – 2y + 8 = 0
⇒ 2y = x + 8
⇒ y = \(\frac { 1 }{ 2 }\)x + 4
(m) = \(\frac { 1 }{ 2 }\)
∴ Slope of the line parallel to it = \(\frac { 1 }{ 2 }\)
∵ It passes through the point (1, 2)
∴ Equation of the line will be
y – y₁ = m(x – x₁) ⇒ y – 2 = \(\frac { 1 }{ 2 }\) (x – 1)
⇒ 2y – 4 = x – 1 ⇒ x – 2y + 4 – 1 = 0
⇒ x – 2y + 3 = 0

Question 15.
Write down the gradient and intercept on the v-axis of the line \(\frac { x }{ 3 }\) + \(\frac { y }{ 4 }\) = 1.
Solution:
The equation of the line is given
\(\frac { x }{ 3 }\) + \(\frac { y }{ 4 }\) = 1
⇒ 4x + 3y = 12 ⇒ 3y = – 4x + 12 – 4
⇒ y = \(\frac { -4 }{ 3 }\)x + 4
∴ Gradient (m) = \(\frac { -4 }{ 3 }\)
and y-intercept (c) = 4

Question 16.
(i) Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b.
(ii) Find the equation of the line through (1, 3) making an intercept of 5 on the y-axis.
Solution:
(i) 2x – by + 5 = 0 ⇒ by = 2x + 5
⇒ y = \(\frac { 2 }{ b }\)x + \(\frac { 5 }{ b }\)
∴ Slope (m₁) = \(\frac { 2 }{ b }\)
ax + 3y = 2 ⇒ 3y = – ax + 2
⇒ y = \(\frac{-a}{3}+\frac{2}{3}\)
∴ Slope (m₂) = \(\frac { -a }{ 3 }\)
∵ These lines are parallel
∴ m₁ = m₂
⇒ \(\frac { 2 }{ b }\) = \(\frac { -a }{ 3 }\)
⇒ – ab = 6
⇒ ab = – 6

(ii) We know that equation of a line is y = mx + c where m is slope and c is y-intercept
It passes through (1,3) and has ^-intercept = 5
∵ It will also pass through (0, 5)
∴ Slop(m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{5-3}{0-1}=\frac{2}{-1}\) = – 2
∴ Equation of the line y = mx + c ⇒ y = – 2x + 5
⇒ 2x + y = 5

Question 17.
(i) Find the value of p, given that the line \(\frac { y }{ 2 }\) = x – p passes through the point (- 4, 4).
(ii) Given that the line \(\frac { y }{ 2 }\) = x – p and the line ax + 5 = 3y are parallel, find the value of a.
Solution:
(i) Equation of the line is given \(\frac { y }{ 2 }\) = x – p
∵ It passes through the point (- 4, 4)
∴ \(\frac { 4 }{ 2 }\) = – 4 – P ⇒ 2 = – 4 – p
⇒ p = – 4 – 2 ⇒ p = – 6
∴ p = – 6

(ii) In equation
\(\frac { y }{ 2 }\) = x – p ⇒ y = 2x – 2p
Slope (m₁) = 2
and in equation ax + 5 = 3y
⇒ y = \(\frac { a }{ 3 }\)x + \(\frac { 5 }{ 3 }\)
Slope (m₂) = \(\frac { a }{ 3 }\)
∵ These two lines are parallel
∴ m₁ = m₂
⇒ 2 = \(\frac { a }{ 3 }\) ⇒ 6 = a
∴ a = 6

Question 18.
A line intersects x-axis at (- 2,0) and cuts off an intercept of 3 from the positive side of y-axis. Write the equation of the line.
Solution:
The line passes through a point (- 2, 0) and has its y-intercept = 3
∴ It will pass through (0, 3)
∴ Slope of the line (m) = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{3-0}{0-(-2)}=\frac{3}{2}\)
∴ Equation of the line will be y = mx + c
y = \(\frac { 3 }{ 2 }\)x + 3 ⇒ 2y = 3x + 6

Question 19.
(a) In the adjoining figure, write down (i) the coordinates of the points A, B and C; (ii) the equation of the line through A, parallel to BC.

(b) In what ratio is the join of A (0,3) and B (4, 1) divided by the x-axis? (AB produced if necessary). Write the co-ordinates of the point where AB intersects the x-axis.
Solution:
(a) In the figure,
(i) Co-ordinates of A are (2, 3), of B are (- 1,2) and of C are (3, 0)
(ii) Equation of a line through A and parallel to BC
Slope of BC (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{0-2}{3+1}=\frac{-2}{4}\) = \(\frac { -1 }{ 2 }\)
∴ Slope of the required line = \(\frac { 1 }{ 2 }\)
∴ y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac { -1 }{ 2 }\) (x – 2)
⇒ 2y – 6 = – x + 2
⇒ x + 2y = 2 + 6 ⇒ x + 2y = 8
⇒ x + 2y – 8 = 0

(b) Let ratio be m₁ : m₂ which divides the join of
A (0, 3) and B (4, 1) by the x-axis at P
∵ P lies on x-axis
∴ Co-ordinate of P be (x, 0)

∴ Co-ordinates of P will be (6, 0)

Question 20.
Write down the equation of the line AB, through (3, 2), perpendicular to the line 2y = 3x + 5. AB meets the x-axis at A and y-axis at B. Write down the coordinates of A and B. Calculate the area of ∆OAB where O is the origin.
Solution:
In the given line 2y = 3x + 5 ⇒ y = \(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 2 }\)
Slope (m₁) = \(\frac { 3 }{ 2 }\)
∴ Slope of the line AB perpendicular to the
given line (m₂) = \(\frac{-1}{m_1}=\frac{-2}{3}\)
∵ It passes through (3, 2)
∴ Equation of the line will be y – y₁ = m(x – x₁)
⇒ y – 2 = \(\frac { -2 }{ 3 }\) (x – 3) ⇒ 3y – 6 = – 2x + 6
⇒ 2x + 3y = 6 + 6 ⇒ 2x + 3y = 12
⇒ 2x + 3y – 12 = 0
∵ This line meets x-axis at A and y-axis at B
∴ Substituting the values of y = 0 and x = 0
If y = 0, then
2x + 0 x y = 12
⇒ 2x = 12
⇒ x = \(\frac { 12 }{ 2 }\) = 6
If x = 0, then
2 x 0 + 3y = 12 ⇒ 0 + 3y = 12
⇒ 3y = 12 ⇒ y = \(\frac { 12 }{ 3 }\) = 4
∴ Co-ordinates of A and B will be A (6,0) and B (0, 4)
∵ O is the origin
∴ Area of AOAB = \(\frac { 1 }{ 2 }\) x OA x OB
= \(\frac { 1 }{ 2 }\) x 6 x 4 = 12 sq. units

Question 21.
(i) Write down the coordinates of the point P that divides the line joining A (- 4, 1) and B (17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the y-axis divide the line AB?
Solution:
(i) Let co-ordinates of point P be (x, y)
∵ P divides the line joining A (- 4, 1) and B(17, 10) in the ratio 1 : 2 i.e., m₁ : m₂ = 1 : 2
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{1 \times 17+2 \times(-4)}{1+2}\)
= \(\frac{17-8}{3}=\frac{9}{3}\) = 3
and y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{1 \times 10+2 \times 1}{1+2}\)
= \(\frac{10+2}{3}=\frac{12}{3}\) = 4
∴ Co-ordinates of P are (3, 4)

(ii) ∵ O is the origin (0, 0)
∴ OP = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(3-0)^2+(4-0)^2}=\sqrt{(3)^2+(4)^2}\)
= \(\sqrt{9+16}=\sqrt{25}\) = 5

(iii) Let the ratio be m₁ : m₂, in which y-axis divides the line segment AB. Let R be the point on y-axis which divides AB in the ratio m₁ : m₂
Co-ordinates of R be (0, y)
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}\)
0 = \(\frac{m_1 \times 17+m_2 \times(-4)}{m_1+m_2}\)
⇒ 17m₁ – 4m₂ = 0 ⇒ 17m₁ = 4m₂
⇒ \(\frac{m_1}{m_2}=\frac{4}{17}\)
∴ Ratio = 4 : 17

Self Evaluation And Revision
(LATEST ICSE QUESTIONS)

Question 1.
(a) The mid-point of the line segment AB shown in the diagram is (4, -3). Write down the coordinates of A, B.

(b) Match the equations A, B, C, D with the lines L₁, L₂, L₃, L₄, whose graphs are roughly drawn in figure below.
A ≡ y = 2x, B ≡ y – 2x + 2 = 0, C ≡ 3x + 2y = 6, D ≡ y = 2.

(c) Write down the equation of the line whose gradient is \(\frac { 3 }{ 2 }\) and it passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3.
Solution:
(a) The mid-point of the line segment AB as shown in the figure is (4, -3). Point A lies on x-axis and point B lies on y-axis

Let the co-ordinates of A be (x, 0) and of B be (0, y)
∴ 4 = \(\frac { x+0 }{ 2 }\) ⇒ x + 0 = 8 ⇒ x = 8
and – 3 = \(\frac { 0+y }{ 2 }\) ⇒ – 6 = y
∴ Co-ordinates of A are (8, 0) and of B are (0, – 6)

(b) We are given four equations of 4 lines
i.e., Equation A = y = 2x
Equation B ≡ y – 2x + 2 = 0
Equation C ≡ 3x + 2y – 6
and Equation D ≡ y – 2
In equation A, x = 0, then y = 0
or it passes through the origin 0 (0, 0)
But line L₃ in the graph, is passing through the origin O
∴ A → L₃
In equation B, y – 2x + 2 = 0 ⇒ y = 2x – 2
Slope (m) = 2 and C = – 2
We see that line L₄ has negative y-intercept and positive slope,
∴ B → L₄
In equation C, 3x + 2y = 6
⇒ 2y = – 3x + 6 ⇒ y = \(\frac { -3 }{ 2 }\)x + 3
Slope (m) = \(\frac { -3 }{ 2 }\) and y-intercept (C) = 3
We see that line L₂ has negative slope and positive y-intercept
∴ C → L₂
In equation D, y = 2
Which is parallel to x-axis at a distance of 2 unit on the positive side of y-axis and we see line L, is parallel to x-axis
∴ D → L₁

(c) The gradient of a line = \(\frac { 3 }{ 2 }\)
and it passes through a point P
∵ P divides the line segment joining A (-2, 6)
and B (3, -4) in the ratio 2 : 3
Let co-ordinates of P be (x, y), then
x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{2 \times(3)+3 \times(-2)}{2+3}\)
= \(\frac { 6-6 }{ 5 }\) = 0
and y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{2 \times(-4)+3 \times 6}{2+3}\)
= \(\frac{-8+18}{5}=\frac{10}{5}\) = 2
∴ Co-ordinates of P are (0, 2)
Now equation of line through P will be
y – y₁ = m(x – x₂) ⇒ y – 2 = \(\frac { 3 }{ 2 }\) (x – 0)
⇒ 2y – 4 = 3x ⇒ 3x – 2y + 4 = 0

Question 2.
(a) Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
(b) (i) The line 4x – 3y + 12 = 0 meets the x- axis at A. Write down the co-ordinates of A.
(ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y +12 = 0.
Solution:
(a) ∵ Point B is on x-axis is and its abscissa is 11
∴ Co-ordiantes of B will be (11, 0)
∴ Distance between A (7, 3) and B (11, 0)
= \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(11-7)^2+(0-3)^2}=\sqrt{(4)^2+(-3)^2}\)
= \(\sqrt{16+9}=\sqrt{25}\)
= 5 units

(b) (i) The line 4x – 3y + 12 = 0, meets x-axis at A
Let co-ordinates of A be (x, 0), then
4 × x – 3 x 0 + 12 = 0 ⇒ 4x + 12 = 0
⇒ 4x = – 12
⇒ x = \(\frac { -12 }{ 4 }\) = – 3
∴ Co-ordinates of A are (- 3, 0)

(ii) In equation 4x – 3y + 12 = 0
⇒ 3y = 4x + 12 ⇒ y = \(\frac { 4 }{ 3 }\)x + 4
Slope (m₁) = \(\frac { 4 }{ 3 }\)
∴ Slope of the line perpendicular to it m₂ = \(\frac{-1}{m_1}\)
= \(\frac { -3 }{ 4 }\)
∵ This line passes through A (- 3, 0)
∴ Equation of the line will be
y – y₁ = m(x – x₁) ⇒ y – 0 = \(\frac { -3 }{ 4 }\) (x + 3)
⇒ 4y = – 3x – 9 ⇒ 3x + 4y + 9 = 0

Question 3.
(a) The centre O, of a circle has the coordinates (4, 5) and one point on the circumference is (8, 10). Find the coordinates of the other end of the diameter of the circle through this point.
(b) Find the equation of a line, which has the y-intercept 4 and is parallel to the line 2x – 3y = 1. Find the coordinates of the point, where it cuts the x-axis.
Solution:
(a) Co-ordinates O, the centre of a circle are (4, 5)
One point on the circle is B (8, 10)

BO is joined and produced to A such that AB
is the diameter of the circle
Let co-ordinates of A be (x, b)
∴ O is the mid-point of AB,
∴ 4 = \(\frac { 8+x }{ 2 }\) ⇒ 8 = 8 + x ⇒ x = 8 – 8 = 0
and 5 = \(\frac { 10+y }{ 2 }\) ⇒ 10 = 10 + y
⇒ y = 10 – 10 = 0
∴ Co-ordinates of A will be (0, 0)

(b) y-intercept of a line (C) = 4
and parallel to the line whose equation is 2x – 3y = 7 ⇒ 3y = 2x + 7
y = \(\frac { 2 }{ 3 }\)x + \(\frac { 7 }{ 3 }\)
∴ Slope (m₁) = \(\frac { 2 }{ 3 }\)
Now the slope of the line which is parallel to
it = m₂ = m₁ = \(\frac { 2 }{ 2 }\)
∴ Equation of the line will be y = mx + c
⇒ y = \(\frac { 2 }{ 3 }\)x + 4 ⇒ 3y = 2x + 12
Let this line cut the x-axis at A
∴ Ordinate of A will be O
Let abscissa of A be x
But it will satisfy the equation
3 x 0 = 2x + 12 ⇒ 0 = 2x + 12
⇒ 2x = – 12 ⇒ x = \(\frac { -12 }{ 2 }\) = – 6
∴ Co-ordinates of A are (- 6, 0)

Question 4.
(a) Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0.
(b) In the given figure line APB meets the x-axis at A, y-axis at B. P is the point (-4, 2) and AP : PB = 1 : 2. Write down the coordinates of A and B.

(c) The centre of a circle of radius 13 units is the point (3, 6). P (7, 9) is a point inside the circle. APB is a chord of the circle such that AP = PB. Calculate the length of AB.
Solution:
(a) In the given equation of a line is + 5y + 15 = 0 ⇒ 5y = – 3x – 15
⇒ y = \(\frac { -3 }{ 5 }\)x – 3
∴ Slope (m₁) = \(\frac { -3 }{ 5 }\)
Now slope of the required line which is parallel to the given line = m₂ = \(\frac { -3 }{ 5 }\) (∵ m₁ = m₂)
∵ This line passes through the point (0, 4)
∴ Equation of the line will be
y – y₁ = m(x – x₁) ⇒ y – 4 = \(\frac { -3 }{ 5 }\) (x – 0)
⇒ 5y – 20 = – 3x + 0 ⇒ 3x + 5y = 0 + 20
⇒ 3x + 5y = 20

(b) P (- 4,2) is a point on the line AB such that P divides it in the ratio 1 : 2 i.e., AP = PB = 1 : 2 and the line intersects x-axis at A and y-axis at B
Let co-ordinates of A be (x, 0) and of B be (0, y)
∴ – 4 = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{1 \times 0+2 \times x}{1+2}=\frac{2 x}{3}\)
∴ x = \(\frac{-4 \times 3}{2}\)
and 2 = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{1 \times y+2 \times 0}{1+2}=\frac{y}{3}\)
∴ y = 2 x 3 = 6
∴ Co-ordinates of A are (- 6, 0) and of B are (0, 6)

(c) Radius of a circle = 13 units
Centre of the circle is O (3, 6)
P (7, 9) is a point inside the circle and APB is a chord of the circle Joint OP and OA

∵ P is the midpoint of chord AB and OP is joined
∴ OP ⊥ AB
Now length of OP = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(7-3)^2+(9-6)^2}\)
= \(\sqrt{(4)^2+(3)^2}=\sqrt{16+9}=\sqrt{25}\)
= 5 units
Now in right angled AOAP,
OA² = OP² + AP² (Pythagoras theorem)
⇒ (13)² = (5)² + AP²
⇒ 169 = 25 + AP² ⇒ AP² = 169 – 25 = 144 = (12)²
∴ AP = 12 units
∵ P is mid-point of AB
∴ AB = 2AP = 2 x 12 = 24 units

Question 5.
(a) Calculate the ratio in which the line joining A (6, 5) and B (4, – 3) is divided by the line y = 2.
(b) In this figure, AB and CD are the lines 2x -y + 6 = 0 and x – 2y = 4 respectively.

(i) Write down the coordinates of A, B, C, D;
(ii) Prove that the triangles OAB and ODC are similar;
(iii) Is figure ABCD cyclic? Give reasons for your answer.
(c) ABCD is a rhombus. The coordinates of A and C are (3, 6) and (- 1, 2) respectively. Write down the equation of BD.
Solution:
(a) Let points A (6, 5) and B (4, – 3) are joined and a line y = 2 divides it in the ratio m₁ : m₂
∵ The point which divides the line segment AB lies on y = 2
∴ 2 = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{m_1(-3)+m_2 \times 5}{m_1+m_2}\)
⇒ 2 = \(\frac{-3 m_1+5 m_2}{m_1+m_2}\)
⇒ 2m₁ + 2m₂ = – 3m₁ + 5m₂
⇒ 2m₁ + 3m₁ = 5m₂ – 2m₂ ⇒ 5m₁ = 3m₂
⇒ \(\frac{m_1}{m_2}=\frac{3}{5}\)
Ratio = 3 : 5

(b) (i) Equation of line AB is 2x – y + 6 = 0 and of CD is x – 2y = 4
Which intersect x-axis at B and D and y-axis at A and C respectively

∵ A lies on y-axis
∴ x-coordinates of A = 0
∴ From the equation 2x – y + 6 = 0
⇒ 2 x 0 – y + 6 = 0
⇒ – y + 6 = 0 ⇒ y = 6
∴ Co-ordinates of A are (0, 6)
∵ B lies on x-axis
∴ Its y-coordinates = 0
From the equation 2x – y + 6 = 0
2x – 0 + 6 = 0 ⇒ 2x = – 6
⇒ x = \(\frac { -6 }{ 2 }\) = – 3
∴ Co-ordinates of B are (-3, 0)
∵ C lies on y-axis
∴ Its x-coordinates = 0
From the equation x – 2y = 4
⇒ 0 – 2y = 4 ⇒ y = \(\frac { 4 }{ -2 }\) = -2
∴ Co-ordinates of C are (0, – 2)
∵ D lies on x-axis
∴ Its y-coordinates = 0
From the equation x – 2y = 4
⇒ x – 2 x 0 = 4 ⇒ x – 0 = 4 ⇒ x = 4
Co-ordinates of D are (4, 0)

(ii) In ∆OAB and ∆ODC,
∠AOB = ∠COD
(Vertically opposite angles or each = 90°)
∵ \(\frac{\mathrm{OA}}{\mathrm{OB}}=\frac{6}{3}=\frac{2}{1}=\frac{\mathrm{OD}}{\mathrm{OC}} \frac{4}{2}=\frac{2}{1}\)
∴ \(\frac{\mathrm{OA}}{\mathrm{OB}}=\frac{\mathrm{OD}}{\mathrm{OC}}\)
∆OAB ~ ∆ODC (SAS axiom)

(iii) ∴ ∠OAB = ∠ODC
and ∠OBA = ∠OCB
But these are angles in the same segment
∴ A, B, C and D are concyclic
Hence ABCD is a cyclic quadrilateral

(c) In a rhombus ABCD,
Co-ordinates of A are (3, 6) and of C are (- 1, 2)
∴ Slop of AC (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-6}{-1-3}\)
= \(\frac { -4 }{ -4 }\) = 1
∵ AC and BD bisect each other at right angles
∴ Slope of BD = \(\frac { -1 }{ m }\) = – 1
and mid-point of AC = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
or \(\left(\frac{3-1}{2}, \frac{6+2}{2}\right) \text { or }\left(\frac{2}{2}, \frac{8}{2}\right)\) or (1, 4)
∴ Equation of the line BD will be,
y – y₁ = m(x – x₁)
y – 4 = – 1 (x – 1)
⇒ y – 4 = – x + 1
⇒ x + y = 1 + 4 ⇒ x + y = 5
⇒ x + y – 5 = 0

Question 6.
(a) A (10, 5), B (6, – 3) and C (2, 1) are the vertices of a triangle ABC. L is the mid-point of AB, and M is the mid-point of AC. Write down the coordinates of L and M. Show that LM = \(\frac { 1 }{ 2 }\) BC.
(b) Write down the equation of the line whose gradient is \(\frac { 3 }{ 2 }\) and which passes through point P that divides the line segment joining A (- 2, 6) and B (3, – 4) in the ratio 2 : 3.
Solution:
(a) Vertices of a A ABC are A (10, 5), B (6. -3) and C (2, 1)
L is the mid point of AB and M is the mid-point of AC
∴ Co-ordinates of L will be

We see that LM = \(\frac { 1 }{ 2 }\) BC

(b) ∵ P divides the line segment joining the points A (-2, 6) and B (3, -4) in the ratio 2 : 3
∵ Co-ordinates of P will be
\(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\) or
\(\left(\frac{2 \times 3+3 \times(-2)}{2+3}, \frac{2 \times(-4)+3 \times 6}{2+3}\right)\) or
\(\left(\frac{6-6}{5}, \frac{-8+18}{5}\right) \text { or }\left(0, \frac{10}{5}\right)\) or (0, 2)
∴ Gradient (slope) of the line = \(\frac { 3 }{ 2 }\)
∴ Equation of the line will be
(y – y₁) = m (x – x₁) ⇒ y – 2 = \(\frac { 3 }{ 2 }\) (x – 0)
⇒ 2y – 4 = 3x ⇒ 3x – 2y + 4 = 0

Question 7.
(a) Find the equation of a line passing through the point (- 2, 3) and having the x-intercept of 4 units.
(b) A (1, 4), B (3, 2), and C (7, 5) are the vertices of a triangle ABC. Find
(i) The coordinates of the centroid G of ∆ABC.
(ii) The equation of a line through G and parallel to AB.
Solution:
∵ x-intercept = 4
∴ Coordinates of the point on x-axis = (4, 0)
Now slope of the line joining the points (-2, 3) and (4, 0)
= \(\frac{y_2-y_1}{x_2-x_1}=\frac{0-3}{4+2}=\frac{-3}{6}=\frac{-1}{2}\)
∴ Equation of the line will be
y – y₁ = m (x – x₁)
⇒ y – 3 = \(\frac { -1 }{ 2 }\) (x + 2)
⇒ 2y – 6 = – x – 2
⇒ x + 2y – 6 + 2 = 0
⇒ x + 2y – 4 = 0

(b) Vertices of a ∆ABC are A (1,4), B (3,2) and C (7, 5)
(i) ∴ Coordinates of the centroid G of ∆ABC
= \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
or \(\left(\frac{1+3+7}{3}, \frac{4+2+5}{3}\right)\) or \(\left(\frac{11}{3}, \frac{11}{3}\right)\)

(ii) Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{3-1}=\frac{-2}{2}\) = – 1
Slope of the line parallel to AB from G = – 1
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – \(\frac { 11 }{ 3 }\) = – 1(x – \(\frac { 11 }{ 3 }\)
\(\frac{3 y-11}{3}\) = – \(\frac{(3 x-11)}{3}\)
⇒ 3y – 11 = – 1 (3x -11)
⇒ 3y – 11 = – 3x + 11
⇒ 3x + 3y – 11 – 11 = 0
⇒ 3x + 3y – 22 = 0

Question 8.
(a) A straight line passes through the points P (- 1, 4) and Q (5, – 2). It intersects the coordinate axes at points A and B. M is the mid-point of the segment AB. Find
(i) The equation of the line.
(ii) The coordinates of A and B.
(iii) The coordinates of M.

(b) Find the value of k for which the lines kx – 5j + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other.
Solution:
(a) A line passes through two points P (- 1, 4) and Q (5, -2) which intersects the axes at A and B. M is the mid-point of AB
Now slope of the line (m) = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{-2-4}{5+1}=\frac{-6}{6}\) = – 1

(i) ∴ Equation of the line will be
y – y₁ = m (x – x₁) ⇒ y – 4 = – 1 (x + 1)
⇒ y – 4 = – x – 1 ⇒ x + y – 4 + 1 = 0
⇒ x + y – 3 = 0

(ii) ∵ A lies on x-axis and B lies on y-axis
If A lies on x-axis then y-coordinates = 0
∴ x + 0 – 3 = 0 ⇒ x = 3
∴ Coordinates of A will be (3, 0)
∵ B lies on y-axis, then x-coordinate = 0
∴ 0 + y – 3 = 0 ⇒ y = 3
∴ Co-ordinates of B will be (0, 3)

(iii) ∵ M is the mid-point of AB
Co-ordinates of M will be
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
or \(\left(\frac{0+3}{2}, \frac{3+0}{2}\right)\) or \(\left(\frac{3}{2}, \frac{3}{2}\right)\) or (1.5, 1.5)

(b) In line kx – 5y + 4 = 0 ⇒5y = kx + 4
⇒ y = \(\frac{k x}{5}+\frac{4}{5}\)
Slope (m₁) = \(\frac { k }{ 5 }\)
and in line 4x – 2y + 5 = 0 ⇒ 2y = 4x + 5
⇒ y = \(\frac { 4 }{ 2 }\)x + \(\frac { 5 }{ 2 }\) ⇒ y = 2x + \(\frac { 5 }{ 2 }\)
∴ Slope (m₂) = 2
∵ The lines are perpendicular to each other
∴ m₁ x m₂ = – 1 ⇒ \(\frac { k }{ 5 }\) x 2 = – 1
⇒ k = \(\frac{-1 \times 5}{2}=\frac{-5}{2}\)

Question 9.
(a) KM is a straight line of 13 units. If K has coordinates (2, 5) and M has co-ordinates (x, – 7), find the possible values of x.
(b) The line joining P (- 4, 5) and Q (3, 2), intersects the y-axis at R. PM and QN are perpendiculars from P and Q on the x-axis. Find :
(i) the ratio PR : RQ.
(ii) the co-ordinates of R.
(iii) the area of quadrilateral PMNQ.
(c) P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.
Solution:
(a) KM is a straight line and length of KM = 13 units coordinates of K are (2, 5) and of M are (x, – 7)
∴ Length of KM = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
13 = \(\sqrt{(x-2)^2+(-7-5)^2}\)
⇒ 13 = \(\sqrt{(x-2)^2+(-12)^2}\)
Squaring both sides
169 = (x – 2)² + (- 12)²
169 = x² – 4x + 4 + 144
⇒ x² – 4x + 148 – 169 = 0
⇒ x² – 4x – 21 = 0
⇒ x² – 7x + 3x – 21 = 0
⇒ x (x – 7) + 3 (x – 7) = 0
⇒ (x – 7) (x + 3) = 0
∴ Either x – 7 = 0, then x = 7
or x + 3 = 0, then x = – 3
∴ Value of x are 7 or – 3

(b) The line joining the points P (- 4, 5) and Q (3, 2) intersects x-axis at R.
PM and QN are the perpendiculars drawn from P and Q on x-axis
Let R divides PQ in the ratio m₁ : m₂
∵ R lies on y-axis
(i) ∴ Its x-coordinates will be 0
∴ 0 = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2} \Rightarrow 0=\frac{m_1 \times 3+m_2 \times(-4)}{m_1+m_2}\)
⇒ 3m₁ – 4m₂ = 0 ⇒ 3m₁ = 4m₂
⇒ \(\frac{m_1}{m_2}=\frac{4}{3}\) ⇒ m₁ : m₂ = 4 : 3
∴ PR : RQ = 4 : 3

(ii) y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{4 \times 2+3 \times 5}{4+3}\)
= \(\frac{8+15}{7}=\frac{23}{7}\)
∴ Co-ordinates of R will be \(\left(0, \frac{23}{7}\right)\)

(iii) Now area of quadrilateral PMNQ or trapezium PMNQ
= \(\frac { 1 }{ 2 }\) (PM + QN) x MN
= \(\frac { 1 }{ 2 }\) (5 + 2) x 7 = \(\frac { 1 }{ 2 }\) x 7 x 7 = \(\frac { 49 }{ 2 }\)
= 24.5 sq. units

(c) The vertices of APQR are P (3, 4), Q (7, – 2) and R (- 2, – 1)
Let L is the mid point of PQ
∴ Co-ordinates of L are \(\left(\frac{3+7}{2}, \frac{4-2}{2}\right)\) or \(\left(\frac{10}{2}, \frac{2}{2}\right)\) or (5, 1)
∴ Slope of RQ = \(\frac{y_2-y_1}{x_2-x_1}=\frac{1+1}{5+2}=\frac{2}{7}\)
∴ Equation of median RL will be
y – y₁ = m (x – x₁)
⇒ y + 1 = \(\frac { 2 }{ 7 }\) (x + 2) ⇒ 7y + 7 = 2x + 4
⇒ 2x – 7y – 7 + 4 = 0
⇒ 2x – 7 – 3 = 0

Question 10.
(a) Use a graph for this question:
The graph of a linear equation in x and y, passes through A (- 1, – 1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and (\(\frac { 1 }{ 2 }\), k)

(b) In the given figure, write,
(i) The coordinates of A, B and C.
(ii) The equation of the line through A and || to BC.

Solution:
(a) Plot the points A (- 1, – 1) and B (2, 5) and join them. On this line two more points are taken P (h, 4) and Q (\(\frac { 1 }{ 2 }\), k)
We see that in P
If y = 4, then x = 1.5
∴ h = 1.5
and in Q, if x = \(\frac { 1 }{ 2 }\)
Then y = 2
∴ k = 2

(b) (i) From the figure, we find that
Co-ordinates of A are (2, 3) of B are (- 1, 2) and of C are (3, 0)

(ii) Now slope of BC
= \(\frac{y_2-y_1}{x_2-x_1}=\frac{0-2}{3+1}=\frac{-2}{4}=\frac{-1}{2}\)
∴ Slope of line drawn from Aparallel to BC = \(\frac { -1 }{ 2 }\)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac { -1 }{ 2 }\) (x – 2)
⇒ 2y – 6 = – x + 2 ⇒ x + 2y = 2 + 6 ⇒ x + 2y = 8

Question 11.
(a) The line segment joining A (2, 3) and B (6, – 5) is intercepted by the x-axis at the point K. Write the ordinate of the point K. Hence find the ratio in which K divides AB.
(b) If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
(c) Find the coordinates of the centroid of a triangle whose vertices are : A (-1, 3), B (1, -1) and C (5, 1).
Solution:
The slope of the line joining the points A (2, 3) and B (6, – 5) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-5-3}{6-2}\)
= \(\frac { -8 }{ 4 }\) = – 2
∵ This line is intercepted by x-axis at K
∴ Its ordinate = 0
Let K divides the line in the ratio m₁ : m₂
∴ x = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}\)
⇒ 0 = \(\frac{m_1 \times-5+m_2 \times 3}{m_1+m_2} \Rightarrow \frac{-5 m_1+3 m_2}{m_1+m_2}\) = 0
⇒ \(-5 m_1+3 m_2=0 \Rightarrow 5 m_1=3 m_2\)
⇒ \(\frac{m_1}{m_2}=\frac{3}{5}\)
Ratio = 3 : 5

(b) In the line y = 3x + 7
Slope (m₁) = 3
and in the line 2y + px = 3 ⇒ 2y = – px + 3
⇒ y = \(\frac{-p}{2} x+\frac{3}{2}\)
∴ Slope (m₂) = \(\frac { – p }{ 2 }\)
∵ Lines are perpendicular to each other
∴ m₁m₂ = – 1 ⇒ 3 x \(\frac { – p }{ 2 }\) = – 1
⇒ P = \(\frac{-1 \times 2}{-3}=\frac{2}{3}\)
Hence p = \(\frac { 2 }{ 3 }\)

(c) Vertices of a AABC are A (- 1, 3), B (1, – 1), C (5, 1)
Let G be the centroid of the triangle ABC
∴ Co-ordinates of G will be
\(\left(\frac{x_1+x_2+x_2}{3}, \frac{y_1+y_2+y_3}{3}\right)\) or
\(\left(\frac{-1+1+5}{3}, \frac{3-1+1}{3}\right) \text { or }\left(\frac{5}{3}, \frac{3}{3}\right) \text { or }\left(\frac{5}{3}, 1\right)\)

Question 12.
(a) The mid-point of the line segment joining (2a, 4) and (- 2, 2b) is (1, 2a + 1). Find the values of a and b.
(b) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).
(c) If the line joining the points A (4, – 5) and B (4, 5) is divided by the point P such that \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{2}{5}\), find the co-ordinates of P.
Solution:
(a) ∵ (1, 2a + 1) is the mid-point of the line joining the points (2a, 4) and (- 2, 2b)
∵ 1 = \(\frac { 2a – 2 }{ 2 }\) ⇒ 2a – 2 = 2
⇒ 2a = 2 + 2 = 4 ⇒ a = \(\frac { 4 }{ 2 }\) = 2
and 2a + 1 = \(\frac { 4+2b }{ 2 }\) ⇒ 4a + 2 = 4 + 2b
⇒ 4 x 2 + 2 – 4 = 2b ⇒ 10 – 4 = 2b
⇒ 2b = 6 ⇒ b = \(\frac { 6 }{ 2 }\) = 3
Hence a = 2, b = 3

(b) In line 3x + 2y = 8 ⇒ 2y = – 3x + 8
⇒ y = \(\frac { -3 }{ 2 }\)x + 4
Slope (m₁) = \(\frac { 3 }{ 2 }\)
∴ Slope of the line which is parallel to the given line
= m₂ = \(\frac { -3 }{ 2 }\) (∵ m₁ = m₂)
∵ It passes through the point (0, 1)
∴ Equation of the line will be
y – y₁ = m(x – x₁)
y – 1 = \(\frac { -3 }{ 2 }\) (x – 0) ⇒ 2y – 2 = – 3x
⇒ 3x + 2y – 2 = 0

(c) Two points are given A (4, – 5), B (4, 5)
P is a point on the line AB such that
\(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{2}{5}\)
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AP}}{\mathrm{AB}-\mathrm{AP}}=\frac{2}{5-2}=\frac{2}{3}\)
or m₁ : m₂ = 2 : 3
Let coordinates of P be (x, y)
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{2 \times 4+3 \times 4}{2+3}\)
= \(\frac{8+12}{5}=\frac{20}{5}\) = 4
y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{2 \times 5+3 \times(-5)}{2+3}\)
= \(\frac{10-15}{5}=\frac{-5}{5}\) = – 1
∴ Coordinates of P are (4, – 1)

Question 13.
(a) If A = (- 4, 3) and B (8, – 6),
(i) Find the length of AB.
(ii) In what ratio is the line joining AB, divided by the x-axis?
(b) Points A and B have coordinates (7, – 3) and (1, 9) respectively. Find :
(i) the slope of AB;
(ii) the equation of the perpendicular bisector of the line segment AB;
(iii) the value of ‘p’ if (- 2, p) lies on it.
Solution:
(a) Co-ordinates of A are (- 4, 3) and of B are (8, – 6)
(i) ∴ Length of AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{[8-(-4)]^2+(-6-3)^2}\)
= \(\sqrt{(8+4)^2+(-6-3)^2}=\sqrt{(12)^2+(-9)^2}\)
= \(\sqrt{144+81}=\sqrt{225}\)
∵ AB is intersected by x-axis
∴ The y-coordinates of the point of intersection = 0
Let this point divides AB in the ratio m₁ : m₂
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}\) ⇒ 0 = \(\frac{m_1 \times 8+m_2 \times(-4)}{m_1+m_2}\)
⇒ \(\frac{8 m_1-4 m_2}{m_1+m_2}=0 \Rightarrow 8 m_1-4 m_2=0\)
⇒ 8m₁ = 4m₂ ⇒ \(\frac{m_1}{m_2}=\frac{4}{8}=\frac{1}{2}\)
Ratio = 1 : 2

(b) Co-ordinates of two points are A (7, -3) and B (1, 9)
(i) Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}=\frac{9-(-3)}{1-7}\)
= \(\frac{9+3}{1-7}=\frac{12}{-6}\)

(ii) ∵ Slope of perpendicular to AB
\(\frac{-1}{m}=-\left(\frac{-1}{2}\right)=\frac{1}{2}\)
Co-ordinates of mid-point of AB
= \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \text { or }\left(\frac{7+1}{2}, \frac{-3+9}{2}\right)\) or (\(\frac { 8 }{ 2 }\), \(\frac { 6 }{ 2 }\)) or (4, 3)
Equation of the perpendicular line will be
y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac { 1 }{ 2 }\) (x – 4) ⇒ 2y – 6 = x – 4
⇒ x – 2y + 6 – 4 = 0 ⇒ x – 2y + 2 = 0

(iii) ∵ Point (- 2, p) lies on the line x – 2y + 2 = 0
∴ It will satisfy it
∴ – 2 – 2p + 2 = 0 ⇒ – 2p = 0 ⇒ p = 0
Hence p = 0

Question 14.
Find the equation of a line with x intercept = 5 and passing through the point (4, – 7).
Solution:
Given : x intercept = 5
∴ Line passes through (5, 0)
∴ Equation is
y = \(\frac { – 7 }{ – 1 }\)(x – 5)
y = 7x – 35 ⇒ 7x – y – 35 = 0

Question 15.
A and B are two points on the x-axis and y-axis respectively. P (2, – 3) is the mid point of AB. Find the
(i) Coordinates of A and B.
(ii) Slope of line AB.
(iii) equation of line AB.

Solution:
As P (2, – 3) is mid-point of AB.
Let coordinates of B be (0, y) and coordinates of A be (x, 0)
∴ By mid point formula \(\left(\frac{0+x}{2}, \frac{y+0}{2}\right)\) is the mid-point of AB. Comparing coordinates of mid-point x
∴ \(\frac { x }{ 2 }\) = 2 ⇒ x = 4
\(\frac { y }{ 2 }\) = – 3 ⇒ y = – 6

(i) ∴ Coordinates of A is (4, 0)
Coordinates of B is (0, -6)

(ii) Slope of line AB
= \(\frac{\text { Difference of } y \text { coordinates }}{\text { Difference of } x \text { coordinates }}\)
= \(\frac{0-(-6)}{4-0}\)
∴ Slope of line AB = \(\frac { 6 }{ 4 }\) = \(\frac { 3 }{ 2 }\)

(iii) Equation of line AB is
y – (- 6) = (Slope of AB) (x – 0)
y + 6 =\(\frac { 3 }{ 2 }\) (x)
2y+ 12 = 3x
⇒ 3x – 2y – 12 = 0

Question 16.
The equation of a line is 3x + 4y – 7 = 0. Find :
(i) the slope of the line.
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
Solution:
Given line 3x + 4y – 1 = 0
(i) Slope of line = – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } y}\)
Slope of line = \(\frac { -3 }{ 4 }\)

(ii) Point of intersection of the lines
x – y + 2 = 0
and 3x + y – 10 = 0,
Adding both we get,
4x – 8 = 0
4x = 8
⇒ x = 2
and putting the value of x in above equation 2 – y + 2 = 0 and y = 4
∴ Point of intersection is (2, 4)
Slope of any line perpendicular to 3x + 4y – 7 = 0 = \(\left(\frac{-4}{3}\right)=\frac{4}{3}\)
∴ Equation of line whose slope is \(\frac { 4 }{ 3 }\) and passing through (2, 4) is given by
y – 4 = \(\frac { 4 }{ 3 }\) (x – 2), [y – y₁ = m (x – x₁)]
⇒ 3 (y – 4) = 4 (x – 2)
⇒ 3y – 12 = 4x – 8
⇒ 4x – 3y – 8 + 12 = 0
⇒ 4x – 3y + 4 = 0

Question 17.
ABC is a triangle and G (4, 3) is the centroid of the triangle. If A = (1, 3), B = (4, b) and C = (a, 1), find ‘a’ and ‘b’ Find the length of side BC.
Solution:

Question 18.
ABCD is a parallelogram when A (x, y), B (5, 8) and C (4, 7) and D (2, -4). Find :
(i) Coordinates of A
(ii) Equation of diagonal BD.
Solution:

(i) Coordinates of O = \(\left(\frac{5+2}{2}, \frac{8-4}{2}\right)\) = (3.5, 2)
For the line AC
3.5 = \(\frac { x+4 }{ 2 }\)
⇒ x + 4 = 7
⇒ x = 7 – 4 = 3
x = 3, y = – 3

2 = \(\frac { y+7 }{ 2 }\)
⇒ y + 7 = 4
⇒ y = 4 – 7 = – 3
Thus, the coordinates of A are (3, -3)

(ii) Equation of diagonal BD is given by
y – 8 = \(\frac{-4-8}{2-5}\) (x – 5)
y – 8 = \(\frac { -12 }{ -3 }\) (x – 5)
⇒ y – 8 = 4x – 20
⇒ 4x – y – 12 = 0

Question 19.
Given equation of line L₁ is y = 4.
(i) Write the slope of line L₂ if L₂ is the bisector of angle O.
(ii) Write the coordinates of point P.
(iii) Find the equation of L₂.

Solution:
Line L₁ is y = 4 (given)
Coordinates of O is (0, 0)
(i) L₂ is the bisector of angle O
∴ θ = \(\frac { 90° }{ 2 }\) = 45°
m = tan θ = tan 45° = 1

(ii) OAPB is square
[∵ OB = BP = 4 as slope of line L₂ is 1]
∴ Coordinates of P is (4, 4)

(iii) Equation of L₂ :
y – o = \(\frac{4-0}{4-0}\) (x – 0)
⇒ y = \(\frac { 4 }{ 4 }\) (x)
⇒ y = x
⇒ x – y = 0

Question 20.
Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii) the length of AB.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1
So, 0 = \(\frac{m \times 8-4 \times 1}{m+1}\) ⇒ 8m – 4 = 0
⇒ m = \(\frac { 4 }{ 8 }\) ⇒ m = \(\frac { 1 }{ 2 }\)
So, required ratio = \(\frac { 1 }{ 2 }\) : 1 or 1 : 2

(ii) Also, y = \(\frac{1 \times(-3)+2 \times 6}{1+2}=\frac{9}{3}\) = 3
So, coordinates of the point of intersection are (0, 3)

(iii) AB = \(\sqrt{(8+4)^2+(-3-6)^2}\)
= \(\sqrt{144+81}\) = 15 units

Question 21.
The line through P (5, 3) intersects y axis at Q.

(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinate of Q.
Solution:
(i) Here θ = 45°
So, slope of the line = tan θ = tan 45° = 1

(ii) Equation of the line using point slope is
y – y₁ = m(x – x₁)
⇒ y – 3 = 1(x – 5) ⇒ y – 3 = x – 5
⇒ y = x – 2 or x – y – 2 = 0

(iii) Let the coordinates of Q be (0, y)
Then m = \(\frac{y_2-y_1}{x_2-x_1}\)
⇒ 1 = \(\frac{3-y}{5-0}\)
⇒ 5 = 3 – y ⇒ y = – 2
So, coordinates of Q are (0, – 2)

Question 22.
(a) AB is a diameter of a circle with centre C = (- 2, 5). If A = (3, – 7), find the coordinates of B.
(b) In ∆ABC, A (3, 5), B (7, 8) and C (1, – 10).
Find the equation of the median through A.
Solution:
(a) AC = \(\sqrt{(3+2)^2+(-7-5)^2}\)
= \(\sqrt{5^2+12^2}=\sqrt{25+144}\) = \(\sqrt{169}\) = 13 units

∵ AB is diameter and C is mid point of AB
Let co-ordinate of B are (x, y)
∴ \(\frac { 3+x }{ 2 }\) = – 2 and \(\frac { y – 7 }{ 2 }\) = 5
3 + x = – 4 and y – 7 = 10
x = – 4 – 3 and y = 10 + 7
x = – 7 and y = 17
∴ B is (- 7, 17)

(b) AD is median
⇒ D is mid point of BC

∴ D is \(\left(\frac{7+1}{2}, \frac{8-10}{2}\right)\)
i.e. (4 , – 1)
Slope of AD(m) = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{5+1}{3-4}=\frac{6}{-1}\) = – 6
∴ Equation of AD
y – y₁ = m (x – x₁)
y + 1 = – 6(x – 4)
y + 1 = – 6x + 24
y + 6x = – 1 + 24
6x + y = 23
6x + y – 23 = 0

Question 23.
In the figure given below, the line segment AB meets X-axis at A and Y-axis at B. The point P (- 3, 4) on AB divides it in the ratio 2 : 3. Find the coordinates of A and B.

Solution:
Let A (x, 0) and B (0, y)

3x = – 15 and 2y = 20
x = – 5 and y = 10
∴ Co-ordinates of A are (- 5, 0) and B are (0, 10)

Question 24.
(a) Calculate the ratio in which the line joining A (-4, 2) and B (3, 6) is divided by point P (x, 3). Also find (i) x (ii) Length of AP.
(b) Find the value of ‘a’ for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line.
Solution:
(a) Let ratio = k : 1

(b) A (a, 3), B(2, 1) and C(5, a) are collinear.
Slope of AB = Slope of BC
\(\frac{1-3}{2-a}=\frac{a-1}{5-2}\)
= \(\frac{-2}{2-a}=\frac{a-1}{3}\)
– 6 = (a – 1) (2 – a) (Cross-multipication)
– 6 = 2o – a² – 2 + a
– 6 = 3a – a² – 2
a² – 3a+ 2 – 6 = 0
a² – 3a – 4 = 0
a² – 4a + a – 4 = 0
a(a – 4) + (a – 4) = 0
(a + 1) (a – 4) = 0
a = – 1, or a = 4
a = – 1 (∵ does not satisfy the equation)
∴ a = 4
Slope of BC = \(\frac{a-1}{5-2}=\frac{4-1}{3}=\frac{3}{3}\) = 1 = m
Equation of BC ; (y – 1) = 1(x – 2)
y – 1 = x – 2 ⇒ x – y = – 1 + 2
x – y = 1
x – y – 1 = 0

Question 25.
Three vertices of a parallelogram ABCD taken in order ae A (3, 6), B (5, 10) and C (3, 2) find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD.
Solution:
Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2)
(i) We need to find the co-ordinates of D We know that the diagonals of a parallelogram bisect each other
Let (x, y) be the co-ordinates of D
∴ Mid-point of diagonal AC = \(\left(\frac{3+3}{2}, \frac{6+2}{2}\right)\) = (3, 4)
And, mid-point of diagonal BD
= \(\left(\frac{5+x}{2}, \frac{10+y}{2}\right)\)

Thus, we have
\(\frac{5+x}{2}=3 \text { and } \frac{10+y}{2}=4\)
⇒ 5 + x = 6 and 10 + y = 8
⇒ x = 1 and y = – 2
∴ Coordinate of D = (1, – 2)

(ii) Length of diagonal BD
= \(\sqrt{(1-5)^2+(-2-10)^2}\)
= \(\sqrt{(4)^2+(-12)^2}\)
= \(\sqrt{16+144}=\sqrt{160}\) units
= 12.65 units

(iii) Equation of the side joining A (3, 6) and D (1, -2) is given by
\(\frac{x-3}{3-1}=\frac{y-6}{6+2}\)
⇒ \(\frac{x-3}{2}=\frac{y-6}{8}\)
⇒ 4(x – 3) ⇒ y – 6 ⇒ 4x – 12 = y – 6
⇒ 4x – y = 6
Thus, the equation of the side joining A (3, 6) and D (1, -2) is 4x – y = 6

Question 26.
In the given figure ABC is a triangle and BC is parallel to the y-axis. AB and AC intersects the y-axis at P and Q respectively.
(i) Write the coordinates of A.
(ii) Find the length of AB and AC.
(iii) Find the ratio in which Q divides AC.
(iv) Find the equation of the line AC.

Solution:
(i) The line intersects the x-axis where, y = 0
The co-ordinates of A are (4, 0)

(ii) Length of AB = \(\sqrt{[4-(-2)]^2+(0-3)^2}\)
= \(\sqrt{36+9}=\sqrt{45}\) = 6.71 units
Length of AC = \(\sqrt{[4-(-2)]^2+(0+4)^2}\)
= \(\sqrt{36+16}=\sqrt{52}\)
= 7.21 units

(iii) Let k be the required ratio which divides the line segment joining the co-ordinates A (4, 0) and C (- 2, – 4)
Let the co-ordinates of Q be x and y
∴ x = \(\frac{k(-2)+1(4)}{k+1}\) and y = \(\frac{k(-4)+0}{k+1}\)
Q lies on the y-axis where x = 0
⇒ \(\frac{-2 k+4}{k+1}\) ⇒ – 2k + 4 = 0
⇒ 2k = 4
The ratio is 2 : 1

(iv) The equation of line AC is
\(\frac{x-4}{4+2}=\frac{y-0}{0+4}\)
⇒ \(\frac{x-4}{6}=\frac{y}{4}\)
⇒ \(\frac{x-4}{3}=\frac{y}{2}\)
⇒ 2(x – 4) = 3y
⇒ 2x – 8 = 3y
⇒ 2x – 3y = 8
The equation of the line AC is 2x – 3y = 8

Question 27.
(a) The slope of a line joining P (6, k) and Q (1 – 3k, 3) is \(\frac { 1 }{ 2 }\). Find
(i) k
(ii) Midpoint of PQ, using the value of ‘A’ found in (i).
(b) A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio 1 : 2.
(i) Find the coordinates of A and B.
(ii) Find the equation of the line through P and perpendicular to AB.

Solution:
(a) (i) Slope of PQ = \(\frac{3-k}{1-3 k-6}\)
\(\frac{1}{2}=\frac{3-k}{-3 k-5}\)
⇒ -3k – 5 = 2(3 – k)
⇒ -3k – 5 = 6 – 2k
⇒ k = 1

(ii) Substitute k in P and Q.
∴ P (6, k) = P (6, 1)
and Q (1 – 3k, 3) = Q (- 2, 3)
Midpoint of PQ = \(\left(\frac{6-2}{2}, \frac{1+3}{2}\right)\)
= \(\left(\frac{4}{2}, \frac{4}{2}\right)\)
= (2, 2)

(b) (i) Since, A lies on the x-axis, let the coordinates of A be (x, 0).
Since B lies on the y-axis, let the coordinates of B be (0, y).
Let m = 1 and n = 2.
Using section formula,
Coordinates of P = \(\left[\frac{1(0)+2(x)}{1+2}, \frac{1 y+2(0)}{1+2}\right]\)
∴ (4, – 1) = \(\left(\frac{2 x}{3}, \frac{y}{3}\right)\)
⇒ \(\frac { 2x }{ 3 }\) = 4 and \(\frac { y }{ 3 }\) = – 1
⇒ x = 6 and y = – 3
So, the coordinates of A are (6, 0) and that of B are (0, -3).

(ii) Slope of AB = \(\frac{0-(-3)}{6-0}\)
= \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
(Slope of line through P)
(Slope of AB) = – 1
⇒ Slope of line through P\(\frac { 1 }{ 2 }\) = – 1
(Since the lines are perpendicular)
⇒ Slope of line through P = – 2
Let the equation of the line through P be y = mx + c
⇒ y = – 2x + c
c is the x-intercept
Put y = 0 and x = c in (i) to find c
y = – 2x + c ⇒ 0 = – 2c + c ⇒ c = 0
So, the equation of the line is y = – 2x

Effective S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(d) can help bridge the gap between theory and application.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(d)

Question 1.
In figure, PA and PB are tangents from P to a circle with centre O. At M, a tangent is drawn cutting PA at K and PB at N. Prove that KN = AK + BN.

Solution:
Given : In a circle with centre O, from a point P outside of it, PA and PB are the tangents to the circle.
At a point M, on the circle, another tangent is drawn which intersects AP at K and BP at N.
To prove : KN = AK + BN
Proof: From K, KA and KM are tangents to
the circle.
∴ KA = KM ….(i)
Similarly from N, NM and NB tangents are drawn.
∴ NB = MN ….(ii)
Adding (i) and (ii)
KM + MN = KA + NB
KN=AK+BN
Hence proved.

Question 2.
In figure, two concentric circles are given. Prove that every chord of the greater circle which is tangent to the smaller circle is bisected at its point of contact.

Solution:
Given : Two concentric circles with centre O.
A chord AB is drawn to the greater circle which touches the smaller circle at P.
To prove : AB is bisected at P i.e. AP = PB
Construction : Join OP, OA and OB.

Proof : ∵ OP is the radius and AB is the tangent
∴ OP ⊥ AB
In right △OAP and △OBP
Hyp. OA = OB (radii of the same circle)
Side OP = OP (common)
∴ △OAP ≅ △OBP (RHS axiom)
∴ AP = PB (c.p.c.t.)
Hence P bisects AB
Hence proved.

Question 3.
Prove that the angle between two tangents drawn from an external point to a circle is supplementary to angle subtended by the line segments joining the points of contact at the centre.
Solution:
Given : A circle with centre O, from a point P outside of it, two tangents PT and PS are drawn TS is joined. Which subtends
∠TOS at the centre.
To prove : ∠TPS + ∠TOS = 180° (or are supplementary)
Construction : Join OP.

Proof: In right angle △OTP,
∠OTP = 90°
∴ ∠TOP + ∠TPO = 90° ….(i)
Similarly in right △OSP,
∠SOP + ∠SPO = 90° ….(ii)
Adding (i) and (ii)
∠TOP + ∠SOP + ∠TPO + ∠SPO = 90° + 90°
⇒ ∠TPS + ∠TOS = 180°
∠TPS and ∠TOS are supplementary
Hence proved.

Question 4.
In figure, a circle touches the side be of a △ABC at X and AB and AC produced at Y and Z respectively. Prove that AY is half the perimeter of △ABC.

Solution:
Given : In the figure a circle touches the side BC of a △ABC externally and AB and AC on producing at Y and Z.
To prove : AY = \(\frac { 1 }{ 2 }\) (AB + BC + CA)
Proof: ∵ From A, AY and AZ are tangents to the circle
∴ AY = AZ
Similarly from B, BY and BX are tangents
∴ BY = BX
and from C, CX and CZ are tangents
∴ CZ = CX
Now AY = AB + BY = AB + BX
AZ = AC + CZ = AC + CX
∴ AY + AZ = AB + BX + CX + AC
⇒ AY + AY=AB + BC + CA (∵ AY = AZ)
⇒ 2AY = AB + BC + CA
⇒ AY = \(\frac { 1 }{ 2 }\) (AB + BC + CA) Hence proved.

Question 5.
In figure, the incircle of a △ABC touches the sides BC, CA and AB at D, E and F respectively. Show that AF + BD + CE = AE + CD + BF = \(\frac { 1 }{ 2 }\) (perimeter of △ABC).

Solution:
Given : Incircle touches the sides BC, CA and AB at D, E and F respectively of the △ABC
To prove : AF + BD + CE = AE + CD + BF = \(\frac { 1 }{ 2 }\) (AB + BC + CA)
Proof: From A, AF and AE are the tangents to the circle
∴ AF = AE ….(i)
Similarly from B, BD and BF are the tangents
∴ BD = BF ….(ii)
and from C, CE and CD are the tangents
∴ CE = CD ….(iii)
Adding (i), (ii) and (iii) we get,
AF + BD + CE = AE + BF + CD
Adding to both sides
AF + BD + CE
2AF + 2BD + 2CE = AE + BF + CD + AF + BD + CE
2(AF + BD + CE) = AE + CE + BD + CD + AF + BF
⇒ 2(AF + BD + CE) = AC + BC + AB
⇒ AF + BD + CD = \(\frac { 1 }{ 2 }\)(AB + BC + CA) Hence proved.

Question 6.
Two circles touch each other externally. Show that the lengths of the tangents, drawn to the two circles from any point on the common tangents, are equal (See figure).

Solution:
Two circles touch each other externally at T.
PT is their common tangent and from P, tangents PA and PB are drawn to the two circles
To prove : PA = PB
Proof: ∵ From P, PA and PT are the tangents to the bigger circle
∴ PA = PT ….(i)
Similarly from P, PT and PB are the tangents drawn to the smaller circle
∴ PT = PB ….(ii)
From (i) and (ii)
PA = PB Hence proved.

Question 7.
Two circles touch externally at A. A common tangent touches them at B and C, and another common tangent at A meets the previous common tangent at P, as is shown in figure. Prove that
(i) PB = PC,
(ii) ∠BAC = 90°

Solution:
Given : Two circles touch each other externally at A.
A common tangent touches them at B and C Another common tangent at A meets BC at P.
To prove :
(i) PB = PC
(ii) ∠BAC = 90°
Proof:
(i) ∵ Through P, PB and PA are tangents to the first circle
∴ PB = PA ….(i)
Similarly through P, PA and PC are tangents to the second circle
∴ PA = PC ….(ii)
From (i) and (ii)
PB = PC

(ii) In △PAB,
PA = PB (proved)
∴ ZPBA = ZPAB …(iii)
Similarly in △PAC
PA = PC
∴ ∠PCA = ∠PAC ….(iv)
Adding (iii) and (iv)
∠PBA + ∠PCA = ∠PAB + ∠PAC = ∠BAC
But ∠PBA + ∠PCA + ∠BAC = 180° (Sum of angles of a triangle)
∴ ∠BAC + ∠BAC = 180°
⇒ 2 ∠BAC = 180°
⇒ ∠BAC = 180° × \(\frac { 1 }{ 2 }\) = 90°
Hence ∠BAC = 90° Hence proved.

Question 8.
Two circles touch internally at A. P is any point on the tangent at A. From P, two tangents PB and PC are drawn to the two circles. Prove that PB = PC.
Solution:
Given : Two circles touch each other at A externally. Through A, a common tangent is drawn. A point P is taken on the tangents.
Through P, PB and PC two tangents are drawn.

To prove : PB = PC
Proof: Through P, two tangents PA and PB are drawn to the first circle
∴ PA = PB ….(i)
Again through P, PA and PC are the tangents
drawn to the second circle
∴ PA = PC ….(ii)
From (i) and (ii)
PB = PC Hence proved.

Question 9.
From a point P, outside a circle, with centre O, tangents PA and PB are drawn as shown in the figure. Prove that :
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of AB.

Solution:
Solution-
Given : In the circle with centre O, from a point P outside of it, PA and PB are tangents drawn
OP, OA, OB and AB are joined
To prove :
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of AB
Proof:
(i) In right AOAP and AOBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∴ △OAP = △OBP (RHS axiom)
∴ ∠AOP = ∠BOP (c.p.c.t.)
and ∠APO = ∠BPO (c.p.c.t.)
Now in △AMP and △BMP,
MP = MP (common)
PA = PB (Tangents to the circle)
∠APM = ∠BPM (proved)
∴ △AMP ≅ △BMP (SAS axiom)
∴ AM = BM (c.p.c.t.)
∠AMP = BMP
But ∠AMP + ∠BMP = 180° (Linear pair)
∴ ∠AMP = ∠BMP = 90°
Hence OP is the perpendicular bisector of AB.
Hence proved.

Question 10.
Prove that the tangents to a circle at the extremities of any chord make equal angles with the chord.
Solution:
Given : In a circle with centre O, AB is the chord. Through A and B tangents PQ and PR are drawn which meet each other at P

To prove : ∠PAB = ∠PBA
Construction : Join OA, OB and OP
Proof : In △OAP and △OBP,
OP = OP (common)
OA = OB (radii of the same circle)
PA = PB (tangents to the circle)
∴ △OAP ≅ △OBP (SSS axiom)
∴ ∠APO = ∠BPO (c.p.c.t.)
Now in △PAM and △PBM,
PM = PM (common)
PA = PB (tangents to the circle)
∠APM = ∠BPM (∠APO = ∠BPO proved)
∴ △PAM ≅ △PBM (SAS axiom)
∴ ∠PAM = ∠PBM
⇒ ∠PAB = ∠PBA Hence proved.

Question 11.
In the figure, two circles touch each other externally at C. Prove that the common tangent at C bisects the other two common tangents.

Solution:
Given : Two circles with centres O and O’ touch each other externally at C.
PQ and RS are common tangents and a third common tangent from C is drawn which intersects PQ and RS and L and M.
To prove : LM bisects PQ and RS.
Proof: From L, LP and LC are the tangents to the circle.
∴ LP = LC ….(i)
Similarly from L, LQ and LC are two tangents.
∴ LC = LQ ….(ii)
From (i) and (ii)
LP = LC = LQ
∴ L is the midpoint of PQ
Similarly we can prove that
M is the midpoint of RS
Hence LM bisects the tangents PQ and RS
Hence proved.

Question 12.
PA and PB are tangents drawn from an external point P to a circle with centre C (See figure). Prove that ∠APB = 2∠CAB.

Solution:
Given : A circle with centre C. PA and PB are two tangents drawn from P to the circle CA is joined
To prove : ∠APB = 2∠CAB
Proof: In △APB,
AP = BP (tangents to the circle from P)
∴ ∠PAB = ∠PBA (opposite angles to equal sides)
But vPAB + ∠PBA + ∠APB = 180° (Sum of angles of a triangle)
⇒ ∠PAB + ∠PAB + ∠APB = 180°
⇒ ∠APB = 180°- 2∠PAB
But ∠PAB = ∠CAP – ∠CAB
= 90° – ∠CAB (∵ CA ⊥ PA)
∴ ∠APB = 180° – 2 (90° – ∠CAB)
⇒ ∠APB = 180° – 180° + 2∠CAB
⇒ ∠APB = 2 ∠CAB
Hence proved.

Continuous practice using S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(c) can lead to a stronger grasp of mathematical concepts.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(c)

Question 1.
(a) In figure, APB is tangent to circle with centre O. If ∠QPB = 50°, find ∠POQ.

(b) In figure, AB and AC are tangents. If AB = 4 cm, find AC.

(c) In the figure, PQ and PR are tangents to circle, centre O. If ∠QPR = 80°, find ∠QOR.

Solution:
(a) In the figure, APB is tangent to the circle
with centre O
∠QPB = 50°
∵ OP is the radius and APB is tangent
∴ OP ⊥ APB
∴ ∠OPB = 90° ⇒ ∠OPQ + ∠QPB = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50°
⇒ OPQ = 40°
But in △OPQ, OP = OQ (radii of the circle)
∴ ∠OPQ = ∠OQP = 40°
∠POQ + ∠OPQ + ∠OQP = 180°
⇒ ∠POQ + 40° + 40° = 180°
⇒ ∠POQ + 80°= 180°
∠POQ = 180° – 80 = 100°

(b) In circle, two tangent AB and AC are drawn, a point A outside the circle
∴ AC = AB = 4 cm

(c) In the figure, a circle with centre O from a point P outside the circle two tangents PQ and PR are drawn and ∠QPR = 80°
∴ ∠QPR and ∠QOR are supplementary ∠QPR + ∠QOR = 180°
⇒ 80° + ∠QOR = 180°
⇒ ∠QOR = 180° – 80° = 100°
∠QOR = 100°

Question 2.
In figure, O is the centre of the circle. Find ∠POS.

Solution:
In the figure, a circle with centre O
From a point P outside of it, tangents PT and
PS are drawn to the circle, and ∠TPO = 30°
In △PTO, OT ⊥ PT
∴ ∠OTP = 90°
∴ ∠TOP + ∠TPO = 90°
⇒ ∠TOP + 30° = 90°
⇒ ∠TOP = 90°- 30° = 60°
∴ OP is the bisector of ∠TOS
∴ ∠TOP = ∠POS = 60°

Question 3.
In figure, PQ is tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate
(i) ∠QAB
(ii) ∠PAD
(iii) ∠CDB

Solution:
In the figure, BD is the diameter of the circle PQ is tangent to the circle at A
∠ADB = 30°, ∠DBC = 60°

(i) ∵ QAP is tangent and AB is chord of the circle
∴ ∠QAB = ∠ADB = 30°

(ii) ∠PAD + ∠DAB + ∠QAB = 180° (Angles of a line)
⇒ ∠PAD + 90° + 30° = 180°
(∵ ∠DAB = 90° angle in semicircle)
⇒ ∠PAD + 120° = 180°
⇒ ∠PAD = 180° – 120° = 60°

(iii) In ABCD,
∠CDB + ∠CBD + ∠BCD = 180° (Sum of angles of a triangle)
⇒ ∠CDB + 60° + 90°= 180°
⇒ ∠CDB + 150° = 180°
⇒ ∠CDB = 180°- 150° = 30°

Question 4.
In figure, PR and PQ are the tangents, each of them being equal to 9 cm, ∠QPR = 60°. Find the length of the chord QR which joins their points of contact.

Solution:
In the figure, PQ and PR the tangents drawn
from P outside the circle such that
PQ = PR = 9 cm and ∠QPR = 60°
QR is joined
In △PQR, ∠QPR = 60°
∵ PQ = PR
∴ ∠PQR = ∠PRQ = 60°
∴ △PQR is an equilateral triangle
PQ = PR = QR = 9 cm

Question 5.
Find the length of the tangent drawn to a circle of radius 3 cm, from a point distant 5 cm from the centre.
Solution:
In a circle of radius 3 cm, point P is 5 cm away from the centre O of the circle
PQ and PR are the tangents drawn from P to the circle

∵ OQ is radius and PQ is tangent
∴ OQ ⊥ QP or ∠OQP = 90°
Now in right angled △OPQ,
OP² = OQ² + PQ² (Pythagoras Theorem)
⇒ (5)² = (3)² + PQ²
⇒ 25 = 9 + PQ²
⇒ PQ² = 25 – 9 = 16 = (4)²
∴ PQ = 4 cm
But PQ = PR
∴ PQ = PR = 4 cm

Question 6.
A circle touches the side BC of △ABC at P and touches AB and AC produced at Q and R respectively. If AQ = 5 cm, find the perimeter of △ABC.

Solution:

∵ From A, AQ and AR are the tangents drawn to the circle
∴ AQ = AR = 5cm ….(i)
Similarly from B, tangent BQ and BP are drawn
∴ BQ = BP ….(ii)
and from C,
CR = CR ….(iii)
Now perimeter of △ABC,
= AB + AC + BC = AB + AC + BP + CP = AB + AC + BQ + CR [From (ii) and (iii)]
= AB + BQ + AC + CR
= AQ + AR = 5 cm + 5 cm [From (i)] = 10 cm

Question 7.
There are two concentric circles of radii 3 cm and 5 cm respectively. Find the length of the chord of the outer circle which touches the inner circle
Solution:
Two circles which are concentric and their centre is O, are of radii 5 cm and 3 cm
i. e. OA = 5 cm and OP = 3 cm
AB is chord of the larger circle which touches the smaller circle at P

∴ OP ⊥ AB ⇒ AP = PB
In right △OAP,
OA² = OP² + AP² (Pythagoras Theorem)
⇒ (5)² = (3)² + AP²
⇒ 25 = 9 + AP²
⇒ AP² = 25 – 9 = 16 = (4)²
∴ AP = 4 cm
Hence AB = 2AP = 2 × 4 = 8 cm

Question 8.
Three circles with centres A, B, C touch each other externally; AB = 4 cm, BC = 6 cm, CA = 7 cm; find their radii.
Solution:
ABC is a triangle and with centre A, B and C, three circles are drawn touching each other externally a t P, Q and R respectively AB = 4 cm, BC = 7 cm and AC = 6 cm

Let radii of circles with centre A, B and C respectively be x, y and z
∴ AB = x + y, BC = y + z, CA = z + x
⇒ x + y = 4 cm, y + z = 6 cm, z + x = 7 cm
∴ AB + BC + CA = x +y +y + z + z + x
⇒ 4 + 6 + 7 = 2(x + y + z)
⇒ x + y + z = \(\frac { 17 }{ 2 }\) = 8.5 cm
Subtracting from x + y + z, we get
x = 8.5 – 4 = 4.5 cm
y = 8.5 – 6 = 2.5 cm
z = 8.5 – 7 = 1.5 cm
Hence their radii are 2.5 cm, 1.5 cm and 4.5 cm

Question 9.
Equal circles, centres O and O’ touch each other at X. OO’ is produced to meet the circle O’ at A. AC is tangent to the circle whose centre is O. O’ is perpendicular to AC. Find the value of

Solution:
Two equal circles with centres O and O’ touch each other externally at X. 00′ is produce to meet the circle O’ at A. Through A, a tangent AC is drawn to the circle with
centre O. O’D ⊥ AC
Let r be the radius of each circle
In △AO’D and △AOC,
∠D = ∠C (each 90°)
∠A = ∠A (common)

Question 10.
In figure, P and Q are centres of two circles of radii 12 cm and 3 cm respectively. A and B are the points of contact of the common tangent XY. Find AB.

Solution:
Two circles with centres P and Q touch externally at R
XY is their common tangent
A and B are their points of contact
Join PA, QB and PQ
From Q, draw QS || XY

∵ PA and QB are perpendicular to XY and
QS || AB
∴ QS = AB
PA = 12 cm, QB = 3 cm, PQ = 12 + 3 = 15 cm
∴ PS = PA – SA = 12 – 3 = 9 cm
Now in right △PSQ,
PQ² = PS² + QS² (Pythagoras Theorem)
⇒ (15)² = (9)² + QS²
⇒ 225 = 81 + QS²
⇒ QS² = 225 – 81 = 144 = (12)²
∴ QS = 12 cm
But AB = QS = 12 cm

Peer review of S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(a) can encourage collaborative learning.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(b)

Question 1.
In the fig., in △ABC, AB = AC, and XY || BC. Prove that BCYX is a cyclic quadrilateral.

Solution:
Given : In the figure, in △ABC, AB = AC. XY || BC
To prove : BCYX is a cyclic quadrilateral
Proof : In △ABC,
XY || BC
∴∠AXY = ∠ABC (Corresponding angles)
But △ABC = ∠ACB (∵AB = AC)
∴∠AXY = ∠ACB
But Ext. ∠AXY is equal to its interior opp. ∠ACB
∴ BCYX is a cyclic quadrilateral Hence proved.

Question 2.
In the figure, In a quad. ABCD, ext. ∠XCD = int. opp ∠A. Prove that the quad. ABCD is a cyclic quad.

Solution:
Given : In the figure, in quad. ABCD
BC is produced to X
Ext. ∠XCD = Int. opp. ∠A
To prove : Quad. ABCD is a cyclic
Proof: ∠XCD + ∠DCB = 180° (Linear pair)
∠A = ∠DCB = 180° (∵ ∠A = ∠XCD)
But these are opposite angles of a quad.
∴ Quad ABCD is a cyclic Hence proved.

Question 3.
In the figure, PQR is an isosceles triangle with PQ equal to PR. A circle passes through Q and R and intersects the sides PQ and PR at points S and T respectively. Prove that QR || ST.

Solution:
Given : In △PQR, PQ = PR
A circle passing through Q and R, intersects PQ and PR at S and T respectively. ST is joined
To prove : ST || QR
Proof: In △PQR
PQ = PR (given)
∴ ∠Q = ∠R ….(i) (Angles opposite to equal sides)
∵ SQRT is a cyclic quadrilateral
∴ Ext. ∠S = Int. opp. ∠R …(ii)
From (i) and (ii)
∠Q = ∠S
But these are corresponding angles
∴ QR || ST Hence proved.

Question 4.
In the figure, AB is the common chord of two circles. If AC and AD are diameters, prove that D, B and C are in a straight line. O₁ and O₂ are the centres of the circles.
Solution:

Given : Two circles with centres O₁ and O₂
intersect each other at A and B
AO₁C and AO₂D are diameters
To prove : D, B and C are in a straight line or D, B and C are collinear
Proof : ∵ AC is the diameter of circle of centre O₁
∴ ∠ABC = 90° (Angle in a semicircle)
Similarly
∠ABD = 90° (Angle in a semicircle)
Adding we get,
∠ABC + ∠ABD = 90° + 90° = 180°
∴ CBD is a straight line
Hence D, B and C are in the same straight line Hence proved.

Question 5.
In figure, AB is the diameter of the circle whose centre is O. AD and BC are perpendiculars to the line XY. CB meets the circle at E. Prove that CE = AD.

Solution:
Given : In a circle with centre O, AB is its diameter AD ⊥ XY and BC ⊥ XY which intersects the circle at E
To prove : CE = AD
Construction : Join A, E.

Proof: ∠AEB = 90° (Angle in a semicircle)
∴ ∠AEC = 90°
(∵ ∠AEB + ∠AEC = 180° linear pair)
But ∠C = ∠D = 90°
(∵ AD and EC are perpendicular to XY)
∴ AECD is a rectangle
∵ Opposite sides of a rectangle are equal
∴ CE =AD Hence proved.

Question 6.
In fig., AB and CD are parallel chords of a circle whose diameter is AC. Prove that AB = CD.

Solution:
Given : In a circle with centre O, AC is its diameter and chord AB || CD
To prove : AB = CD
Construction : Join OB and OD

Proof: In AAOB and ACOD,
OA = OC
OB = OD (radii of the same circle)
∠BAO = ∠OCD (Alternate angles)
∴ △AOB ≅ △COD (SSA axiom)
∴ AB = CD Hence proved.

Question 7.
In figure, APB and CQD are straight lines through the points of intersection of two circles. Prove
(i) AC || BD,
(ii) ∠CPD = ∠AQB (SC)

Solution:

Given : Two circles intersect each other at P and Q
Lines APB and CQD are drawn from the point of intersection respectively PQ, AC, BD and AQ, QB, CP and PD are joined
To prove :
(i) AC || BD (ii) ∠CPD = ∠AQB
Proof:
(i) ∵ APQC is a cyclic quad.
∴ Ext. ∠BPQ = Int. opp. ∠C ….(i)
∵ PBDQ is a cyclic quad.
∴ ∠BPQ + ∠D = 180° (sum of opp. angles)
⇒ ∠C + ∠D = 180° {from (∠)}
But these are co-interior angles
∴ AC || BQ

(ii) In △AQB and △CPD,
∠PAQ = PCQ (Angles in the same segment)
∠PBQ = ∠PDQ (Angles in the same segment)
or ∠BAQ = ∠ABQ
∠ABQ = ∠PDC
∴ △AQB ~ △APD (AA axiom)
∴ Third angle = Third angle
⇒ ∠AQB = ∠CPD Hence proved.

Question 8.
Prove that the circle drawn with any side of a rhombus as a diameter, passes through the point of intersection of its diagonals.
Solution:
(a) Given: ABCD is a rhombus whose diagonals AC and BD intersect each other at O. A circle with AB as diameter is drawn.

To prove : The circle passes through O
Proof : Let the circle drawn on AB as diameter does not passes through O, let it intersect AC at P
Join PB
The ∠APB = 90° (Angle in a semicircle)
But ∠AOB = 90°
(∵ The diagonals bisect each other at right angles)
∴ ∠APB = ∠AOB
But it is not possible because ∠APB is the exterior angle of AOPB and an exterior angle of a triangle is always greater than its interior opposite angle
∴ Our supposition is wrong
Hence the circle will pass through O Hence proved.

Question 9.
If two non-parallel sides of a trapezium are equal, it is cyclic.
OR
An isosceles trapezium is always cyclic.

Solution:
Given : An isosceles trapezium ABCD in
which AD || BC and AB = DC
To prove : ABCD is cyclic
Construction : Draw AE and DF perpendicular on BC
Proof: In right △ABE and △DCF
Hyp. AB = DC (given)
Side AE = DF (Distance between two parallel lines)
∴ △ABE ≅ △DCF (RHS axiom)
∴ ∠B = ∠C (c.p.c.t.)
Now AD || BC
∴ ∠DAB + ∠B = 180° (Co-interior angles)
⇒ ∠DAB + ∠C = 180° (∵ ∠B = ∠C proved)
But these are sum of opposite angles of a quad.
∴ ABCD is a cyclic. Hence proved.

Question 10.
Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles.

Solution:
Given : PQRS is a cyclic quadrilateral
∠A, ∠B, ∠C and ∠D are angles in the four segment so formed exterior to the cyclic quadrilateral
To prove : ∠A + ∠B + ∠C + ∠D = 6 right angles
Construction : Join AS and AR
Proof: In cyclic quad. ASDP,
∠PAS + ∠D = 2 rt. angles ….(i) (Sum of opposite angles)
Similarly in cyclic quad. ARBQ,
∠RAQ + ∠B = 2 rt. angles ….(ii)
and in cyclic quad. ARCS,
∠SAR + ∠C = 2rt. angles ….(iii)
Adding (i), (ii) and (iii)
∠PAS + ∠D + ∠RAQ + ∠B + ∠SAR + ∠C = 2 + 2 + 2 = 6rt. angles
⇒ PAS + ∠SAR + ∠RAQ + ∠B + ∠C + ∠D = 6 rt. angles
⇒ ∠A + ∠B + ∠C + ∠D = 6 rt. angles Hence proved.

Question 11.
(i) In figure, O is the circumcentre of △ABC and OD ⊥ BC. Prove that ∠BOD = ∠A.

Solution:
Given : O is the circumcentre of △ABC
OD ⊥ BC. OB and OC are joined
To prove : ∠BOD = ∠A
Proof: Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle
∴ ∠BOC = 2 ∠BAC ….(i)
In △OBC, OD ⊥ BC
OB = OC (radii of the same circle)
∴ OD bisects ∠BOC
⇒ ∠BOD = \(\frac { 1 }{ 2 }\) ∠BOC ….(ii)
From (i) and (ii)
∠BOD = \(\frac { 1 }{ 2 }\) ∠BAC = ∠BAC
Hence ∠BOD = ∠A
Hence proved.

Question 12.
ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC in the points E and F respectively. Prove that EF || DC.

Solution:
Given : ABCD is a cyclic quadrilateral
A circle passing through A and B meet AD
and BC at E and F respectively
EF is joined
To prove : EF || DC
Proof: ∵ ABCD is a cyclic quad.
∴ ∠1 + ∠3 = 180° ….(i) (Sum of opposite angles)
Similarly ABFE is a cyclic quadrilateral
∴ ∠1 + ∠2 = 180° ….(ii)
From (i) and (ii)
∠1 + ∠3 = ∠1 + ∠2 ⇒ ∠3 = ∠2
But these are corresponding angles
∴ EF || DC Hence proved.

Regular engagement with S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(a) can boost students’ confidence in the subject.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(a)

Question 1.
Fill up the blanks :
(i) O is the centre of the circle, ∠BAC = 46°, ∠ABC ………………………
(ii) If ∠BAC = 70° and ∠DAC = 40°, then ∠BCD = …………..
(iii) O is the centre of the circle. If ∠x = 120°, then ∠y = …………….
(iv) BC is a diameter of the circle and ∠CAD = 65°, then ∠BCD = …………..

Solution:
(i) AB is diameter of the circle
∴ ∠ACB = 90° (Angle in a semicircle)
∴ ∠BAC + ∠ABC = 90°
⇒ 46° + ∠ABC = 90°
⇒ ∠ABC = 90°- 46° = 44°

(ii) AC is diameter of the circle
∴∠ADC = ∠ABC = 90° (Angle in a semicircle)
∴∠BAC + ∠BCA = 90°
⇒ 70° +∠BCA = 90°
⇒ ∠BCA = 90° – 70° = 20°
and ∠DAC + ∠DCA = 90°
⇒ 40° + ∠DCA = 90°
⇒ ∠DAC = 90° – 40° = 50°
∴ ∠BCD = ∠BCA + ∠ACD = 20° + 50° = 70°

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴ ∠AOB = 2 ∠ACB
⇒ ∠x = 2 ∠y ⇒ 2 ∠y = 120°
⇒ ∠y = \(\frac{120^{\circ}}{2}\) = 60°
∴ ∠y = 60°

(iv) BC is the diameter of the circle
∴ ∠BAC = 90° (Angle in a semicircle)
⇒ ∠BAD + ∠DAC = 90°
⇒ ∠BAD + 65° = 90°
⇒ ∠BAD = 90°- 65° = 25°
But ∠BCD = ∠BAD (Angles in the same segment)

Question 2.
In the figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.

Solution:
In the figure, O is the centre of the circle ∠OAB = 30° and ∠OCB = 40°
Join OB

In △OAB, OA = OB (radii of the semicircle)
∴ ∠OBA = ∠OAB = 30°
Similarly in △OBC, OB = OC
∴ ∠OBC = ∠OCB = 40°
∴ ∠ABC = ∠OBA + ∠OBC = 30° + 40° = 70°
Now arc AB subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle
∴ ∠AOC = 2∠ABC = 2 × 70° = 140°
∴ ∠AOC = 140°

Question 3.
In Fig., O is the centre of the circle. Given ∠AOB = 80°.Calculate the value of ∠OAB and ∠OAC.

Solution:
In the figure, O is the centre and BOC is the diameter of the circle ∠AOB = 80°
AC and AB are joined

Now arc AB subtends
∠AOB at the centre and ∠ACB at the remaining part of the circle
∴ ∠AOB = ∠ACB
⇒ 80° = 2∠ACB ⇒ ∠ACB = \(\frac{80^{\circ}}{2}\) = 40°
But in △OAC, OA = OC (radii of the semicircle)
∴ ∠OAC = ∠OCA = ∠ACB = 40°
But ∠BAC = 90° (Angle in the semicircle)
⇒ ∠OAC + ∠OAB = 90°
⇒ 40° + ∠OAB = 90°
⇒ ∠OAB = 90°- 40° = 50°
Hence ∠OAB = 50° and ∠OAC = 40°

Question 4.
In figure, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°. Find
(i) ∠ACB
(ii) ∠OBC
(iii) ∠OAB
(iv) ∠CBA

Solution:
In the figure, O is the centre of the circle ∠AOB = 140°, ∠OAC = 50°
JoinAB

∴ Reflex ∠AOB = 360° – 140° = 220°
Now major arc AB subtends
Ref. ∠AOB at the centre and ∠ACB at the remaining part of the circle
(i) ∴Ref. ∠AOB = 2∠ACB
⇒ 220° = 2∠ACB ⇒ ∠ACB = \(\frac { 1 }{ 2 }\) × 220°
⇒ ∠ACB =110°

(ii) In quad. OACB,
∠BOA + ∠OAC + ∠ACB + ∠OBC = 360°
⇒ 140° + 50° + 110° + ∠OBC = 360°
⇒ 300° + ∠OBC = 360°
⇒ ∠OBC = 360° – 300° = 60°

(iii) ∠OAB + ∠OBA + ∠AOB = 180°
⇒ ∠OAB + ∠OBA + 140°= 180°
⇒ 2∠OAB = 180° – 140° = 40°
∴ ∠OAB = \(\frac{40^{\circ}}{2}\) = 20°

(iv) In △ABC,
∠BAC + ∠CBA + ∠ACB = 180°
⇒ 30° + ∠CBA + 110° = 180°
(∵ ∠BAC = ∠OAC – ∠OAB = 50° – 20° = 30°)
⇒ 140° +∠CBA = 180°
⇒ ∠CBA = 180° – 140° = 40°

Question 5.
In figure, if ∠ABC = 50° and ∠BDC = 40°, calculate
(i) ∠CDA
(ii) ∠BAC
(iii) ∠BCA

Solution:
(i) In the figure, ∠ABC = 50°, ∠BDC = 40°
AB is the diameter of the circle
∵ ∠BDA = 90° (Angle in a semicircle)
∴ ∠CDA = ∠BDA – ∠BDC = 90°- 40° = 50°

(ii) ∠BCA = 90° (Angle in a semicircle)
∴ ∠BAC + ∠ABC = 90°
⇒ ∠BAC + 50° = 90° ⇒ ∠BAC = 90° – 50°
⇒ ∠BAC = 40°
Hence (i) ∠CDA = 50°, (ii) BAC = 40° and (iii) ∠BCA = 90°

Question 6.
In figure, AC is a diameter of the given circle and ∠BCD = 75°. Calculate the size of : (i) ∠ABC, (ii) ∠EAF

Solution:
In the figure, AC is the diameter of the circle and ∠BCD = 75°
(i) ∠ABC = 90° (Angle in a semicircle)
(ii) ∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quad.)
⇒ 75° + ∠BAD = 180°
⇒ ∠BAD = 180° – 75° = 105°
But ∠EAF = ∠BAD = 105° (Vertically opposite angles)
∴ ∠EAF = 105°

Question 7.
In fig., ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E. If ∠ADE = 70° and ∠OBA = 45°, calculate ∠BAC and ∠OCA.

Solution:
A quadrilateral ABCD which is inscribed ina circle with centre O. CD is produced to E ∠ADE = 70° and ∠OBA = 45°
Join OA, OB, OC, AC
ABCD is a cyclic quadrilateral
∴ Ext. ∠ADE = Interior opposite ∠ABC
∴ ∠ABC = 70° (∠ADE = 70°)

Arc AC subtends ∠AOC at the centre and ∠ABC on the remaining part of the circle
∴ ∠AOC = 2∠ABC = 2 × 70°= 140°
Now in △OAC, OA = OC (radii of the same circle)
∴ ∠OCA = ∠OAC
But ∠OCA + ∠OAC + ∠AOC = 180° (sum of angles of a △)
⇒ ∠OCA + ∠OCA + 140° = 180°
⇒ 2∠OCA = 180° – 140° = 40°
∴ ∠OCA = \(\frac{40^{\circ}}{2}\) = 20°
and ∠OAC = 20°
Now ∠BAC = ∠BAO + ∠OAC
= 45° + 20° = 65° (∵ ∠BAO = ∠ABO = 45°)
Hence (i) ∠BAC = 65° and (ii) ∠OCA = 20°

Question 8.
In figure, O is the centre of the circle. Calculate the values of x and y.

Solution:
In the figure, ABCD is a cyclic quadrilateral in which O is the centre of the circle
∠AOC = 120°
But ∠AOC + ref. ∠AOC = 360°(Angles at a point)

Ref. ∠AOC = 360° – ∠AOC = 360° – 120° = 240°
Now arc ADC subtends ∠AOC at the centre and ∠ABC at the remaining part of a cyclic
∴ ∠AOC = 2∠ABC ⇒ 120° = 2x
⇒ x = \(\frac{120^{\circ}}{2}\) = 60°
But ∠B + ∠D = 180°
(Sum of opposite angles of a cyclic quadrilateral)
⇒ x + y = 180°
⇒ 60° + y = 180°
⇒ y = 180°- 60°
⇒ y = 120°
Hence x = 60° and y – 120°

Question 9.
In fig. AB || CD and O is the centre of the circle. If ∠ADC = 25°, find ∠AEB. Give reasons in support of your answer.

Solution:
In the figure, AB || CD O is the centre of the circle and ∠ADC = 25°
Join OA and OB

∵ AB || CD
∴ ∠BAD = ∠ADC = 25° (Alternate angles)
In △OAD, OA = OD (radii of the same circle)
∴ ∠OAD = ∠ADO or ∠ADC = 25°
∴ ∠OAB = ∠OAD + ∠BAD = 25° + 25° = 50°
But in △OAB,
OA = OB (radii of the same circle)
∴ ∠OAB = ∠OBA = 50°
But ∠OAB + ∠OBA + ∠AOB = 180° (Angles of a triangle)
⇒ 50° + 50° + ∠AOB = 180°
⇒ ∠AOB+ 100°= 180°
⇒ ∠AOB = 180° – 100° = 80°
Now arc AB, subtends ∠AOB at the centre and ∠AEB at the remaining part of the circle
∴ ∠AOB = 2∠AEB
⇒ ∠AEB = \(\frac { 1 }{ 2 }\) \(\left(\mathrm{AOB}=\frac{1}{2} \times 80^{\circ}=40^{\circ}\right)\)
Hence ∠AEB = 40°

Question 10.
In the fig, AB || CD. If ∠BCD = 100° and ∠BAC = 40°, calculate
(i) ∠CAD
(ii) ∠CBD
(ii) ∠BCA

Solution:
In the figure ABCD is a cyclic quadrilateral
in which AB || DC
∠BCD = 100° and ∠BAC = 40°

∴ AB || DC and AC is its transversal
∴ ∠BAC = ∠ACD = 40° (Alternate angles)
But ∠BCD = 100°
∴ ∠BCA = ∠BCD – ∠ACD =100° – 40° = 60°
∵ ABCD is a cyclic quadrilateral
∴ ∠BAD + ∠BCD = 180° (Sum of opp. angles)
⇒ ∠BAD + 100°= 180°
⇒ ∠BAD = 180° – 100° = 80°
⇒ ∠BAC + ∠CAD = 80°
⇒ 40° + ∠CAD = 80°
⇒ ∠CAD = 80° – 40° = 40°
But ∠CBD = ∠CAD (Angles in the same segment)
= 40°
Hence (i) ∠CAD = 40°, (ii) ∠CBD = 40°, (iii) ∠BCA = 60°

Question 11.
In figure, ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. Given that ∠BEF = 80°, find ∠ABC.

Solution:
ABCD is a parallelogram
and a circle pass through A and D, intersects
AB at E and DC at F
EF is joined.
∠BEF = 80°
∵ ADFE is a cyclic quadrilateral
∴ Ext. ∠BEF = Int. opp. ∠ADF
∴ ∠ADF = 80° or ∠ADC = 80°
But in||gm ABCD
∠ADC = ∠ABC (opp. angles of a ||gm)
∴ ∠ABC = 80°

Question 12.
In figure, if ∠BCD = 125° and AD is the diameter of the circle, calculate
(i) ∠DAB
(ii) ∠ADB

Solution:
In the figure, AD is the diameter of the circle and ABCD is a cyclic quadrilateral ∠BCD = 125°
Join BD

(i) ∵ ABCD is a cyclic quadrilateral
∴ ∠BAD + ∠BCD = 180° (Sum of opposite angles)
⇒ ∠BAD + 125° = 180°
⇒ ∠BAD = 180° – 125° = 55°

(ii) Now in △ABD,
∠ABD = 90° (Angle in a semicircle)
∴ ∠BAD + ∠ADB = 90°
⇒ 55° + ∠ADB = 90°
⇒ ∠ADB = 90°- 55° = 35°
∴ ∠ADB = 35°
Hence (i) ∠BAD or ∠DAB = 55° and
(ii) ∠ADB = 35°

Question 13.
In the figure, given a cyclic trapezium ABCD in which AD is parallel to BC and ∠B = 70°, find
(i) ∠BAD
(ii) ∠BCD

Soluiton:
In the figure,
ABCD is a cyclic trapezium in which AD || BC and ZB = 70°
(i) ∵ AD || BC
∴ ∠ABC + ∠BAD = 180° (Sum of co-interior angles)
70° + ∠BAD = 180° => ∠BAD =180°- 70°
∴ ∠BAD = 110°

(ii) In cyclic trapezium,
∠BAD + ∠BCD = 180° (Sum of opposite angles)
110° + ∠BCD = 180°
⇒ ∠BCD = 180° – 110° = 70°
Hence ∠BAD =110°, ∠BCD = 70°

Question 14.
In the figure AB is a diameter of the circle APBR, as shown in the figure. APQ and RBQ are straight lines; ∠A = 35°, ∠Q = 25°. Find :
(i) ∠PRB
(ii) ∠PBR
(iii) ∠BPR

Solution:
In the figure, AB is the diameter of the circle APQ and RBQ are straight lines ∠A = 35° and ∠Q = 25°

(i) ∠PRB and ∠PAB are in the same segment
∴ ∠PRB = ∠PAB = 35°

(ii) In △PBQ,
Ext. ∠PBR = ∠BPQ + ∠Q
= 90° + 25° = 115°

(iii) In △BRP,
∠BRP + ∠PBR + ∠BPR = 180° (Sum of angles of a triangle)
⇒ 35° +115° + ∠BPR = 180°
⇒ 150° + ∠BPR = 180°
⇒ ∠BPR = 180°- 150° = 30°

Question 15.
In the figure, it is given that O is the centre of the circle and ∠AOC = 130°. Find ∠ABC.

Solution:
In the figure, O is the centre of the circle and ∠AOC = 130°
∴ Reflex ∠AOC = 360° – 130° = 230°
Now major arc AC subtends reflex ∠AOC at the centre and ∠ABC at the remaining part of the circle
∴ Reflex ∠AOC = 2∠ABC
⇒ 230° = 2∠ABC ⇒ ∠ABC = \(\frac{230^{\circ}}{2}\) = 115°

Question 16.
In the figure, AB is a diameter of the circle. If ∠BCD = 140°, find ∠DBA.

Solution:
In the figure, AOB is the diameter of the circle
∠BCD = 140°
∵ ABCD is a cyclic quadrilateral
∴ ∠BCD + ∠BAD = 180° (Sum of opposite angles)
⇒ 140° + ∠BAD = 180°
⇒ ∠BAD = 180° – 140°
∠BAD = 40°
Now in △ADB,
∠ADB = 90° (Angles in a semicircle)
∴ ∠BAD + ∠DBA = 90°
⇒ 40° + ∠DBA = 90°
⇒ ∠DBA = 90°- 40° = 50°

Question 17.
In the figure, O is the centre of the circle, ABD is a st. line and ∠CBD = 65°. Find
(i) ∠AEC,
(ii) ∠AOC (marked x°)

Solution:
In the figure, ABCE is a cyclic quadrilateral in which AB is produced to D ∠CBD = 65°
∵ ABCE is a cyclic quadrilateral
∴ Ext. ∠CBD = Interior opposite ∠AEC
⇒ ∠AEC = ∠CBD = 65°
∵ Arc ABC subtends ∠AOC at the centre and ∠AEC at the remaining part of the circle
∴ ∠AOC = 2∠AEC = 2 × 65° = 130°
But ∠AOC + reflex ∠AOC = 360° (Angles at a point)
⇒ 130° + x° = 360° ⇒ x° = 360° – 130° = 230°

Question 18.
In figure, the diagonals of a cyclic quadrilateral ABCD intersect in P and the area of the triangle APB is 24 cm². If AB = 8 cm, and CD = 5 cm, calculate the area of the ACPD.

Solution:
In the figure,
Two diagonals of a cyclic quadrilateral ABCD intersect each other at P inside the circle. AB = 8 cm, CD = 5 cm and area (AAPB) = 24 cm²
∵ Diagonals AC and BD intersect each other at P
∴ AP.PC = BP.PD
⇒ \(\frac{\text { AP }}{\text { BP }}\) = \(\frac{\text { DP }}{\text { CP }}\) ⇒ \(\frac{\text { AP }}{\text { DP }}\) = \(\frac{\text { BP }}{\text { CP }}\) (By alternend0)
and ∠APB = ∠CPD (Vertically opposite angles)
∴△APB ~ △CPD (SAS axiom)

Question 19.
In figure, O is the centre of the circle. If ∠BAD = 30°, find the values of p, q and r.

Solution:
In the figure,
(i) O is the centre of the circle and ∠BAD = 30°
∵ ABCD is a cyclic quadrilateral
∴ ∠BAD + ∠BCD = 180° (Sum of opposite angles)
⇒ 30 °+ p = 180°
⇒ p = 180°- 30° = 150°
Arc BCD subtends ∠BOD at the centre and ∠BAD at the remaining part of the circle
∴ ∠BOD = 2∠BAD
q = 2 × 30° = 60°
∴ ∠BAD and ∠BED are in the same segment
∴ ∠BED = ∠BAD
⇒ r = 30°
Hence p = 150°, q = 60° and r = 30°

Question 20.
In figure, if ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°, calculate
(i) ∠BCD
(ii) ∠ADB and show that AC is a diameter.

Solution:
(i) In the figure, ABCD is a cyclic quadrilateral in which its diagonals AC and BD intersect each other at P
∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
∵ ABCD is a cyclic quadrilateral ∠BAD + ∠BCD= 180° (Sum of opposite angles)
⇒ 65° + ∠BCD = 180°
⇒ ∠BCD = 180° – 65° = 115°

(ii) In △ABD,
∠BAD + ∠ABD + ∠ADB = 180° (Sum of angles of a triangle)
⇒ 65° + 70° + ∠ADB = 180°
⇒ 135° + ∠ADB = 180°
⇒ ∠ADB = 180° – 135° = 45°
∵ ∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°
∴ ADC is a semicircle
Hence AC is the diameter of the circle

Question 21.
In figure, AB is a diameter of the circle whose centre is O. Given that ∠ECD = ∠EDC = 32°. Calculate :
(i) ∠CEF
(ii) ∠COF

Solution
In the figure, AOB is the diameter of the circle with centre O
∠ECD = ∠EDC = 32°

(i) In △CDE,
Ext. ∠CEF = ∠ECD + ∠EDC = 32° + 32° = 64°
(ii) Arc CF subtends ∠COF at the centre and ∠CDE or ∠CEF at the remaining part of the circle
∴ ∠COF = 2∠CDF = 2 × 32° = 64°

Question 22.
In figure, ABCD is a cyclic quadrilateral in which ∠DAC = 27°, ∠DBA = 50°, ∠ADB = 33°. Calculate :
(i) ∠DBC,
(ii) ∠DCB,
(iii) ∠CAB

Solution:
In the figure, ABCD is a cyclic quadrilateral AC and BD are joined
∠DAC = 27°, ∠DBA = 50°, ∠ADB = 33°
∠ADB = ∠ACB (Angles in the same segment)
∴ ∠ACB = 33°
Similarly ∠ABD = ∠ACD (Angles in the same segment)
∴ ∠ACB = 50°
∴ ∠DCB = ∠ACB + ∠ACD = 33° + 50°= 83°
In cyclic quad. ABCD
∠DCB + ∠BAD = 180°
⇒ 83° + ∠BAD = 180°
⇒ ∠BAD = 180°- 83° = 97°
⇒ ∠BAC + ∠DAC = 97°
⇒ ∠BAC + 27° = 97°
⇒ ∠BAC = 97° – 27° = 70°
or ∠CAB = 70°
∠DBC = ∠DAC (Angles in the same segment)
= 27°
Hence (i) ∠DBC = 27°, (ii) ∠DCB = 83°, ∠CAB = 70°

Question 23.
The diagram (see figure) shows a pentagon ABCDF inscribed in a circle, centre O. Given AB = BC = CD and ∠ABC = 132°. Calculate the value of
(i) ∠AEB
(ii) ∠AED
(ill) ∠COD
Solution:
In the figure, ABCD E is a pentagon inscribed
in the circle with centre O
AB = BC = CD and ∠ABC = 132°
Join BE and CE

(i) In cyclic quad. AECB,
∠AEC + ∠ABC = 180° (Sum of opposite angles)

⇒ ∠AEC + 132°= 180°
⇒ AEC = 180° – 132° = 48°
∵ AB = BC
∠AEB = ∠BEC
= \(\frac{48^{\circ}}{2}\) = 24°

(ii) ∵ AB = BC = CD
∴ ∠AEB = ∠BEC = ∠CED = 24°
∴ ∠AED = ∠AEB + ∠AEC + ∠CED
= 24° + 24° + 24° = 72°

(iii) Arc CD subtends ∠COD at the centre and ∠CED at the remaining part of the circle
∴ ∠COD = 2∠CED = 2 × 24° = 48°

Question 24.
In the given circle with diameter AB, find the value of x.

Solution:
In the given figure, AB is the diameter of the circle with centre O
∠ACD = 30° and ∠DAB = x
Join BC

∠ACB = 90° (Angle in a semicircle)
∴ ∠BCD = ∠ACB – ∠ACD = 90° – 30° = 60°
But ∠BAD = ∠BCD (Angles in the same segment)
∴ x = 60°

Question 25.
In figure, PQ is the diameter of the circle whose centre is O. Given ∠ROS = 42°, Calculate ∠RTS.

Solution:
In the figure, PQ is the diameter of the circle whose centre is O, ∠ROS = 42°
∠PRQ = 90° (Angle in a semicircle)
But ∠QRT + ∠PRQ = 180° (Linear pair)
⇒ ∠QRT + 90°= 180°
⇒ ∠QRT = 180°- 90° = 90°
Arc RS subtends ∠ROS at the centre and ∠RQT at the remaining part of the circle
∴ ∠ROS = 2∠RQT
⇒ ∠RQT = \(\frac { 1 }{ 2 }\)∠ROS = \(\frac { 1 }{ 2 }\) × 42° = 21°
Now in △RQT
∠QRT + ∠RQT +∠RTS = 180° (Sum of angle of a triangle)
⇒ 90°+ 21° + ∠RTS = 180°
⇒ 111° + ∠RTS = 180°
⇒ ∠RTS = 180°- 111° = 69°

Question 26.
In figure, PR is the diameter of the circle, PQ = 7 cm, QR = 6 cm and RS = 2 cm. Calculate the perimeter of the cyclic quadrilateral PQRS.

Solution:
In the figure, PR is diameter of the circle PQRS is the cyclic quad, in which PQ = 7 cm, QR = 6 cm, RS = 2 cm
In △PQR, ∠Q = 90° (Angle in a semicircle)
∴ PR² = PQ² + QR² (Pythagoras Theorem)
= (7)² + (6)² = 49 + 36 = 85
Similarly, in △PSR, ∠S = 90°
PR² = PS² + RS²
85 = PS² + (2)² ⇒ PS² = 85 – 4 = 81 = (9)²
∴ PS = 9 cm
Now perimeter of PQRS
= PQ + QR + RS + PS
= 7 + 6 + 2 + 9 = 24 cm

Question 27.
In the fig., AB = AC = CD, ∠ADC = 38°. Calculate : (i) ∠ABC, (ii) ∠BEC.

Solution:
In the given figure,
ABCE is a cyclic quadrilateral whose sides AE and BC are produced to meet at D and AB = AC = CD and ∠ADC = 38°
Join BE
In △ACD, AC = CD (given)
∴ ∠CAD = ∠ADC = 38°
In △ABC,
∵ AB = AC (given)
∴ ∠ABC = ∠ACB
But in △ACD,
Ext. ∠ACB = ∠CAD + ∠ADC = 38° + 38° = 76°
But ∠ABC = ∠ACB = 76°

(ii) In △ABC,
∠BAC =180°- (∠ABC + ∠ACB)
= 180° – (76° + 76°) = 180° – 152° = 28°
But ∠BAC = ∠BEC (Angles in the same segment)
∴ ∠BEC = 28°

Access to comprehensive OP Malhotra Class 10 ICSE Solutions Chapter 11 Coordinate Geometry Ex 11(b) encourages independent learning.

S Chand Class 10 ICSE Maths Solutions Chapter 11 Coordinate Geometry Ex 11(b)

Question 1.
Find the slope of a line whose inclination to the positive direction of x-axis in anticlockwise direction is given as :
(a) 30°
(b) 45°
(c) 60°
(d) 15°
(e) 75°
Solution:
If the inclination of a line to the positive direction is θ then slope of that line will be tan θ. Therefore
(a) tan 30° = \(\frac{1}{\sqrt{3}}\)
(b) tan 45° = 1
(c) tan 60° = \(\sqrt{3}\)
(d) tan 15° = 0.2679
(From the table of tan θ)
(e) tan 75° = 3.7231
(From the table of tan θ)

Question 2.
Find the slope and inclination of the line through each pair of the following points:
(a) (1, 2) and (5, 6)
(b) (0, 0) and (\(\sqrt{3}\), 3)
Solution:
(a) Points are (1, 2) and (5, 6)
∴ Slope (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{5-1}=\frac{4}{4}\) = 1
Angle of inclination = 45° (∵ tan 45° = 1)

(b) Points are (0, 0) and (\(\sqrt{3}\), 3)
∴ Slope (m) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{\sqrt{3}-0}=\frac{3}{\sqrt{3}}\)
= \(\frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3 \sqrt{3}}{3}=\sqrt{3} .\)
Angle of inclination = 60° (∵ tan 60° = \(\sqrt{3}\))

Question 3.
The side BC of an equilateral ∆ABC is parallel to the x-axis. What are the slopes of its sides?
Solution:
In an equilateral ∆ABC, each angle = 60°
and side BC || x-axis

Sides AB and AC are produced to meet the x-axis is at L and M
∴ ∠L = ∠B = 60°
(∵ Each angle of the triangle is 60°)
and ∠AMX = 120°
∴ Slope of AB = tan 60° = \(\sqrt{3}\)
and slope of AC = tan 120° = tan (180° – 60°) = tan 60° = – \(\sqrt{3}\)
and slope of BC = tan 0° = 0 (∵ BC || x-axis)

Question 4.
In a regular hexagon ABCDEF, AB || ED || the x-axis. What are the slopes of its sides?
Solution:
ABCDEF is a regular hexagon in which side AB and ED are parallel to x-axis.
Now slope of AB = tan 0° = 0

Slope of BC = tan 60° = \(\sqrt{3}\)
Slope of CD = tan 120° = – \(\sqrt{3}\)
Slope of DE = tan 0° = 0 (∵ ED || x-axis)
Slope of EF = tan 60° = \(\sqrt{3}\)
and slope of FA = tan 120° = – \(\sqrt{3}\)

Question 5.
Find y if the slope of the line joining (- 8, 11), (2, y) is \(\frac { -4 }{ 3 }\).
Solution:
We know that slope (m) = \(\frac{y_2-y_1}{x_2-x_1}\)
⇒ \(\frac{-4}{3}=\frac{y-11}{2-(-8)} \Rightarrow \frac{-4}{3}=\frac{y-11}{2+8}\)
⇒ \(\frac{-4}{3}=\frac{y-11}{10}\) ⇒ 3y – 33 = – 40
⇒ 3y = – 40 + 33 ⇒ 3y = – 7
∴ y = \(\frac { -7 }{ 3 }\)

Question 6.
Find the value of a, if the line passing through (-5, -8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a).
Solution:
Slope (m₁) of the line passing through the points (-5, -8) and (3, 0)
= \(\frac{y_2-y_1}{x_2-x_1}=\frac{0-(-8)}{3-(-5)}\)
= \(\frac{8}{3+5}=\frac{8}{8}\) = 1
and slope (m₂) of the line passing through the points (6, 3) and (4, a) = \(\frac{a-3}{4-6}=\frac{a-3}{-2}\)
∵ The two lines are parallel
∴ Their slopes are equal
⇒ m₁ = m₂
⇒ 1 = \(\frac { a – 3 }{ – 2 }\) ⇒ – 2 = a – 3
⇒ a = – 2 + 3 = 1
Hence a = 1

Question 7.
Find the slope of a line perpendicular to the line whose slope is :
(a) 3
(b) 1
(c) 5
(d) – 5
(e) 0
(f) Infinite
Solution:
We know that the slope of a line is m, then slope of the line perpendicular to the given line will be \(\frac { -1 }{ m }\)
(a) Slope of the line (m) = \(\frac { 1 }{ 3 }\)
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = \(\frac { -3 }{ 1 }\) = – 3

(b) Slope of the line (m) = \(\frac { -5 }{ 6 }\)
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = – (\(\frac { -6 }{ 5 }\)) = \(\frac { 6 }{ 5 }\)

(c) Slope of the line (m) = 5
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = \(\frac { -1 }{ 5 }\)

(d) Slope of the line (m) = -5\(\frac { 1 }{ 7 }\) = \(\frac { -36 }{ 7 }\)
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = – (\(\frac { -7 }{ 36 }\)) = \(\frac { 7 }{ 36 }\)

(c) Slope of the line (m) = 0
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = \(\frac { -1 }{ 0 }\) = infinite

(d) Slope of the line (m) = infinite
∴ Slope of its perpendicular = \(\frac { -1 }{ m }\) = \(\frac { -1 }{ infinite }\) = 0

Question 8.
Find the slope of a line perpendicular to the line which passes through the pair of the following points :
(a) (0, 8) and (-5, 2);
(b) (1, -11) and (5, 2)
(c) (-k, h) and (b, -f)
(d) (x₁, y₁) and (x₂, y₂)
Solution:
(a) Points are (0, 8) and (-5, 2)
∴ Slope of the line joining these points (m)
= \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{2-8}{-5-0}=\frac{-6}{-5}=\frac{6}{5}\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\) = \(\frac { -5 }{ 6 }\)

(b) Points are (1, -11) and (5, 2)
Slope of the line joining these points
= m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-11)}{5-1}=\frac{2+11}{5-1}=\frac{13}{4}\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\) = \(\frac { -4 }{ 13 }\)

(c) Points are (-k, h) and (b, -j)
Slope of the line joining these points
= m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-f-h}{b-(-k)}=\frac{-(f+h)}{b+k}\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\) = \(\frac{-(b+k)}{-(f+h)}=\frac{b+k}{f+h}\)

(d) Points are (x₁, y₁) and (x₂, y₂)
Slope of the line joining these points = m
= \(\frac{y_2-y_1}{x_2-x_1}\)
∴ Slope of the line perpendicular to it = \(\frac { -1 }{ m }\) = – \(\frac{x_2-x_1}{y_2-y_1}\)

Question 9.
In rectangle ABCD, the slope of AB = \(\frac { 5 }{ 6 }\). State the slope of
(a) BC
(b) CD
(c) DA
Solution:
In rectangle ABCD,
Slope of AB = \(\frac { 5 }{ 6 }\), Therefore

Question 10.
In parallelogram ABCD, slope of AB = – 2, slope of BC = \(\frac { 3 }{ 5 }\). State the slope of
(a) AD
(b) CD
(c) the altitude of AD
(d) the altitude of CD
Solution:
In parallelogram ABCD, AB || CD and DA || BC
Slope of AB = – 2 and slope of BC = \(\frac { 3 }{ 5 }\)
(a) ∵ DA || BC
∴ Slope of AD = Slope of BC = \(\frac { 3 }{ 5 }\)

(b) ∵ CD || AB
∴ Slope of CD = Slope of AB = – 2

(c) Slope of altitude of AD = – \(\left(\frac{1}{\text { Slope of } A D}\right)\)

(d) Slope of the altitude of CD = \(\frac{-1}{\text { Slope of } C D}\)
= – (\(\frac { 1 }{ -2 }\)) = \(\frac { 1 }{ 2 }\)

Question 11.
The vertices of a AABC are A (1, 1), B (7, 3) and C (3, 6). State the slope of the altitude to
(a) AB
(b) BC
(c) AC.
Solution:
Vertices of a ∆ABC are A (1, 1), B (7, 3) and C (3, 6)

(a) ∴ Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{3-1}{7-1}=\frac{2}{6}=\frac{1}{3}\)
and slope of its altitude = \(\frac{-1}{\text { Slope of } \mathrm{AB}}\)
= \(\frac { -3 }{ 1 }\) = – 3
Similarly,

(b) Slope of BC = \(\frac{6-3}{3-7}=\frac{3}{-4}\)
and slope of its altitude BC = \(\frac{-1}{\text { Slope of } \mathrm{BC}}\)
= – \(\left(\frac{-4}{3}\right)=\frac{4}{3}\)

(c) Slope of AC = \(\frac{6-1}{3-1}=\frac{5}{2}\)
∴ slope of its altitude AC = \(\frac{-1}{\text { Slope of } \mathrm{AC}}\)
= \(\frac { -2 }{ 5 }\)

Question 12.
The line joining (-5, 7) and (0, -2) is perpendicular to the line joining (1, -3) and (4, x). Find x.
Solution:
Slope of line joining the points (-5, 7) and (0, – 2) = m₁ = \(\frac{y_2-y_1}{x_2-x_1}=\frac{-2-7}{0-(-5)}=\frac{-9}{5}\)
and slope of the line joining the points
(1, – 3) and (4, x) = m₂ = \(\frac{x-(-3)}{4-1}=\frac{x+3}{3}\)
∵ These lines are perpendicular to each other
∴ m₁ x m₂ = – 1
⇒ \(\frac{-9}{5} \times \frac{x+3}{3}\) = 1
⇒ \(\frac{-3(x+3)}{5}=-1 \Rightarrow x+3=-1 \times \frac{5}{-3}=\frac{5}{3}\)
⇒ x = \(\frac{5}{3}-3=\frac{5-9}{3}=\frac{-4}{3}\)
Hence x = = \(\frac { -4 }{ 3 }\)

Question 13.
The vertices of a quad. PMQS are P (0, 0), M (3, 2), Q (7, 7) and S (4, 5). Show that PMQS is a parallelogram.
Solution:
Vertices of a quad PMQS are P(0,0), M (3, 2) , Q (7, 7) and S (4, 5)
Slope of PM (m₁) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-0}{3-0}=\frac{2}{3}\)
Similarly,
Slope of MQ = (m₂) = \(\frac{7-2}{7-3}=\frac{5}{4}\)
Slope of QS (m₃) = \(\frac{5-7}{4-7}=\frac{-2}{-3}=\frac{2}{3}\)
and slope of SP = \(\frac{5-0}{4-0}=\frac{5}{4}\)
∵ Slope of PM = Slope of QS = \(\frac { 2 }{ 3 }\)
∴ PM || QS … (i)
Similarly slope of MQ = slope of SP = \(\frac { 5 }{ 4 }\)
∴ MQ || SP … (ii)
∴ From (i) and (ii)
PMQS is a parallelogram

Question 14.
Show that P (a, b), Q (a + 3, b + 4),R (a – 1, b + 7), S (a – 4, b + 3) are the vertices of a square. What is the area of the square?
Solution:
Co-ordinates of P = (a, b), Q (a – 3, b + 4). R(a – 1, b + 1), S(a – 4, b + 3)

∵ All the sides PQ, QR, RS and SP are equal and both diagonals PR and QS are also equal
∴ PQRS is a square
Now area of square PQRS = (side)² = (PR)² = (5)² = 25 square units.

Question 15.
Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle.
Solution:
Vertices of a AABC are A (0, 4), B (1, 2) and C (3, 3)
Now slope of AB (m₁) = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{1-0}\)
= \(\frac { -2 }{ 1 }\) = – 2
Similarly,
Slope of BC (m₂) = \(\frac{3-2}{3-1}=\frac{1}{2}\)
and slope of CA (m₃) = \(\frac{4-3}{0-3}=\frac{1}{-3}\)
∵ m₁ x m₂ = – 2 x \(\frac { 1 }{ 2 }\) = – 1
∴ AB and BC are perpendicular to each other
∴ ∆ABC is a right angled triangle.

Question 16.
If the points (x, 1), (1, 2) and (0, y + 1) are collinear, show that \(\frac { 1 }{ x }\) + \(\frac { 1 }{ y }\) = 1.
Solution:
Let points are A (x, 1), B (1, 2) and C (0, y + 1)
∴ Slope of AB = \(\frac{y_2-y_1}{x_2-x_1}=\frac{2-1}{1-x}=\frac{1}{1-x}\)
∵ Similarly, slope of BC = \(\frac{y+1-2}{0-1}=\frac{y-1}{-1}\)
∴ A, B and C are collinear
∴ Slope of AB = slope of BC
⇒ \(\frac{1}{1-x}=\frac{y-1}{-1}\) ⇒ (1 – x) (y – 1) = – 1
⇒ y – 1 – xy + x = – 1
⇒ x + y = -1 + 1 + xy ⇒ y + x = xy
Dividing by xy
\(\frac{y}{x y}+\frac{x}{x y}=\frac{x y}{x y} \Rightarrow \frac{1}{x}+\frac{1}{y}\) = 1
Hence \(\frac { 1 }{ x }\) + \(\frac { 1 }{ y }\) = 1

The availability of Class 10 ICSE Maths Solutions S Chand Chapter 15 Three Dimensional Solids Ex 15(e) encourages students to tackle difficult exercises.

S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(e)

Question 1.
The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform circular cross section. If the length of the wire is 36 cm, find its radius.
Solution:
Diameter of metallic sphere = 6 cm
∴ Radius (R) = \(\frac { 6 }{ 2 }\) = 3 cm
and volume = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 3 × 3 × 3 cm³ = \(\frac { 792 }{ 7 }\) cm³
Length of wire (h) = 36 cm
Let r be the radius of circular wire, then πr²h = \(\frac { 792 }{ 7 }\)
⇒ \(\frac { 22 }{ 7 }\) r² × 36 = \(\frac { 792 }{ 7 }\) ⇒ r² = \(\frac{792 \times 7}{7 \times 22 \times 36}\)
⇒ r² = 1 = (1)² ⇒ r = 1 cm
∴ Radius = 1 cm

Question 2.
How many lead balls, each of radius 1 cm, can be made from a sphere whose radius is 8 cm ?
Solution:
Radius of sphere (R) = 8 cm
∴ Volume = \(\frac { 4 }{ 3 }\) πR³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 8 × 8 × 8 cm³
= \(\frac { 45056 }{ 21 }\) cm³
Radius of one small lead ball = 1 cm
∴ Volume of one lead ball = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 1 × 1 × 1 = \(\frac { 88 }{ 21 }\) cm³
∴ Number of ball = \(\frac { 45056 }{ 21 }\) ÷ \(\frac { 88 }{ 21 }\)
= \(\frac { 45056 }{ 21 }\) × \(\frac { 88 }{ 21 }\) = 512

Question 3.
A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 7 cm and height 3 cm. Find the number of cones so formed.
Solution:
Diameter of sphere = 21 cm
∴ Radius (R) = \(\frac { 21 }{ 2 }\) cm
and volume = \(\frac { 4 }{ 3 }\)πR³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 2 }\) × \(\frac { 21 }{ 2 }\) × \(\frac { 21 }{ 2 }\) cm ³ = 4851 cm³
Diameter of a smaller cone = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm
and height (h) = 3 cm
∴ Volume of one cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × 3 cm³ = \(\frac { 77 }{ 2 }\) cm³
∴ Number of cones formed = 4851 ÷ \(\frac { 77 }{ 2 }\)
= \(\frac{4851 \times 2}{77}\) = 63 × 2 = 126 cones

Question 4.
A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm, and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.
Solution:
Radius of disc (R) = 12 cm
and height (H) = 2.5 mm = 0.25 cm
∴ Volume = πr²h = \(\frac { 22 }{ 7 }\) × 12 × 12 × 0.25 cm³
= \(\frac { 22 }{ 7 }\) × 144 × \(\frac { 1 }{ 4 }\) = \(\frac { 792 }{ 7 }\) cm³
Let radius of sphere r then
\(\frac { 4 }{ 3 }\)πr³ = \(\frac { 792 }{ 7 }\) ⇒ \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\)r³ = \(\frac { 792 }{ 7 }\)
⇒ r³ = \(\frac { 792 }{ 7 }\) ⇒ \(\frac{3 \times 7}{4 \times 22}\) = 9 × 3 = 27 = (3)³
∴ r = 3
∴ Radius of the sphere = 3 cm

Question 5.
A hollow sphere of internal and external diameters 6 cm and 10 cm respectively is melted and recast into a cone of base diameter 14 cm. Find the height of the cone.
Solution:
External radius of hollow sphere (R)
= \(\frac { 10 }{ 2 }\) cm = 5 cm
and internal radius (r) = \(\frac { 6 }{ 2 }\) = 3 cm
∴ Volume of the metal used = \(\frac { 4 }{ 3 }\) (R³ – r³)
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) (5³ – 3³) cm³
= \(\frac { 88 }{ 21 }\)(125 – 27) = \(\frac { 88 }{ 21 }\) × 98 cm³
= \(\frac{88 \times 14}{3}\) = \(\frac{1232}{3}\) cm³
∴ Volume of cone = \(\frac{1232}{3}\) cm³
and diameter = 14 cm
∴ Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
Let h be the height of the cone
∴ \(\frac { 1 }{ 3 }\)πr₁²h = \(\frac { 1232 }{ 3 }\)
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × h = \(\frac { 1232 }{ 3 }\)
⇒ \(\frac { 154 }{ 3 }\) h = \(\frac { 1232 }{ 3 }\) ⇒ h = \(\frac{1232 \times 3}{3 \times 154}\)
∴ h = 8
∴ Height of the cone = 8 cm

Question 6.
A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal tube is \(\frac { 1 }{ 2 }\) cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of cone correct to 1 decimal place. \(\left(\text { Take } \pi=\frac{22}{7}\right)\)
Solution:
Internal radius of hollow metallic cylindrical tube (r) = 3 cm
Height (h) = 21 cm
Thickness of metal = \(\frac { 1 }{ 2 }\) cm
∴ Outer radius (R) = 3 + \(\frac { 1 }{ 2 }\) = \(\frac { 7 }{ 2 }\) cm
and outer height (H) = 21
∴ Volume = πR²H – πr²h
= π [R²H – r²h] = \(\frac { 22 }{ 7 }\)

Question 7.
Three solid glass balls of radius 1, 6 and 8 cm, respectively are melted into a solid sphere. Find its radius.
Solution:
Radius of first ball (r₁) = 1 cm
Radius of second ball (r₂) = 6 cm
and radius of third ball (r₃) = 8 cm
∴ Volume of three balls
= \(\frac { 4 }{ 3 }\)πr₁³ + \(\frac { 4 }{ 3 }\)πr₂³ + \(\frac { 4 }{ 3 }\)πr₃³ = \(\frac { 4 }{ 3 }\)π (r₁³ + r₂³ + ₃³)
= \(\frac { 4 }{ 3 }\)π [(1)³ + (6)³ + (8)³] cm³
= \(\frac { 4 }{ 3 }\)π [1 + 216 + 512] = \(\frac { 4 }{ 3 }\)π × 729 cm³
Let the radius of the sphere = R
∴ \(\frac { 4 }{ 3 }\)πR³ = \(\frac { 4 }{ 3 }\)π × 729 R³
= \(\frac { 4 }{ 3 }\)π × 729 × \(\frac{3}{4 \times \pi}\)
⇒ R³ = 729 = (9)³ ⇒ R = 9
∴ Radius of the sphere = 9 cm

Question 8.
A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are necessary to empty the bowl ?
Solution:
Internal radius of hemispherical bowl (R) = 9 cm
Volume of liquid filled in it = \(\frac { 2 }{ 3 }\)πR³
= \(\frac { 2 }{ 3 }\)π × (9)³ cm³ = \(\frac { 2 }{ 3 }\)π 729 = 2 × 243π cm³
= 486π cm³
Radius of cylindrical bottle (l) = \(\frac { 3 }{ 2 }\) cm
and height (h) = 4 cm
∴ Volume of one bottle = πr²h
= π × \(\frac { 3 }{ 2 }\) × \(\frac { 3 }{ 2 }\) × 4 = 9π cm³
∴ Number of bottles filled with liquid
= \(\frac{486 \pi}{9 \pi}\) = 54 bottles

Question 9.
How many spherical lead shots, each 4.2 cm in diameter, can be obtained from a rectangular solid of lead of dimensions 66 cm × 42 cm × 21 cm. (Use π = 22/7)
Solution:
Dimensions of a rectangular solid of lead = 66 cm × 42 cm × 21 cm
∴ Volume = 66 × 42 × 21 cm³
Diameter of one spherical lead shot (d) = 4.2 cm
∴ Radius (r) = \(\frac { 4.2 }{ 2 }\) =2.1 cm
∴ Volume of one shot = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (2.1)³ cm³
= \(\frac{4 \times 22}{21}\) × 2.1 × 2.1 × 2.1 cm³
= \(\frac { 88 }{ 21 }\) × \(\frac { 21 }{ 10 }\) × \(\frac { 21 }{ 10 }\) × \(\frac { 21 }{ 10 }\) cm³ = \(\frac { 38808 }{ 1000 }\) cm³
∴ Number of shots = \(\frac{\text { Volume of rect soild }}{\text { Volume of one shot }}\)
= \(\frac{66 \times 42 \times 21 \times 1000}{38808}\)
= \(\frac{3 \times 1000}{2}\) = 3 × 500 = 1500

Question 10.
The radius of a solid iron sphere is 8 cm. Eight rings of iron plate of external radius 6\(\frac { 2 }{ 3 }\) cm, thickness 3 cm are made by melting this sphere. Find the internal diameter of each ring.
Solution:

Radius of solid sphere (R) = 8 cm
∴ Volume = \(\frac { 4 }{ 3 }\)πR³ = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 8 × 8 × 8 cm³
= \(\frac { 45056 }{ 21 }\) cm³
outer radius of ring(r₁) = 6\(\frac { 2 }{ 3 }\) = \(\frac { 20 }{ 3 }\) cm

Let inner radius of the ring = r₂
Thickness of the ring (h) = 3 cm
∴ Volume of ring = π[R² – r²] × h

∴ Internal radius = 4 cm
and internal diameter = 2 × 4 = 8 cm

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
The figure alongside shows the cross-section of an ice-cream cone consisting of a cone surmounted by a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 10.5 cm. The outer shell ABCDEF is shaded and not filled with ice-cream AE = DC = 0.5 cm, AB || EF and BC || FD. Calculate:
(i) The volume of ice-cream in the cone (the unshaded portion including the hemisphere) in cm³.
(ii) The volume of the outer shell (the shaded portion ) in cm³. Give your answer correct to the nearest cm³.

Solution:
Diameter of hemisphere = 7 cm
Radius of hemisphere AC (R) = \(\frac { 7 }{ 2 }\) = 3.5 cm
Height of conical part NB = 10.5 cm
AE = DC = 0.5 cm, AB || EF and CB || DF
Let height NF = h
Shaded portion is not filled with ice-cream Radius of inner cone = 3.5 – 0.5 = 3.0 cm
In right △ADF and △ACB
∠A = ∠A (each 90°)
∠ADF = ∠ACB (corresponding angles)
∴ △ADF ~ △ACB (AA axiom)
∴ \(\frac { AD }{ AC }\) = \(\frac { EF }{ CB }\) ⇒ \(\frac { 3 }{ 3.5 }\) = \(\frac { h }{ 10.5 }\)
⇒ h = \(\frac{3 \times 10.5}{3.5}\) = 9 cm

(i) Volume of ice-cream = volume of inner cone + volume of hemisphere
= \(\frac { 1 }{ 3 }\)πr²h + \(\frac { 2 }{ 3 }\)πR³
=\(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 3 × 3 × 9 + \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) × \(\frac { 7 }{ 2 }\) cm³ (nearest to cm³)

(ii) volume of outer shell (shaded portion) = volume of outer cone – volume of inner cone
= \(\frac { 1 }{ 3 }\)πR²H – \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5) (3.5) (3.5) – \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 3 × 3 × 9 = 134.75 – \(\frac { 594 }{ 7 }\) = 134.75 – 84.857
= 49.893 cm³ = 50 cm³ (nearest to cm³)

Question 2.
The surface area of a solid metallic sphere is 1256 cm³. It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate :
(i) the radius of the solid sphere,
(ii) the number of cones recast. (π = 3.14)
Solution:
Surface area of a metallic sphere = 1256 cm²
(i) Let R be the radius
∴ 4πR² = 1256 ⇒ 4 × 3.14 R² = 1256
⇒ R² = \(\frac{1256}{4 \times 3.14}\) = \(\frac{1256 \times 100}{4 \times 314}\) = 100 = (10)²
∴ R = 10 cm

(ii) Volume of the sphere = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) (3.14) × (10)³ cm³
= \(\frac{4 \times 314 \times 1000}{3 \times 100}\) cm³ = \(\frac { 12560 }{ 3 }\) cm³
Radius of cone (r) = 2.5 cm
and height (h) = 8 cm
∴ volume of cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) (3.14) × (2.5)² × 8 cm³
= \(\frac{3.14 \times 6.25 \times 8}{3}\) = \(\frac{157}{3}\) cm³
Number of cones formed = \(\frac { 12560 }{ 3 }\) ÷ \(\frac { 157 }{ 3 }\)
= \(\frac { 12560 }{ 3 }\) × \(\frac { 3 }{ 157 }\) = 80

Question 3.
An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².
Solution:
Diameter of the base = 168 m
∴ Radius (r) = = 84 m
Total height of the tent = 85 m
Height of cylinderical portion (h₁) = 50 m

∴ Height of conical portion = 85 – 50 = 35 m
Total curved surface = curved surface area of cylinder + curved surface area of cone = 2πrh₁ + πrl
= 2πrh₁ + πr\(\left(\sqrt{r^2+h^2}\right)\) \((∴l \sqrt{r^2+h^2})\)
= 2 × \(\frac { 22 }{ 7 }\) × 84 × 50 + \(\frac { 22 }{ 7 }\) × 84 × \(\sqrt{\left(84^2+35^2\right)}\)
= 26400 + 264 × \((\sqrt{7056+1225})\)
= 26400 + 264 × \(\sqrt{8281}\)
= 26400 + 264 × 91 = 26400 + 24024 = 50424 m²
Extra area of 20% for stitching etc.
= 50424 m²
Extra area of 20% for stitching etc.
= 50424 + 10084.8 m² = 60508.8 m²
= 60509 m² (nearest to m²)

Question 4.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Solution:
Internal diameter of hollow sphere = 4 cm
and external diameter = 8 cm

∴ Internal radius (r) = \(\frac { 4 }{ 2 }\) = 2 cm
and external radius (R) = \(\frac { 8 }{ 2 }\) = 4 cm
∴ Volume of the hollow sphere
= \(\frac { 4 }{ 3 }\) π (R³ – r³) = \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) [4³ – 2³]
= \(\frac { 88 }{ 21 }\) × (64 – 8) = \(\frac { 88 }{ 21 }\) × 56 cm³ = \(\frac { 704 }{ 3 }\) cm³
Base diameter of cone = 8 cm
∴ Radius (r₁) = \(\frac { 8 }{ 2 }\) = 4 cm
Let h be the height of the cone
Then volume of cone = \(\frac { 1 }{ 3 }\) πr²h

∴ \(\frac { 1 }{ 3 }\) πr²h = \(\frac { 704 }{ 3 }\)
\(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 4 × 4 × h = \(\frac { 704 }{ 3 }\)
\(\frac { 352 }{ 21 }\) h = \(\frac { 704 }{ 3 }\) ⇒ h = \(\frac{704 \times 21}{3 \times 352}\) = 14
∴ Height of the cone = 14 cm

Question 5.
A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top which is open is 2.5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, \(\frac { 2 }{ 5 }\) of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Radius of the cone = 2.5 cm
Height (h) = 11 cm

∴ Volume of the cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 2.5 × 2.5 × 11 cm³
= \(\frac { 1512.5 }{ 21 }\) cm³
Radius of spherical lead = 0.25 cm
∴ Volume of one lead = \(\frac { 4 }{ 3 }\) πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (0.25)³ cm³
= \(\frac { 88 }{ 21 }\) × 0.015625 cm³ = \(\frac { 1.375 }{ 21 }\) cm³
volume of water flows out of the vessel after
putting the shots = \(\frac { 2 }{ 5 }\) 0f \(\frac { 1512.5 }{ 21 }\) = \(\frac { 605.0 }{ 21 }\) cm³
∴ Number of shots = \(\frac { 605.0 }{ 21 }\) ÷ \(\frac { 1.375 }{ 21 }\)
= \(\frac { 605.0 }{ 21 }\) × \(\frac { 1.375 }{ 21 }\) = 440

Question 6.
A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of sand. If the height of the conical heap is 24 cm, find:
(i) its radius and
(ii) its slant height (leave your answer in square root form)
Solution:
Radius of the cylindrical bucket (r) = 18 cm
Height (h) = 32 cm
∴ Volume of sand filled in it = πr²h
= π (18)² × 32 = π × 18 × 18 × 32 cm³
= 10368 π cm³

(i) Volume of sand in a conical shape heap = 10368π cm³
Height (h₁)=24 cm
∴ Let r₁, be the radius, then
\(\frac { 1 }{ 3 }\) πr₁²h = 10368 π
⇒ \(\frac { 1 }{ 3 }\) πr₁² × 24 = 10368π ⇒ 8πr₁² = 10638π
⇒ r₁² = \(\frac { 10368π }{ 8π }\) =1296 = (36)²
∴ r₁ = 36 cm

(ii) and slant height = \(\sqrt{r_1^2+h_1^2}\)
= \(\sqrt{(36)^2+(24)^2}\) cm
= \(\sqrt{1296+576}\) cm = \(\sqrt{1872}\) cm

Question 7.
A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained.
Solution:
Radius of a sphere (R) = 10.5 cm
∴ Volume = \(\frac { 4 }{ 3 }\)πR³ = \(\frac { 4 }{ 3 }\)π (10.5)³ cm³
= \(\frac { 4 }{ 3 }\)π × 1157.625 cm³
Radius of each cone (r) = 3.5 cm
and height (h) = 3 cm
∴ Volume of one cone = \(\frac { 1 }{ 3 }\)πr³h
= \(\frac { 1 }{ 3 }\)π (3.5)² × 3 cm³
= π × 12.25 = 12.25 = 12.25π cm³
∴ Number of cone will be \(\frac{4 \pi \times 1157.625}{3 \times 12.25 \pi}\)
= 126 cones

Question 8.
A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid copes submerged ?
Solution:
Diameter of a conical vessel = 16.8 cm
∴ Radius (R) = \(\frac { 16.8 }{ 2 }\) = 8.4 cm
and height (H) = 20 cm
∴ volume of water filled

in it = \(\frac { 1 }{ 3 }\)πR²H
= \(\frac { 1 }{ 3 }\)π × (8.4)² × 20 cm³
= \(\frac { 1 }{ 3 }\)π × 8.4 × 8.4 × 20 cm³ = 470.4π cm³
By dropping two soild cones in it, volume of water overflows = \(\frac { 1 }{ 3 }\) of 470.4π = 156.8π cm³
∴ volume of two small solid cones = 156.8π cm³
and volume of one soild cone = \(\frac{156.8 \pi}{2}\) cm³
= 78.4π cm³ = 78.4 × \(\frac { 22 }{ 7 }\) = 246.4 cm³

Question 9.
The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained ?
Solution:
Surface area of a metallic solid sphere = 616 cm²
Let radius of the sphere = R
Then 4πR² = 616
⇒ 4 × \(\frac { 22 }{ 7 }\)R² = 616
⇒ R² = \(\frac{616 \times 7}{4 \times 22}\) = 7 × 7 = 49 cm
∴R² = 49 = (7)²
⇒ R = 7 cm
∴ volume = \(\frac { 4 }{ 3 }\)πR³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 7 cm³ = \(\frac { 4312 }{ 3 }\)cm³
Diameter of each smaller spheres = 3.5 cm
Radius (r) = \(\frac { 3.5 }{ 2 }\) = 1.75 cm
volume of one smaller sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (1.75)³ cm³
= \(\frac { 88 }{ 21 }\) × 5.359375 cm³
∴ Number of small spheres will be
= \(\frac { 4312 }{ 3 }\) ÷ \(\frac{88 \times 5.359375}{21}\)
= \(\frac { 4312 }{ 3 }\) × \(\frac{21}{88 \times 5.359375}\)
= \(\frac{49 \times 7 \times 1000000}{5359375}\) = \(\frac{1000000}{15625}\) = 64

Question 10.
The volume of a conical tent is 1232 m³ and the area of the base floor is 154 m². Calculate the :
(i) radius of the floor;
(ii) height of the tent;
(iii) length of canvas required to cover this conical tent, if its width is 2 m.
Solution:
Volume of conical tent = 1232 m³
and area of its floor = 154 m²
Let r be the radius and h be the height of the tent
(i) Then πr² = 154 ⇒ \(\frac { 22 }{ 7 }\)r² = 154
⇒ r² = \(\frac{154 \times 7}{22}\) = 7 × 7 = (7)²
∴ r = 7 m

(ii) Volume = \(\frac { 1 }{ 3 }\)πr²h
⇒ \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × h = 1232
⇒ \(\frac { 154 }{ 3 }\)h = 1232 ⇒ h = \(\frac{1232 \times 3}{154}\) = 24
(i) ∴ Radius = 7 m and (ii) height = 24 cm

(iii) Slant height (l) = \(\sqrt{r^2+h^2}\) = \(\sqrt{(7)^2+(24)^2}\) = \(\sqrt{49+576}\) = \(\sqrt{625}\) = 25
∴ l = 25 m
∴ Curved surface area = πrl
= \(\frac { 22 }{ 7 }\) × 7 × 25 = 550 m²
width of canvas = 2 m
∴ Length of canvas = \(\frac { Area }{ Width }\) = \(\frac { 550 }{ 2 }\) = 275 m

Question 11.
The given figure represents a hemisphere surmounted by a conical block of wood. The diameter of their bases is 6 cm each and the slant height of the cone is 5 cm. Calculate :
(i) the height of the cone ;
(ii) the volume of the solid.
Solution:
(i) Diameter of cone = 6 cm

Radius (r) = \(\frac { 6 }{ 2 }\) = 3 cm
and slant height (l) = 5 cm
Let h be the height of the cone height = \(\sqrt{l^2-r^2}\) = \(\sqrt{(5)^2-(3)^3}\)
= \(\sqrt{25-9}\) = \(\sqrt{16}\) = 4 cm

(ii) Volume = \(\frac { 1 }{ 3 }\)πr²h + \(\frac { 2 }{ 3 }\)πr³ = \(\frac { 1 }{ 3 }\)πr²(h + 2r)
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3)² (4 + 6)
=\(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 9 × 10
= \(\frac { 660 }{ 7 }\) = 94.29 cm³

Question 12.
A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone.
Solution:
Radius of hemispherical bowl = \(\frac { 7.2 }{ 2 }\) = 3.6 cm
volume of hemispherical bowl = \(\frac { 2 }{ 3 }\)πr³
= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 3.6 × 3.6 × 3.6 = 97.755 cm³
∴ Volume of chocolate sauce = 97.755 cm³
It is to poured in cone of radius = 4.8 cm
Volume of inverted cone = \(\frac { 1 }{ 3 }\)πr²h
97.755 cm³ = \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 4.8 × 4.8 × h
⇒ h = \(\frac{97.755 \times 7 \times 3}{4.8 \times 4.8 \times 22}\) = \(\frac{2052.855}{506.88}\)
⇒ h = 4.05 cm

Question 13.
A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed.
Solution:
Volume of a cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\)π × (5)² × 8
= \(\frac { 1 }{ 3 }\)π × 25 × 8 = \(\frac{200 \pi}{3}\) cm³
Volume of a sphere = \(\frac { 1 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\) × π × (0.5)³
= \(\frac { 4 }{ 3 }\) × π × 0.125 = \(\frac { 0.5 }{ 3 }\) π = \(\frac { π }{ 6 }\) cm³
Number of spheres formed = \(\frac{200 \pi}{3}\) × \(\frac{6}{\pi}\) = 400

Question 14.
A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones.
Solution:
Volume of metal in the sphere

Question 15.
A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Solution:
volume of sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4 }{ 3 }\)π × (15)³
volume of cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\)π × (2.5)² × 8
∴ No. of cones = \(\frac{\text { Vol. of spheres }}{\text { Vol.of cone }}\)

Question 16.
The surface area of a solid metallic sphere is 2464 cm² . It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate:
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = 22/7)
Solution:
(i) Surface area = 47πr²= 2464 cm² (given)
r² = \(\frac{2464}{4 \pi}\)
r² = \(\frac{2464 \times 7}{4 \times 22}\) = 196 ⇒ r = \(\sqrt{196}\)
⇒ r = \(\sqrt{14 \times 14}\)
r = 14 cm

(ii) Volume of sphere = \(\frac { 4 }{ 3 }\)πr³ = \(\frac { 4 }{ 3 }\)π (14)³
Volume of cone = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\)π (3.5)² × 7

Question 17.
Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Solution:
Radius of small sphere r = 2 cm
Radius of big sphere R = 4 cm
Volume of small sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac { 4π }{ 3 }\) × (2)³ = \(\frac { 32π }{ 3 }\) cm³
Volume of big sphere = \(\frac { 4 }{ 3 }\)πr³
= \(\frac{256 \pi}{3}\) cm³
Volume of both sphere
= \(\frac{32 \pi}{3}\) + \(\frac{256 \pi}{3}\) = \(\frac{288 \pi}{3}\)cm³
Volume of the cone = \(\frac { 1 }{ 3 }\)πR₁²h
We need to find R₁
h = 8 cm (Given)
Volume of the cone = \(\frac { 1 }{ 3 }\)πR₁² × (8)
Volume of the cone = Volume of both the sphere
\(\frac { 1 }{ 3 }\)πR₁² × (8) = \(\frac{288 \pi}{3}\)
⇒ R₁² × (8) = 288
⇒ R₁² = \(\frac { 288 }{ 8 }\)
⇒ R₁² = 36 ⇒ R₁² = (6)²
⇒ R₁ = 6 cm

Question 18.
A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones.
Solution:
Let the number of cones be n,
Let radius of the sphere be r_s, radius of a cone be r_c and h be the height of the cone. Volume of sphere = n(Volume of a metallic cone)

Hence, the number of cones is 72.

Students appreciate clear and concise OP Malhotra Class 10 ICSE Solutions Chapter 11 Coordinate Geometry Ex 11(a) that guide them through exercises.

S Chand Class 10 ICSE Maths Solutions Chapter 11 Coordinate Geometry Ex 11(a)

Question 1.
Find the mid-points of lines joining
(a) (5, 8), (9, 11)
(b) (0, 0), (8, – 5)
(c) (-7, 0), (0, 10)
(d) (-4, 3), (6, -7)
Solution:
We know that the co-ordinates of a mid-point of a line whose vertices are \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\). Therefore
(a) Mid point of the line joining (5, 8) and (9, 11) will be = \(\left(\frac{5+9}{2}, \frac{8+11}{2}\right)\)

(b) will be = Mid point of the line joining (0, 0) and (8, 5) will be = \(\left(\frac{0+8}{2}, \frac{0-5}{2}\right) \text { or }\left(4, \frac{-5}{2}\right)\)

(c) Mid point of the line joining (-7, 0) and (0, 10) will be = \(\left(\frac{-7+0}{2}, \frac{0+10}{2}\right) \text { or }\left(\frac{-7}{2}, \frac{10}{2}\right)\) or \(\left(\frac{-7}{2}, \frac{10}{2}\right)\) or \(\left(\frac{-7}{2}, 5\right)\)

(d) Mid point of the line joining (-4, 3) and (6,-7) will be = \(\left(\frac{-4+6}{2}, \frac{3-7}{2}\right) \text { or }\left(\frac{2}{2}, \frac{-4}{2}\right)\) or (1,-2)

Question 2.
Find the mid-points of the sides of a triangle whose vertices are A (1, -1), B (4, -1), C (4, 3).
Solution:
Let D, E and F be the mid-points of the sides
BC, CA and AB of the triangle ABC
∴ Co-ordinates of D, the mid-point of BC will
be \(\left(\frac{4+4}{2}, \frac{-1+3}{2}\right) \text { or }\left(\frac{8}{2}, \frac{2}{2}\right)\) or (4, 1)

Co-ordinates of E, the mid-point of CA will
be \(\left(\frac{4+1}{2}, \frac{3-1}{2}\right) \text { or }\left(\frac{5}{2}, \frac{2}{2}\right) \text { or }\left(\frac{5}{2}, 1\right)\)
and co-ordinates of F, the mid-point of AB
will be = \(\left(\frac{1+4}{2}, \frac{-1-1}{2}\right) \text { or }\left(\frac{5}{2}, \frac{-2}{2}\right)\) or \(\left(\frac{5}{2},-1\right)\)
Co-ordinates of mid-point of AB, BC and CA
are (\(\frac { 5 }{ 2 }\), – 1), (4, 1), (\(\frac { 5 }{ 2 }\), 1)

Question 3.
Find the centre of a circle if the end points of a diameter are A (-5, 7) and C (3, -11).
Solution:
Let O be the centre of the circle

∴ O will be the mid point of the diameter AC,
let co-ordinates of O be (x, y), then
x = \(\frac{x_1+x_2}{2}, y=\frac{y_1+y_2}{2}\)
∴ x = \(\frac{-5+3}{2}, y=\frac{7-11}{2} \Rightarrow x=\frac{-2}{2}, y=\frac{-4}{2}\)
⇒ x = – 1, y = – 2
∴ Co-ordinates of O are (- 1, – 2)

Question 4.
If M is the mid-point of AB, find the co-ordinates of :
(a) A if the coordinates of M and B are M (2, 8) and B (-4, 19) and
(b) B if the co-ordinates of A and M are A (-1, 2), M (-2, 4).
Solution:
M is mid-point of AB
(a) Co-ordinates of M are (2, 8) and of B are (-4, 19)
Let co-ordinates of A be (x₁, y₁)
∴ Co-ordinates of M will be
= \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
(2, 8) = \(\left(\frac{x_1-4}{2}, \frac{y_1+19}{2}\right)\)
Comparing, we get
\(\frac{x_1-4}{2}\) = 2 ⇒ x₁ – 4 = 4 ⇒ x₁ = 4 + 4 ⇒ x₁ = 8
and \(\frac{y_1+19}{2}\) = 8 ⇒ y₁ + 19 = 16 ⇒ y₁ = 16 – 19 = – 3
∴ Co-ordinates of A are (8, -3).

(b) Let co-ordinates of B be (x₂, y₂)
But co-ordinates of M are (-2, 4) and of A are (-1,2)
∵ M is the mid point of AB and let co-ordinates of
M be \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
∴ (-2, 4) = \(\left(\frac{-1+x_2}{2}, \frac{2+y_2}{2}\right)\)
Comparing, we get
\(\frac{-1+x_2}{2}\) – 2 ⇒ – 1 + x₂ = – 4
⇒ x = – 4 + 1 = – 3
Ax = – 3 and \(\frac{2+y_2}{2}\) = 4 ⇒ 2 + y₂ = 8
⇒ y₂ = 8 – 2 = 6
∴ Co-ordinates of B are, (- 3, 6)

Question 5.
Find the coordinates of the point which divides internally the join of the points
(a) (8, 9) and (-7, 4) in the ratio 2 : 3
(b) (1, -2) and (4, 7) in the ratio 1 : 2.
Solution:
If a point (x, y) divides a line segment having
its end points (x₁, y₁) and (x₂, y₂) in the ratio
of m₁ : m₂, then
x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}\) and y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}\)

(a) Let co-ordinates of the point P be (x, y) which
divides the join of points A (8, 9) and B (- 7, 4) in the ratio 2 : 3, then
m₁ = 2 and m₂ = 3
∴ x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{2 \times(-7)+3 \times 8}{2+3}\)
= \(\frac{-14+24}{5}=\frac{10}{5}=2\)
y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{2 \times 4+3 \times 9}{2+3}=\frac{8+27}{5}\)
= \(\frac { 35 }{ 5 }\) = 7
∴ Co-ordinate of the required point P will be (2, 7)

(b) Let co-ordinates of the point P be (x, y) which divides the join of points A (1, -2) and B (4, 7) in the ratio 1 : 2, then
m₁ = 1, m₂ = 2

∴ Co-ordinates of the required point will be (2, 1)

Question 6.
Find the co-ordinates of the points of trisection of the line joining the points (2, 3) and (6, 5).
Solution:
Let P and Q are the points which trisect the line segment joining the points A (2, 3) and B (6, 5)

Let co-ordinates of P be (x’, y’) and of Q be be (x”, y”), then
In case of P, m₁ = 1, m₂ = 2

∴ Co-ordinates of P will be (\(\frac { 10 }{ 3 }\), \(\frac { 11 }{ 3 }\))
In case of Q, m₁ = 2, m₂ = 1

∴ Co-ordinates of Q will be (\(\frac { 14 }{ 3 }\), \(\frac { 13 }{ 3 }\))

Question 7.
In what ratio is the line joining the points
(a) (2, -3) and (5, 6) divided by the x-axis;
(b) (3, -6) and (-6, 8) divided by they-axis?
Solution:
(a) Let the ratio in which x-axis divides the join of points (2, -3) and (5, 6), be m₁ : m₂
Let the co-ordinates of point at x-axis be (x, 0)

(b) Let the ratio in which y-axis divides the join of points (3, -6) and (-6, 8) be m₁ : m₂ and let the co-ordinates of point at y-axis be (0, y)

Question 8.
Find the centroid of the triangle whose angular points are (-4, 6), (2, -2) and (2, 5) respectively.
Solution:
Let G be the centroid of the triangle whose vertices are A (-4, 6), B (2, -2) and C (2, 5)
∴ Co-ordinates of centroid will be
\(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
or \(\left(\frac{-4+2+2}{3}, \frac{6-2+5}{3}\right)\)
or \(\left(\frac{0}{3}, \frac{9}{3}\right)\) or (0, 3)

Question 9.
If (x₁, y₁) = (2, 3); x₂ = 3 and y₃ = – 2 and G is (0, 0), find y₂ and x₃.
Solution:
∵ G is the centroid of the triangle
∴ Co-ordinates of G will be
\(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
But co-ordinates of centroid are (0, 0)

Question 10.
The mid-points of the sides of a triangle are at (1, 4), (4, 8) and (5, 6). Find the coordinates of the vertices of the triangle.
Solution:
Let D, E and F are the mid-points of a ∆ABC
Let vertices of A, B and C be (x₁, y₁), (x₂, y₂)
and (x₃, y₃) and co-ordinates of D, E and F
are (1, 4), (4, 8) and (5, 6) respectively..
∴ 1 = \(\frac{x_1+x_2}{2}\) ⇒ x₁ + x₂ = 2 … (i)
4 = \(\frac{x_2+x_3}{2}\) ⇒ x₂ + x₃ = 8 … (ii)
5 = \(\frac{x_3+x_1}{2}\) ⇒ x₃ + x₁ = 10 … (iii)
Adding we get, 2 (x₁ + x₂ + x₂) = 20
⇒ x₁ + x₂ + x₃ = \(\frac { 20 }{ 2 }\) = 10 … (iv)
Subtracting (ii), (iii) and (i) from (iv) term by term, we get
x₁ = 10 – 8 = 2
x₂ = 10 – 10 = 0
x₃ = 10 – 2 = 8
Similarly
4 = \(\frac{y_1+y_2}{2}\) ⇒ y₁ + y₂ = 8 … (v)
8 = \(\frac{y_2+y_3}{2}\) ⇒ y₂ + y₃ = 16 … (vi)
6 = \(\frac{y_3+y_1}{2}\) ⇒ y₃ + y₁ = 12 … (vii)
Adding we get,
2 (y₁ + y₂ + y₃) = 8 + 16 + 12 = 36
⇒ y₁ + y₂ + y₃ = \(\frac { 36 }{ 2 }\) = 18 (viii)
Subtracting (vi), (vii) and (v) from (viii) term by term, we get
y₁ = 18 – 16 = 2
y₂ = 18 – 12 = 6
y₃ = 18 – 8 = 10
∴ Co-ordinates of A, B and C are (2, 2), (0, 6), (8, 10)

Question 11.
The vertices of a triangle are A (-2, 2), B (4, 4) and C (8, 2). Find the length of the medians to the sides (i) AB, (ii) AC and (iii) BC.
Solution:
Let D, E and F be the mid-points of the sides
AB, BC and CA of the ∆ABC. AE, BF and CD are joined.
The vertices of ∆ABC are A (- 2, 2), B (4, 4) and C (8, 2)

∵ D is mid-point of AB
∴ Co-ordinates of D will be
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) or \(\left(\frac{-2+4}{2}, \frac{2+4}{2}\right)\) or \(\left(\frac{2}{2}, \frac{6}{2}\right)\) or (1, 3)
Similarly co-ordinates of E will be \(\left(\frac{4+8}{2}, \frac{4+2}{2}\right)\) or (6, 3) and of F will be
\(\left(\frac{8-2}{2}, \frac{2+2}{2}\right)\) or (3, 2)
Now length of AE = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(-2-6)^2+(2-3)^2}=\sqrt{(-8)^2+(-1)^2}\)
= \(\sqrt{64+1}=\sqrt{65}\)
Similarly,
Length of BF = \(\sqrt{(4-3)^2+(4-2)^2}\)
= \(\sqrt{(1)^2+(2)^2}=\sqrt{1+4}=\sqrt{5}\)
and length of CD = \(\sqrt{(8-1)^2+(2-3)^2}\)
= \(\sqrt{(7)^2+(-1)^2}=\sqrt{49+1}=\sqrt{50}\)
= \(\sqrt{25 \times 2}=5 \sqrt{2}\)
∴ Length of median to AB, BC and CA are 5\(\sqrt{2}\), \(\sqrt{5}\) and \(\sqrt{65}\).

Question 12.
Calculate the coordinates of the point P which divides the line joining A (- 1,3), B (5, 9) in the ratio 1 : 2.
Solution:
Let the co-ordinates of the point P be (x, y) and m₁ : m₂ = 1 : 2

Question 13
The mid-point of the line joining A (2,p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Solution:
Let P (3, 5) be the mid-point of line joining the points A (2, p) and B (q, 4), then
∵ x = \(\frac{x_1+x_2}{2}\) and y = \(\frac{y_1+y_2}{2}\)
∴ 3 = \(\frac{2+q}{2}\) and 5 = \(\frac{p+4}{2}\)
⇒ 2 + q = 6 and p + 4 = 10
⇒ q = 6 – 2 = 4 and p = 10 – 4 = 6
∴ p = 6, 9 = 4

Question 14.
(a) A is the point on the y-axis whose ordinate is 5 and B is the point (-3, 1). Calculate the length of AB.
(b) The mid-point of the line joining (a, 2) and (3, 6) is (2, b). Find the numerical values of a and b.
Solution:
(a) ∵ Point A is on y-axis
∴ its abscissa i.e., x co-ordinates will be 0
∴ Co-ordinates of A will be (0, 5)
and co-ordinates of B are (-3, 1)
∴ Length of AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(-3-0)^2+(1-5)^2}\)
\(\sqrt{(-3)^2+(-4)^2}\) = 9 + 16 = 25 = (5)²
∴ AB = 5

(b) Let point P (2, b) be the mid-point of the line joining A (a, 2) and b (3, 6)
But x = \(\frac{x_1+x_2}{2} \text { and } y=\frac{y_1+y_2}{2}\)
∴ 2 = \(\frac{a+3}{2}\) ⇒ a + 3 = 4 ⇒ a = 4 – 3 = 1
and b = \(\frac{2+6}{2}\) ⇒ 2b = 8 ⇒ b = \(\frac { 8 }{ 2 }\) = 4
∴ a = 1, b = 4

Question 15.
The line joining A (2, 3) and B (6, – 5) meets the x-axis at P. Write down the y-coordinates of P. Hence find the ratio AP : PB.
Solution:
A (2, 3) and B (6, -5) are the points which form a line AB
∵ P lies on x-axis
∴ coordinates of P will be 0.
Let P divides AB in the ratio m₁ : m₂, then
y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2} \Rightarrow 0=\frac{m_1(-5)+m_2(3)}{m_1+m_2}\)
⇒ \(\frac{-5 m_1+3 m_2}{m_1+m_2}\) = 0 ⇒ – 5₁ + 3m₂ = 0
⇒ 3m₂ = 5m₁
⇒ \(\frac{m_1}{m_2}=\frac{3}{5}\)
∴ Ratio = 3 : 5 or AP : PB = 3 : 5

Question 16.
In the given figure P (2, 3) is the mid-point of the line AB. Write down the coordinates of A and B.

Solution:
In the figure, point A lies on x-axis
and B lies on.y-axis
∴ y-coordinates of A will be 0
and x-coordinates B will be 0
Let the co-ordinates of A and B be (x, 0) and (0, y) respectively.
∴ P (2, 3) is the mid-point of AB.
∴ 2 = \(\frac{x_1+x_2}{2}=\frac{x+0}{2}=\frac{x}{2}\)
∴ x = 2 x 2 = 4
and 3 = \(\frac{y_1+y_2}{2}=\frac{0+y}{2}=\frac{y}{2}\)
∴ y = 2 x 3 = 6
∴ Co-ordinates of A and B will be (4, 0) and (0, 6)

Question 17.
The line segment joining A (2, 3) and B (6, -5) is intersected by the x-axis at a point K. Write down the ordinate of the point K. Hence find the ratio in which K divides AB.
Solution:
∵ The line joining the points A (2, 3) and B (6, -5) intersects x-axis at K
∴ Ordinate or y-coordinates of K will be 0
Let K divides AB in the ratio m₁ : m₂
∴ y = \(\frac{m_1 y_2+m_2 y_1}{m_1+m_2} \Rightarrow 0=\frac{m_1(-5)+m_2(3)}{m_1+m_2}\)
⇒ \(\frac{-5 m_1+3 m_2}{m_1+m_2}\) = 0 ⇒ – 5m₁ + 3m₂ = 0
⇒ 3m₂ = 5m₁ ⇒ \(\frac{m_1}{m_2}=\frac{3}{5}\)
∴ Ratio = 3 : 5

Question 18.
(a) Coordinates of A and B are (-3, a) and (1, a + 4). The mid-point of AB is (-1, 1). Find the value of a.
(b) Calculate the ratio in which the line segment joining (3., 4) and (-2, 1) is divided by they-axis.
Solution:
(a) Co-ordinates of A are (- 3, a) and of B (1, a + 4)
Let mid-point of AB is P (- 1, 1)
∴ y = \(\frac{y_1+y_2}{2}\) ⇒ 1 = \(\frac{a+a+4}{2}\) ⇒ 2a + 4 = 2
⇒ 2a = 2 – 4 ⇒ 2a = – 2
∴ a = \(\frac { -2 }{ 2 }\) = – 1

(b) Let a point P divides the line segment joining the points A (3, 4) and B (-2, 1) in the ratio w₁ : m₂
∵ P lies on the y-axis
∴ Its abscissa (x-coordiante) will be 0
∴ But x = \(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}\)
⇒ 0 = \(\frac{m_1(-2)+m_2(3)}{m_1+m_2}=\frac{-2 m_1+3 m_2}{m_1+m_2}\)
⇒ – 2m₁+ 3m₂ = 0 ⇒ 3m₂ = 2m₁.
⇒ \(\frac{m_1}{m_2}=\frac{3}{2}\)
∴ Ratio = 3:2

Question 19.
(a) P divides the distance between A (-2, 1) and B (1,4) in the ratio 2 : 1. Calculate the coordinates of the point P.
(b) Prove that the point A (-5, 4), B (-1, -2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the coordinates of D, so that ABCD is a square.
Solution:
(a) P divides the line segment AB whose vertices are A (-2, 1) and B (1, 4) in the ratio 2 : 1 Let co-ordinates of P be (x₁ y)
Here m₁ : m₂ = 2 : 1

Co-ordinates of P will be (0, 3)

(b) Co-ordinates of A are (-5, 4), of B are (-1, -2) and of C are (5, 2)

AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{[-1-(-5)]^2+(-2-4)^2}\)
= \(\sqrt{(-1+5)^2+(-6)^2}=\sqrt{(4)^2+(-6)^2}\)
= \(\sqrt{16+36}=\sqrt{52}\)
∴ AB² = 52
Similarly
BC² = [5 – (-1)]² + [2 – (-2)]²
= (5 + 1)² + (2 + 2)² = (6 )² + (4)² = 36 + 16 = 52
and AC² = [5 – (-5)]² + (2 – 4)² = (5 + 5)² + (2 – 4)²
= (10)² + (-2)² = 100 + 4 = 104
∵ AB = AC and
AB² + BC² = AC²
∴∆ABC is an isosceles right triangle whose ∠B = 90°
∵ ABCD is a square
Let co-ordinates of D be (a, b)
Join BD which intersects AC at O
∵ Diagonals of a square bisect each other
∴ O is mid-point of AC as well as BD
Now co-ordinates of O will be
\(\left(\frac{5-5}{2}, \frac{2+4}{2}\right) \text { or }\left(\frac{0}{2}, \frac{6}{2}\right)\) or (0, 3)
∵ O is mid-point of BD, then
0 = \(\frac{-1+a}{2}\) ⇒ – 1 + a = 0 ⇒ a = 1
and 3 = \(\frac{-2+b}{2}\) ⇒ – 2 + 6 = 6 ⇒ 6 = 6 + 2 = 8
∴ Co-ordinates of D are (1, 8)

Question 20.
A (2, 2), B (-2, 4), C (2, 6) are the vertices of a triangle ABC. Prove that ABC is an isosceles triangle.
Solution:
Vertices of a AABC are A (2, 2), B (-2, 4). C (2, 6)
Now AB² = (x₂ – x₁)² + (y₂ – y₁)²
= (-2 – 2)² + (4 – 2)² = (- 4)² + (2)²
= 16 + 4 = 20
BC² = [2 – (-2)]² + (6 – 4)² = (2 + 2)² + (2)²
= (4)² + (2)² = 16 + 4 = 20
AC² = (2 – 2)² + (2 – 6)²
= 0² + (-4)²
= 0 + 16 = 16
∵ AB² = BC² ⇒ AB = BC
∴ ∆ABC is an isosceles triangle.