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S Chand Class 10 ICSE Maths Solutions Chapter 12 Similar Triangles Ex 12(c)
Question 1.
In ∆ABC, AB = 6 cm and AC = 3 cm. If M is the mid-point of AB, and a straight line through M parallel to BC cuts AC in N, what is the length of AN?
Solution:
In ∆ABC, AB = 6 cm, AC = 3 cm
M is mid-point of AB
∴ AM = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 6 = 3 cm
MN || BC is drawn
∴ N is mid-point of AC
∴ AN = \(\frac { 1 }{ 2 }\) AC = \(\frac { 1 }{ 2 }\) x 3 = \(\frac { 3 }{ 2 }\) = 1.5 cm
Question 2.
Draw parallelogram ABCD with the following data:
AB = 6 cm, AD = 5 cm and ∠DAB = 45°
Let AC and DB meet in O and let E be the mid-point of BC. Join OE. Prove that
(i) OE || AB, (ii) OE = \(\frac { 1 }{ 2 }\) AB.
Solution:
Given : In parallelogram ABCD AB = 6 cm, AD = 5 cm
Diagonals AC and BD intersect each other at O
E is mid-point of BC, OE is joined
To prove :
(i) OE || AB, (ii) OE = \(\frac { 1 }{ 2 }\) AB
Proof:
In ∆ABC,
O is mid-point of AC
(∵ Diagonals of a parallelogram bisect each other)
and E is the mid-point of BC (given)
∴ OE || AB and OE = \(\frac { 1 }{ 2 }\) AB
Hence proved.
Question 3.
In the ∆ABC, D is the mid-point of AB, E is the mid-point of AC. Calculate :
(i) DE, if BC = 6 cm
(ii) ∠ADE, if ∠DBC = 140°
Solution:
In the figure, ABC is a triangle, D is mid point of AE and E is mid point of AC
(i) If BC = 6 cm, find DE
(ii) If ∠DBC = 140°, find ∠ADE
∵ D and E are the mid points of ∆ABC
(i) ∴ DE || BC and DE = \(\frac { 1 }{ 2 }\) BC
and ∆ABC ~ ∆ADE
∴ \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BC}} \Rightarrow \frac{1}{2} \frac{\mathrm{AB}}{\mathrm{AB}}=\frac{\mathrm{DE}}{6}\)
⇒ \(\frac{\mathrm{DE}}{6}=\frac{1}{2} \Rightarrow \mathrm{DE}=\frac{6}{2}\) = 3 cm
(ii) ∠DBC = 140°
But ∠ADE = ∠DBC (corresponding angles)
= 140° (∵ ∠DBC = 140° given)
∴ ∠ADE = 140°
Question 4.
ABCD is a trapezium with AB parallel to DC. If AB = 10 cm, AD = BC = 4 cm and ∠DAB = ∠CBA = 60°, calculate
(i) the length of CD,
(ii) the distance between AB and CD.
Solution:
In trapezium ABCD,
AB || DC
AD = BC = 4 cm
∠DAB = ∠CBA = 60°
To find :
(i) the length of CD
(ii) distance between AB and CD
Draw DL and CM perpendiculars on AB In ∆ADL and ABCM
∠L = ∠M (each 90°)
∠A = ∠B (each 60°)
DA = CB (each 4 cm)
∴ ∆ADL ≅ ABCM
∴ AL = MB
Now in right ∆ADL
Since \(\frac { DL }{ AD }\) ⇒ sin 60° = \(\frac { DL }{ 4 }\)
⇒ \(\frac{\sqrt{3}}{2}=\frac{\mathrm{DL}}{4} \Rightarrow \mathrm{DL}=\frac{4 \times \sqrt{3}}{2}\) = 2\(\sqrt{3}\) cm
and tan 60° = \(\frac{\mathrm{DL}}{\mathrm{AL}}=\sqrt{3}=\frac{2 \sqrt{3}}{\mathrm{AL}}\)
⇒ AL = \(\frac{2 \sqrt{3}}{\sqrt{3}}\) = 2
∴ MB = AL = 2 cm
But CD = LM = AB – AL – MB
= 10 – 2 – 2
= 10 – 4
= 6 cm
Question 5.
Find the unknown marked lengths in centimetres in the following figures.
Solution:
(i) In ||gm ABCD, AB || CD
Diagonals AC and BD bisects each other at K
Now in ∆ACD,
∵ E is mid-point of AD and K is mid-point of AD
(∵ DE = EA = 6 cm and AK = KC – 8 cm)
∴ EK or EKF || CD or AB
\(\frac{\mathrm{AE}}{\mathrm{AD}}=\frac{\mathrm{AK}}{\mathrm{AC}}=\frac{\mathrm{EK}}{\mathrm{CD}}\)
⇒ \(\frac{6}{6+6}=\frac{8}{8+8}=\frac{5}{y}\)
⇒ \(\frac{5}{y}=\frac{6}{12} \Rightarrow \frac{5}{y}=\frac{1}{2}\) ⇒ y = 5 x 2 = 10
∴ y = 10 cm
Similarly in ∆ACB,
KE || AB
∴ \(\frac{\mathrm{CK}}{\mathrm{CA}}=\frac{\mathrm{KF}}{\mathrm{AB}} \Rightarrow \frac{8}{16}=\frac{x}{10}\)
⇒ x = \(\frac { 8×10 }{ 16 }\) = 5
Hence x = 5 cm, y= 10 cm
(ii) In the figure, PQRS is a parallelogram
OL = LT = 6, ST = 4, LS = 3
PQ = y and PL = x
We have to find the value of x and y
In ∆QTR,
∵ L is mid-point of QT (∵ QL = LT = 6 cm)
and PLS || QR
∴ S is mid-point of TR
∴ TS = SR ⇒ SR = 4
But PQ = SR (opposite sides of ||gm)
∴ PQ = 4 or y = 4
In APLQ and ASLT
LQ = LT (each = 6 cm)
PQ = TS (each = 4 cm)
∠PLQ = ∠SLT (vertically opposite angles)
∴ ∆PLQ = ∆SLT (SAS axiom)
∴ PL = LS ⇒ x = 3
Hence x = 3, y = 4
(iii) In ∆ABD,
AL = LD = 3
∴ L is mid-point of AD
∵ LM || BD
∴ M is mid-point of AB
∴ ∆ALM ~ ∆ADB
∴ \(\frac{\mathrm{AL}}{\mathrm{AD}}=\frac{\mathrm{LM}}{\mathrm{BD}} \Rightarrow \frac{3}{6}=\frac{q}{p} \Rightarrow \frac{q}{p}=\frac{1}{2}\)
⇒ 2q = p … (i)
Similarly in ABCD,
CP = PD (each =11)
∴ P is mid-point of CD
and PQ || BD
∴ ∆CPQ ~ ∆CDB
∴ \(\frac{\mathrm{CP}}{\mathrm{CD}}=\frac{\mathrm{PQ}}{\mathrm{DB}} \Rightarrow \frac{11}{22}=\frac{8}{p}\)
⇒ P = \(\frac { 22×8 }{ 11 }\) = 16
and from (i) 2q = p
2q = 16 ⇒ q = \(\frac { 16 }{ 2 }\) = 8
∴ p = 16, q = 8
Question 6.
ABC is an isosceles triangle. AB = AC = 10 cm, BC = 12 cm. PQRS is a rectangle drawn inside the isosceles triangle. Given PQ = SR = y cm, prove that
x = 6 – \(\frac { 3y }{ 4 }\).
Solution:
Given : ∆ABC is an isosceles triangle in which AB = AC = 10 cm, BC = 12 cm
PQRS is a rectangle inside the isosceles triangle PQ = SR = y cm and PS = QR = 2x cm
To prove : x = 6 – \(\frac { 3y }{ 4 }\)
Construction : From A, draw AL ⊥ BC which bisects BC at L
Proof: ∵ L is mid-point of BC
∴ BL = LC = \(\frac { 12 }{ 2 }\) = 6 cm
Now in right ∆ABL
AB² = AL² + BL²
⇒ (10)² = AL² + (6)²
⇒ AL² = (10)² – (6)²
⇒ AL² = 100 – 36 = 64 = (8)²
∴ AL = 8 cm
Now in ABPQ and ABAL
∠Q = ∠L (each 90°)
∠B = ∠B (common)
∴ ∆BPQ ~ ∆BAL
∴ \(\frac{\mathrm{PQ}}{\mathrm{AL}}=\frac{\mathrm{BQ}}{\mathrm{BC}} \Rightarrow \frac{y}{8}=\frac{6-x}{6}\)
⇒ 6y = 48 – 8x ⇒ 8x = 48 – 6y
⇒ x = \(\frac{48-6 y}{8}=6-\frac{3 y}{4}\)
Hence x = 6 – \(\frac { 3y }{ 4 }\)
Hence proved.
Question 7.
In the figure, PQ || ST. Prove that
(a) As PQR and STR are similar.
(b) PR.RT = QR.RS.
Solution:
Given : In the figure, PQ || ST
To prove :
(i) APQR and STR are similar
(ii) PR.RT = QR.RS
Proof:
(a) In APQR and ASTR
∠PRQ = ∠SRT (vertically opposite angles)
∠QPR = ∠RST (alternate angles)
∴ ∆PQR ~ ∆STR (AA axiom)
(b) \(\frac{\mathrm{PR}}{\mathrm{RS}}=\frac{\mathrm{RQ}}{\mathrm{RT}}\) (sides are proportional)
⇒ PR x RT = QR x RS
(By cross multiplication)
Hence proved.
Question 8.
In the figure ABCD is a trapezium. AB || DC and the diagonals AC, BD intersect in E. Prove that AE.DE = BE.CE.
Solution:
Given : ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at E
To prove : AE.DE = BE.CE
Proof: In ∆AEB and ACED
∠AEB = ∠CED (vertically opposite angles)
∠EAB = ∠ECD (alternate angles)
and ∠EBA = ∠EDC
∴ ∆AEB ~ ∆CED (AAA axiom)
∴ \(\frac{\mathrm{AE}}{\mathrm{CE}}=\frac{\mathrm{BE}}{\mathrm{DE}}\) (sides are proportional)
∴ AE.DE = BE.CE (By cross multiplication)
Hence proved.
Question 9.
Perpendiculars AL, BM are drawn from the vertices A, B of a triangle ABC to meet BC, AC at L, M. By proving the triangles ALC, BMC similar, or otherwise, prove that CM.CA = CL.CB.
Solution:
Given : In ∆ABC
AL ⊥ BC and BM ⊥ AC are drawn
To prove :
(i) ∆ALC ~ ∆BMC
(ii) CM.CA = CL.CB
Proof: In ∆ALC and ∆BMC
∠L = ∠M (each 90°)
∠C = ∠C (common)
∴ ∆ALC ~ ∆BMC (AA axiom)
∴ \(\frac{\mathrm{CM}}{\mathrm{CL}}=\frac{\mathrm{CB}}{\mathrm{CA}}\) (sides are proportional)
⇒ CM.CA = CB.CL (By cross multiplication)
⇒ CM.CA = CL.CB
Hence proved.
Question 10.
If a perpendicular is drawn from the right angle of a right-angled triangle to the hypotenuse, prove that the triangles on each side of the perpendicular are similar to the whole triangle and to each other.
Solution:
Given : ∆ABC is a right angled triangle
AD ⊥ BC is drawn
To prove :
(i) ∆ABD ~ ∆ABC
(ii) ∆ACD = ∆ABC
Proof:
(i) In ∆ABD and ∆ABC
∠B = ∠B (common)
∠ADB = ∠BAC (each 90°)
∴ ∆ABD ~ ∆ABC (AA axiom)
(ii) Similarly in ∆ACD and ∆ABC
∠C = ∠C (common)
∠ADC = ∠BAC (each 90°)
∴ ∆ACD ~ ∆ABC (AA axiom)
Hence proved.
Question 11.
In the figure, it is given that the angle BAD = the angle CAE and the angle B = the angle D. Prove that
(i) \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)
(ii) the angle ADB = the angle AEC. (SC)
Solution:
Given : In the given figure, ∠BAD = ∠CAE
∠B = ∠D.
To prove :
(i) \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)
(ii) ∠AAD = ∠AEC
Proof: ∠BAD = ∠CAE
Adding ∠CAD both sides,
∠BAD + ∠CAD = ∠CAD + ∠CAE
⇒ ∠BAC = ∠DAE
Now in ∆ABC and ∆ADE
∠BAC = ∠DAE (proved)
∠B = ∠D (given)
(i) ∴ ∆ABC ~ ∆ADE (AA axiom)
∴ \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)
or \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)
(ii) Now in ∆ABD and ∆AEC
∠BAD = ∠CAE (given)
\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\) ⇒ A (proved)
∴ ∆ABD ~ ∆AEC (SAS axiom)
∴ ∠ADB = ∠AEC
Hence proved.
Question 12.
In the figure, ABCD and AEFG are squares. Prove that
(i) AF : AG = AC : AD;
(ii) triangles ACF and ADG are similar. (SC)
Solution:
Given : In the figure ABCD and AEFG are two squares
To prove :
(i) AF : AG = AC : AD;
(ii) triangles ACF and ADG are similar
Proof : ∵ AC and AF are the diagonals of two squares ABCD and AEFG respectively
∴ ∆ADC ~ ∆AGF (A)
Now ∠BAF = ∠B AC – ∠FAC
∠GAC = ∠GAF – ∠FAC
But ∠BAC = ∠GAF = 45°
(Diagonals bisects the opposite angles of a square)
∠BAF = 45° – ∠FAC … (i)
and ∠GAC = 45° – ∠FAC … (ii)
From (i) and (ii)
∴ ∠BAF = ∠GAC
Similarly ∠DAG = ∠DAC – ∠GAC = 45° – ∠GAC
and ∠FAC = ∠BAC – ∠BAF = 45° -∠BAF
But ∠GAC = ∠BAF (proved)
∴ ∠DAG = ∠FAC
Now in ∆ADG and ∆ACF
∠DAG = ∠FAC (proved)
∴ From (A)
∆ADG ~ ∆ACF (SAS axiom)
∴ \(\frac{\mathrm{AG}}{\mathrm{AF}}=\frac{\mathrm{AD}}{\mathrm{AC}}\) (sides are proportional)
⇒ \(\frac{A F}{A G}=\frac{A C}{A D}\)
⇒ AF : AG = AC : AD
Hence proved.
Question 13.
In a square ABCD, the bisector of the angle BAC cuts BD at X and BC at Y. Prove that the triangles ACY, ABX are similar. (SC)
Solution:
Given : In a square ABCD, diagonals AC and BD bisect each other at O
Bisector of ∠CAB meets BD is X and BC is Y
To prove : ∆ACY ~ ∆ABX
Proof: In ∆ACY and ∆ABX
∠CAY =∠BAY
(∵ AXY is the bisector of ∠CAB)
∠ACY = ∠ABX (each 45°)
∴ ∆ACY ~ ∆ABX (∆A axiom)
Hence proved.
Question 14.
On one of the longer sides PQ of a rectangle PQRS, a point X is taken such that SX² = PX.PQ. Prove that ∆s PXS, XSR are similar. (SC)
Solution:
Given : In rectangle PQRS, X is a point on PQ (longer side) such that SX² = PX.PQ
To prove : ∆PXS ~ ∆XSR
Proof : SX² = PX.PQ
⇒ \(\frac{S X}{P Q}=\frac{P X}{S X} \Rightarrow \frac{S X}{S R}=\frac{P X}{S X}\)
(∵ SR = PQ opposite sides of rectangle)
Now in APXS and AXSR
\(\frac{S X}{S R}=\frac{P X}{S X}\) (proved)
∠PXS = ∠RSX (alternate angles)
∴ ∆PXS ~ ∆XSR (SAS axiom)
Hence proved.
Question 15.
In the figure, D is a point on the side BC of ∆ABC such that ∠ADC = ∠BAC.
Prove that \(\frac{B C}{C A}=\frac{C A}{C D}\).
Solution:
Given : In ∆ABC, D is a point on BC such that ∠ADC = ∠B AC
To prove : \(\frac{B C}{C A}=\frac{C A}{C D}\)
Proof: In ∆ABC and ∆ADC
∠BAC = ∠ADC (given)
∠C = ∠C (common)
∴ ∆ABC ~ ∆ADC (AA axiom)
∴ \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{CD}} \Rightarrow \frac{\mathrm{BC}}{\mathrm{CA}}=\frac{\mathrm{CA}}{\mathrm{CD}}\)
Hence proved.
Question 16.
In the figure, PQR is a triangle in which PQ = PR and Z is a point on the side PR such that QR² = PR x RZ. Prove that QZ = QR.
Solution:
Given : In ∆PQR, PQ = PR
Z is a point on PR such that
QR² = PR x RZ
To prove : QZ = QR
Proof: QR² = PR x RZ
Hence proved.
Question 17.
In the figure, segments AD and BE are perpendicular to the sides BC and AC respectively. Such that
(i) ∆ADC ~ ∆BEC;
(ii) CA.CE = CB.CD;
(iii) ∆ABC ~ ∆DEC;
(iv) CD.AB = CA.DE.
Solution:
Given : In ∆ABC, AD and BE are perpendicular to the sides BC and AC respectively
To prove :
(i) ∆ADC ~ ∆BEC
(ii) CA.CE = CB.CD
(iii) ∆ABC ~ ∆DEC
(iv) CD.AB = CA.DE
Proof:
(i) In ∆ADC and ∆BEC
∠C = ∠C (common)
∠ADC = ∠BEC (each 90°)
∴ ∆ADC ~ ∆BEC (AA axiom)
∴ \(\frac{C A}{C B}=\frac{C D}{C E}\) (sides are portional)
(ii) ∴ CA.CE = CB.CD
Again in ∆ABC and ∆DEC
∴ \(\frac{C A}{C D}=\frac{A B}{D E}\)
(corresponding sides are proportional)
(iv) ⇒ CA.DE = CD.AB
or CD.AB = CA.DE
Hence proved.
Question 18.
In the figure, ABCD is a quadrilateral P, Q, R, S are the points of trisection of the sides AB, BC, CD and DA respectively and are adjacent to A and C, prove that PQRS is a parallelogram.
Solution:
Given : In quadrilateral ABCD P, Q, R and S are the points of trisection of the sides AB, BC, CD and DA respectively and are adjacent to A and C
To prove : PQRS is a parallelogram
Construction : Join diagonals AC
Proof: In ∆ABC
AP = \(\frac { 1 }{ 3 }\) AB and CQ = \(\frac { 1 }{ 3 }\) BC or \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{CQ}}{\mathrm{BC}}\)
= \(\frac { 1 }{ 3 }\)
∴ PQ || AC and PQ = \(\frac { 2 }{ 3 }\) AC … (i)
Similarly in ∆ADC,
AS = \(\frac { 1 }{ 3 }\) AD and CR = \(\frac { 1 }{ 3 }\) CD or \(\frac{\mathrm{AS}}{\mathrm{AD}}=\frac{\mathrm{CR}}{\mathrm{CD}}\) = \(\frac { 2 }{ 3 }\)
∴ RS || AC and RS = \(\frac { 2 }{ 3 }\) AC … (ii)
From (i) and (ii)
PQRS is a parallelogram
Hence proved.
Question 19.
In the figure, if AD ⊥ BC, BD = 4, AD = 6, CD = 9, prove that triangles ADB and CDA are similar. Also prove that ∠BAC is a right angle.
Solution:
Given : In ∆ABC, AD ⊥ BC
BD = 4, AD = 6, CD = 9
To prove :
(i) ∆ADB ~ ∆CDA
(ii) ∠BAC is a right angle
Proof: In ∆ADB and ∆CDA
∠ADB = ∠ADC (each 90°)
\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{6}{4}, \frac{\mathrm{CD}}{\mathrm{AD}}=\frac{9}{6}\)
⇒ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{3}{2} \text { and } \frac{\mathrm{CD}}{\mathrm{AD}}=\frac{3}{2}\)
∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{CD}}{\mathrm{AD}}\) (each \(\frac { 2 }{ 3 }\))
∴ ∆ADB ~ ACDA (SAS axiom)
∴ ∠ABD = ∠CAD and ∠BAD = ∠ACD
∴ ∠BAD + ∠CAB = ∠ABD + ∠ACD
⇒ ∠BAC = ∠ABD + ∠ACD
But ∠BAC + ∠ABD + ∠ACD = 180°
∴ ∠BAC = 90°
Hence proved.
Question 20.
If two triangle are equiangular, prove that the ratio of the corresponding sides is same as the ratio of corresponding altitudes.
Solution:
Given : ∆ABC ~ ∆DEF
AL ⊥ BC and DM ⊥ EF
Question 21.
The diagonal BD of a parallelogram ABCD intersect AE at a point F, where E is any point on side BC. Prove that DF.EF = FB.FA.
Solution:
Given : In parallelogram ABCD
BD is diagonal and E is any point on BC
AE is joined which intersect BD at F
To prove : DF.EF = FB.FA
Proof: In ∆AFD and ABFE
∠AFD = ∠BFE (vertically opposite angles)
∠ADF = ∠FBE (Alternate angles)
∴ ∆AFD ~ ∆BFE (AA axiom)
∴ \(\frac{D F}{F B}=\frac{F A}{F E}\) (Sides are proportional)
∴ DF.EF = FB.FA (By cross multiplication)
Hence proved.
Question 22.
Any point X inside ADEF is joined to its vertices. From a point P in DX, PQ is drawn || DE meeting XE at Q and QR is drawn || EF meeting XF in R. Prove that PR || DF.
Solution:
Given : X is any point in ADEF
XD, XE and SF are joined.
P is a point on DX, PQ || DE is drawn meeting XE in Q and QR || EF isdrawn meeting XF in R
To prove : PR || DF
Proof: ∵ P is any point on DX and PQ || DE
∴ \(\frac{X P}{P D}=\frac{X Q}{Q E}\)
∵ QR || EF
∴ \(\frac{\mathrm{XQ}}{\mathrm{QE}}=\frac{\mathrm{XR}}{\mathrm{RF}}\)
From (i) and (ii)
\(\frac{X P}{P D}=\frac{X R}{R F}\)
∴ PR || DF
Hence proved.