The availability of S Chand Class 10 Maths Solutions ICSE Chapter 12 Similar Triangles Ex 12(e) encourages students to tackle difficult exercises.
S Chand Class 10 ICSE Maths Solutions Chapter 12 Similar Triangles Ex 12(e)
Question 1.
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 2.
A triangle, whose area is 12 cm, 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 3.
∆ABC with sides AB = 10 cm, BC = 12 cm and AC = 16 cm is enlarged to the triangle A’B’C’ such that the largest side of the triangle A’B’C’ is 24 cm. Find the Scale (enlargement) factor and use it to find the lengths of other sides of the linage triangle A’B’C’.
Solution:
Sides of given ∆ABC, AB = 10 cm, BC = 12 cm and AC = 16 cm
Longest side AC = 16 cm
But longest side of enlarged ∆A’B’C’ = 24 cm
i.e. A’C’ = 24 cm
Let k be the scale factor
∴ k = \(\frac { 24 }{ 16 }\) = \(\frac { 3 }{ 2 }\)
Now length of side A’B’ = 10 cm x \(\frac { 3 }{ 2 }\) = 15 cm
and length of side B’C’ = 12 cm x \(\frac { 3 }{ 2 }\) = 18 cm
Question 4.
A ∆PQR right angled at Q has PQ = 8 cm and QR = 15 cm. It is enlarged to a triangle P’Q’R’ such that the image of P is P’, image of Q is Q’, image of R is R’ and length of P’R’ = 42.5 cm. Find :
(i) the scale (enlargement factor);
(ii) the lengths of P’Q’ and Q’R’;
(iii) the area of ∆PQR and hence area of image triangle P’Q’R’.
Solution:
In right angled ∆PQR, ∠Q = 90°
PQ = 8 cm and QR = 15 cm
∴ PR = \(\sqrt{\mathrm{PQ}^2+\mathrm{QR}^2}=\sqrt{(8)^2+(15)^2}\)
= \(\sqrt{64+225}=\sqrt{289}\) = 17 cm
∆PQR is enlarged into AP’Q’R’ in which . image of P is P’ of Q is Q’ and of R is R’
Length of P’R’ = 42.5 cm
(i) Let k be the scale factor of enlargement 42.5 = k x PR = k x 17
∴ k = \(\frac{42.5}{17}=2.5=\frac{25}{10}=\frac{5}{2}\)
(ii) Now length of P’Q’ = \(\frac { 5 }{ 2 }\) x PQ = \(\frac { 5 }{ 2 }\) x 8 = 20 cm
and length of Q’R’ = \(\frac { 5 }{ 2 }\) QR = \(\frac { 5 }{ 2 }\) x 15 = \(\frac { 75 }{ 2 }\)
= 37.5 cm
(iii) Now area of ∆PQR = \(\frac { 1 }{ 2 }\) x PQ x QR
∴ \(\frac { 1 }{ 2 }\) x 8 x 15 cm² = 60 cm²
∴ area of ∆P’Q’R’ = k² x area of ∆PQR
= (\(\frac { 5 }{ 2 }\))² x 60 cm² = \(\frac { 25 }{ 4 }\) x 60 = 375 cm²
Question 5.
On a map scale 1 cm to the km an island has an area of 3.5 cm². Find the actual area.
Solution:
Scale of map = 1 cm : 1 km or 1 : 100000
Now area of island on the map = 3.5 cm²
∴ Actual area = k² x area of the map
= (100000)² x 3.5 cm²
= 100000 x 100000 x 3.5 cm²
= \(\frac{100000 \times 100000 \times 3.5}{100000 \times 100000}\) km²
= 3.5 km²
Question 6.
The scale of a map is 5 m to 1 cm. (i) What area in m² is represented by a square of side 3 cm on the map ? (ii) What is the length of the side of a square on the map which represents an area of 50625 m² ?
Solution:
Scale of map = 5 m to 1 cm
or 500 : 1 ⇒ k = \(\frac { 500 }{ 1 }\)
Side of a square on the map = 3 cm
∴ Its area = (side)² = (3 cm)² = 9 cm²
(i) ∴ Actual area of the square = k² x area of square on the map = (500)² x 9 cm²
= 500 x 500 x 9 cm²
= \(\frac{500 \times 500 \times 9}{100 \times 100}\) = 225 m²
(ii) Actual area of a square = 50625 m²
Area on the map = \(\frac { 1 }{ k² }\) x actual area
= \(\frac{1}{(500)^2}\) x 50625
= \(\frac{50625 \times 100 \times 100}{500 \times 500}\) cm² = 2025 cm²
∴ Side – \(\sqrt{2025}\) = 45 cm
Question 7.
A ground plan of a house is made on the scale 1 cm to 15 m. Find the scale factor of the plan. Find the length and breadth on the plan of a room 18 m by 12 m. What area on the ground is represented by 1 cm² on the plan ?
Solution:
Scale of a ground plan of a house = 1 cm : 15 m
Scale factor (k) = 1 cm : 1500 cm
= 1 : 1500 = \(\frac { 1 }{ 1500 }\)
Now length of a room = 18 m
and breadth = 12 m
∴ Length on plan = 18 x \(\frac { 1 }{ 1500 }\) m
= \(\frac{18 \times 100}{1500}\) cm = \(\frac { 6 }{ 5 }\) cm = 1.2 cm
and breadth = 12 x \(\frac { 1 }{ 1500 }\) m = \(\frac { 1200 }{ 1500 }\) m = \(\frac { 1200 }{ 1500 }\) cm
= \(\frac { 4 }{ 5 }\) cm = 0.8 cm
Area on the plan = 1 cm³
Area on the ground = 1 x (1500)² cm²
= \(\frac{1 \times 1500 \times 1500}{100 \times 100}\) = 225 m²
Question 8.
On a map of 1 cm to the 4 kni, an estate occupies an area of 9.37 cm². Find the actual area to the nearest km².
Solution:
Scale on the map = 1 cm to 4 km = 1 cm : 400000 cm
= 1 : 400000
⇒ k = \(\frac { 1 }{ 400000 }\)
Area of an estate on the map = 9.37 cm²
Actual area = k² (area on the map)
= (400000)² x 9.37 cm²
= \(\frac{937 \times(400000)^2}{100}\)
= \(\frac{937 \times 400000 \times 400000}{100 \times 100000 \times 100000}\)
= \(\frac { 14992 }{ 100 }\) = 149.92 km²
= 150 km² (nearest km²)
Question 9.
The map of a rectangular field drawn to a scale of 1 : 20000 measures 20 cm by 15 cm. Calculate the actual area of the Field in sq. km (km²).
Solution:
Scale of a map = 1 : 20000
Measure of field on the map = 20 cm x 15 cm = 300 cm²
Actual area = k² (Area of the field on the map)
= (20000)² x 300 cm²
= 20000 x 20000 x 300 cm²
= \(\frac{20000 \times 20000 \times 300}{100000 \times 100000}\)
= 12 km²
Question 10.
The scale of a map is 1 : 50,000. In the map, a triangular plot ABC of land has the following dimensions : AB = 2 cm, BC = 3.5 cm and ∠ABC = 90°. Calculate: (i) the actual length of BC,in km of the land (ii) the area of the plot in sq. km.
Solution:
Scale on the map = 1 : 50000
On the map in right angled ∆ABC, AB = 2
cm, BC = 3.5 cm and ∠ABC = 90°
∴ Area = \(\frac { 1 }{ 2 }\) x AB x BC = \(\frac { 1 }{ 2 }\) x 2 x 3.5 cm²
= 3.5 cm²
(i) Now actual length of BC = 3.5 cm x 50000
= \(\frac{3.5 \times 50000}{100000}\)
= \(\frac{35 \times 50000}{10 \times 100000}=\frac{35 \times 5}{100}\)
= \(\frac { 175 }{ 100 }\)
and actual area = 3.5 cm² x (50000)²
= \(\frac{3.5 \times 50000 \times 50000}{100000 \times 100000}\)km²
= \(\frac{3.5 \times 25}{100}=\frac{3.5}{4}\) = 0.875 k²
Question 11.
The dimensions of the model of a multistorey building are 1 m by 60 cm by 1.20 m. If the scale factor is 1 : 50, find the actual dimensions of the building. Also, find :
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq. cm;
(ii) the space (volume) inside a room of the model, if the space inside the corres-ponding room of the building is 90 m³.
Solution:
Scale factor of the model = 1 : 50
Dimensions of a model of a multistorey building are 1 m x 60 cm x 1.20 m
or 100 cm x 60 cm x 120 cm
∴ Actual length = 100 x 50 cm = \(\frac{100 \times 50}{100}\) m
= 50 m
Breadth = 60 x 50 cm
= \(\frac{60 \times 50}{100}\) = 30 m
and height = 120 x 50 cm
= \(\frac{120 \times 50}{100}\) = 60 m
∴ Actual dimensions = 50 m x 30 m x 60 m
(i) Floor area of a room on the model = 50 sq. cm
∴ Actual area = 50 x k² = 50 x (50 x 50) cm²
= \(\frac{50 \times 2500}{100 \times 100}\) m² = 12.5 m²
(ii) Actual volume of the room space = 90 m³
∴ Volume on the model = \(\frac{90}{(50)^3}\) m³
= \(\frac{90}{50 \times 50 \times 50}\) m³
= \(\frac{90 \times 100 \times 100 \times 100}{50 \times 50 \times 50}\) cm³
= \(\frac { 90000 }{ 125 }\) cm² = 720 cm³
Self Evaluation And Revision
(LATEST ICSE QUESTIONS)
Question 1.
In the figure, LM is parallel to BC AB = 6 cm, AL = 2 cm and AC = 9 cm. Calculate :
(i) The length of CM.
(ii) The value of the ratio
= \(\frac{\text { Area of triangle ALM }}{\text { Area of trapezium LBCM }}\).
Solution:
In ∆ABC, LM || BC
AB = 6 cm, AL = 2 cm, AC = 9 cm
∴ LB = AB – AL = 6 – 2 = 4 cm
Let AM = x, then MC = AC – AM = (9 – x) cm
Now in ∆ABC,
LM || BC
(i) ∴ \(\frac{\mathrm{AL}}{\mathrm{LB}}=\frac{\mathrm{AM}}{\mathrm{MC}} \Rightarrow \frac{2}{4}=\frac{x}{9-x}\)
⇒ 4x = 18 – 2x ⇒ 4x + 2x = 18
⇒ 6x = 18 ⇒ x = \(\frac { 18 }{ 6 }\) = 3
∴ MC = 9 – x = 9 – 3 = 6cm
(ii) In ∆ALM and ∆ABC
∵ LM || BC
∴ ∆ALM ~ ∆ABC
Question 2.
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:
Scale of the map = 1 : 200000
It means that 1 cm on the map will represent 200000 cm² on the ground
(i) ∴ 1 cm = 200000 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 3.
In the figure, PQRS is a parallelogram, PQ = 16 cm, QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.
(i) Prove that triangle RLQ is similar to triangle PLN. Hence find PN.
(ii) Name a triangle similar to triangle RLM. Evaluate RM as a fraction.
Solution:
In ||gm PQRS,
PQ = 16 cm, QR = 10 cm
L is a point on PR such that RL : LP = 2 : 3
QL is produced to meet RS at M and PS on producing at N
(i) In ∆RLQ and ∆PLN,
∠RLQ = ∠PLN
(Vertically opposite angles)
∠LQR = ∠LNP (Altenate angles)
∴ ∆RLQ ~ ∆PLN (AA axiom)
∴ \(\frac{R Q}{P N}=\frac{R L}{L P}\)
⇒ \(\frac{10}{P N}=\frac{2}{3} \Rightarrow P N=\frac{10 \times 3}{2}\) = 15 cm
(ii) In ∆RLM and ∆LPQ,
∠RLM = ∠LPQ
(Vertically opposite angles)
∠LRM = ∠LPQ (Altenate angles)
∴ ∆RLM ~ ∆LOQ (AA axiom)
∴ \(\frac{\mathrm{RM}}{\mathrm{PQ}}=\frac{\mathrm{RL}}{\mathrm{LP}}\)
⇒ \(\frac{\mathrm{RM}}{16}=\frac{2}{3} \Rightarrow \mathrm{RM}=\frac{16 \times 2}{3}\)
⇒ RM = \(\frac { 32 }{ 3 }\)cm
Question 4.
On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm, angle ABC = 90°, Calculate:
(i) the actual length of AB in km;
(ii) the area of the plot in sq km.
Solution:
Scale on the map is 1 : 250000
(i) i.e. 1 cm on the map = 250000 cm on the ground = \(\frac{250000}{100 \times 1000}\) = 2.5 km
But length of AB on the map = 3 cm
∴ Actual length = 3 x 2.5 = 7.5 km
(ii) and length of BC = 4 cm on the map
Actual length = 4 x 2.5 = 10 km
Area of plot in sq. km = \(\frac { 1 }{ 2 }\) x Base x Height
= \(\frac { 1 }{ 2 }\) x 7.5 x 10 km² = 37.5 km²
Question 5.
On a map drawn to a scale of 1 : 250000, a rectangular plot of land ABCD has the following measurements : AB = 12 cm and BC = 16 cm. A, B, C, D are all 90° each. Calculate : (i) the diagonal distance of the plot in km. (ii) the area of the plot in sq. km.
Solution:
Scale of the map = 1 : 250000
i.e. 1 cm on the map = 250000 cm on the ground = \(\frac{250000}{100 \times 1000}\) = 2.5 km
(i) Length of plot on the map (l) = 12 cm and breadth (b) = 16 cm
Length of diagonal AC = \(\sqrt{A B^2+B C^2}\)
= \(\sqrt{l^2+b^2}=\sqrt{(12)^2+(16)^2}\)
= \(\sqrt{144+256}=\sqrt{400}\) = 20 cm
∴ Actual length of diagonal = 20 x 2.5 km = 50 km
(ii) Area of plot = l x b = 12 cm x 16 cm = 192 cm²
∴ Actual area = 192 x (2.5)² km²
= 192 x \(\frac { 25 }{ 10 }\) x \(\frac { 25 }{ 10 }\) km²
= 192 x \(\frac { 5 }{ 2 }\) x \(\frac { 5 }{ 2 }\) = 1200 km²
Question 6.
In the figure, P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.
(i) Calculate the ratio PQ : AC giving reasons for your answer;
(ii) In ∆ARC, ∠ARC = 90° and in ∆PQS, ∠PSQ = 90°.
Given QS = 6 cm, calculate the length of AR.
Solution:
In the figure,
In ∆ABC, Pisa point on AB such that AP : PB = 4 : 3
PQ || AC
(i) Now in ∆PQB and ∆ABC,
∠PBQ = ∠ABC (common)
∠PQB = ∠ACB (corresponding angles)
∴ ∆PQB ~ ∆ABC (AA axiom)
∴ \(\frac{P Q}{A C}=\frac{P B}{A B}\)
= \(\frac{\mathrm{PB}}{\mathrm{AP}+\mathrm{PB}}=\frac{3}{4+3}=\frac{3}{7}\)
∴ Ratio = 3 : 7
(ii) In ∆ARC and ∆PQS,
∠ARC = ∠PSQ (each = 90°)
∠SPQ = ∠RCA (alternate angles)
∴ ∆ARC ~ ∆PQS (AA axiom)
∴ \(\frac{\mathrm{AR}}{\mathrm{QS}}=\frac{\mathrm{AC}}{\mathrm{PQ}}\)
⇒ \(\frac{\mathrm{AR}}{6}=\frac{7}{3}\) (QS = 6 cm given)
⇒ AR = \(\frac { 7 }{ 3 }\) x 6 = 14 cm
Question 7.
In the figure, BC is parallel to DE. Area of triangle ABC = 25 cm², area of trapezium BCED = 24 m². DE = 14 cm. Calculate the length of BC.
Solution:
In ∆ADE,
∵ BC || DE
∴ ∆ABC ~ ∆ADE
∴ \(\frac { area (∆ABC) }{ area(∆ADE) }\) = \(\frac{B C^2}{D E^2}\)
{The areas of similar triangles are proportional to the square of their corresponding sides}
⇒ \(\frac { area (∆ABC) }{ area(∆ABC)+area(trap BCED) }\) = \(\frac{\mathrm{BC}^2}{\mathrm{DE}^2}\)
⇒ \(\frac{25}{25+24}=\frac{\mathrm{BC}^2}{(14)^2} \Rightarrow \frac{25}{49}=\frac{\mathrm{BC}^2}{(14)^2}\)
⇒ \(\frac{\mathrm{BC}^2}{(14)^2}=\frac{(5)^2}{(7)^2} \Rightarrow \frac{\mathrm{BC}}{14}=\frac{5}{7}\)
⇒ BC = \(\frac { 5 }{ 7 }\) x 14 = 10 cm
Question 8.
(a) In the figure, the medians BD and CE of a triangle ABC meet at G. Prove that
(i) ∆EGD ~ ∆CGB, and
(ii) BG = 2GD, from (i) above
(b) In the right angled triangle OPR, PM is an altitude. Given that QR = 8 cm, and MQ = 3.5 cm, calculate the value of PR.
Solution:
(a) In ∆ABC, BD and CE are the medians of side AC and AB
Join ED
∵ D and E are the mid-points of AC and AB
∴ ED || BC and ED = \(\frac { 1 }{ 2 }\) BC … (i)
Now in ∆EGD and ∆CGB,
(b) In right angled ∆QPR,
Question 9.
In figure, in a triangle PQR, L and M are two points on the base QR such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP.
Prove that :
(i) ∆PQL ~ ∆RPM
(ii) QL.RM = PL.PM
(iii) PQ² = QR.QL
Solution:
In ∆PQR, L and M are two points on the base QR such that
∠LPQ = ∠QRP and ∠RPM = ∠RQP
(i) Now in ∆PQL and ARPM,
∠PQL = ∠RPM (given)
∠LPQ = ∠MRP (given)
∴ ∆PQL ~ ∆RPM (AA axiom)
(ii) ∴ \(\frac{\mathrm{QL}}{\mathrm{PM}}=\frac{\mathrm{PL}}{\mathrm{RM}}\)
⇒ QL – RM = PL.PM
(iii) Again in ∆LQP and ∆PQR,
∠Q = ∠Q (common)
∠LPQ = ∠PRQ (given)
∴ ∆LQP ~ ∆PQR (AA axiom)
∴ \(\frac{P Q}{Q R}=\frac{Q L}{P Q}\) ⇒ PQ² = QR.QL
Hence proved.
Question 10.
In the figure, DE || BC.
(i) Prove that ∆ADE and ∆ABC are similar.
(ii) Given that AD = \(\frac { 1 }{ 2 }\) BD, calculate DE, if BC = 4.5 cm.
Solution:
(i) In ∆ABC, DE || BC
In ∆ADE and ∆ABC,
∠A = ∠A (common)
∠ADE = ∠ABC (corresponding angles)
∠AED = ∠ACB (corresponding angles)
∴ ∆ADE ~ ∆ABC (AAA axiom)
(ii) ∵ AD = \(\frac { 1 }{ 2 }\) BD
∴ 2AD = BD
⇒ 2AD + AD = AD + DB
⇒ 3AD = AB ⇒ \(\frac { AD }{ AB }\) = \(\frac { 1 }{ 3 }\)
∵ ∆ADE ~ ∆ABC
∴ \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BC}} \Rightarrow \frac{1}{3}=\frac{\mathrm{DE}}{4.5}\) {∵ BC = 4.5 cm}
⇒ DE = 4.5 x \(\frac { 1 }{ 3 }\) = 1.5 on
Question 11.
In figure, AB and DE are perpendiculars to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm. Calculate AD.
Solution:
In ∆ABC, AB ⊥ BC and DE ⊥BC
AB = 9 cm, DE = 3 cm and AC = 24 cm
In ∆ABC and ∆DEC,
∠B = ∠E (each 90°)
∠C = ∠C (common)
∴ ∆ABC ~ ∆DEC (AA axiom)
∴ \(\frac { CA }{ CD }\) = \(\frac { AB }{ DE }\)
⇒ \(\frac { 24 }{ CD }\) = \(\frac { 9 }{ 3 }\)
⇒ CD = \(\frac{24 \times 3}{9}\) = 8 cm
∴ AD = AC
Question 12.
In the figure, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB =120 cm², find the area of ∆QOA.
Solution:
In the figure
PB ⊥ AB at B and QA ⊥ AB at A.
PQ is joined which intersects AB at O
PO = 6 cm, QO = 9 cm
Question 13.
In the figure, ABC is a triangle DE is parallel to BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}\).
(i) Determine the ratios \(\frac{\mathrm{AD}}{\mathrm{DB}}, \frac{\mathrm{DE}}{\mathrm{BC}}\)
(ii) Prove that ∆DEF is similar to ∆CBF.
Hence, find \(\frac {EF}{FB}\).
(iii) What is the ratio of the areas of ∆DFE and ∆BFC.
Solution:
In the figure, in ∆ABC
DE || BC
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}\)
(i) \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}=\frac{3}{3+2}=\frac{3}{5}\)
In ∆ABC,
∵ DE || BC
Question 14.
In AABC, AP : PB = 2 : 3. PO is parallel to BC and extended to Q, so that CQ is parallel to BA. Find :
(i) area ∆APO : area ∆ABC
(ii) area ∆APO : area ∆CQO
Solution:
In ∆ABC
PO || BC
AP : PB = 2 : 3
⇒ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{2}{3}\)
PO is produced to Q
and CQ || BA
∵ CQ = BP = 3
(i) In ∆APO and ∆ABC,
∠A = ∠A (common)
∠APO = ∠ABC (corresponding angles)
∴ ∆APO ~ ∆ABC (AA axiom)
∴ \(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{AO}}{\mathrm{AC}}=\frac{\mathrm{PO}}{\mathrm{BC}}\) … (i)
But \(\frac { AP }{ PB }\) = \(\frac { 2 }{ 3 }\)
∴ \(\frac{A P}{A B}=\frac{A P}{A P+P B}=\frac{2}{2+3}=\frac{2}{5}\) … (ii)
From (i) and (ii)
\(\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{AO}}{\mathrm{AC}}=\frac{\mathrm{PO}}{\mathrm{BC}}=\frac{2}{5}\)
∆APO and ∆ABC are similar
∴ \(\frac { area (∆APO) }{ area(∆ABC) }\) = \(\frac{A P^2}{A B^2}\)
{Areas of similar triangles is proportional to the square of their corresponding sides}
= \(\frac{(2)^2}{(5)^2}=\frac{4}{25}\)
∴ area (∆APO): area (∆ABC) = 4 : 25
(ii) In ∆APO and ∆CQO,
Question 15.
In the given figure, ABC and CEF are two triangles where BA is parallel to CE and AF : AC = 5:8.
(i) Prove that ∆ADF ~ ∆CEF
(ii) Find AD if CE = 6 cm
(iii) If DF is parallel to BC, find : area of ∆ADF : area of ∆ABC.
Solution:
Given : In the figure, in ∆ABC and CEF, BA || CE and AF : AC = 5 : 8
To prove :
(i) ∆ADF ~ ∆CEF
(ii) Find AD if CE = 6 cm
(iii) If DF || BC, then find area ∆ADF : area ∆ABC
Proof:
(i) In ∆ADF and ∆CEF
∠AFD = ∠EFC (Vertically opp. angles)
∠ADF = ∠FEC (Alt. angles)
∴ ∆ADF ~ ∆CEF (AA axiom)
Question 16.
The model of a building is constructed with scale factor 1 : 30.
(i) If the height of the model is 80 cm, find the actual height of the building in metres. ,
(ii) If the actual volume of a tank at the top of the building if 27 m³, find the volume of the tank on the top of the model.
Solution:
Scale factor = 1 : 30
\(\frac{\text { Height of model }}{\text { Height of building }}=\frac{1}{30}\)
(i) Height of model = 80 cm
Let the actual height of the building = k.
\(\frac { 80 }{ k }\) = \(\frac { 1 }{ 30 }\) ⇒ k = 80 x 30 cm
⇒ k – 2400 cm = 24 m
(ii) \(\frac{\text { Volume of model }}{\text { Volume of tank }}=\left(\frac{1}{30}\right)^3\)
⇒ \(\frac{\text { Volume of model }}{27}=\frac{1}{27000}\)
⇒ Volume of model = \(\frac{27}{7000}=\frac{1}{1000}\)m³
= \(\frac { 1 }{ 1000 }\) x 100 x 100 x 100 cm³
= 1000 cm³
Question 17.
In the given figure ABC is a triangle with ∠EDB = ∠ACB. Prove that ∆ABC ~ ∆EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of ∆BED = 9 cm². Calculate the (i) length of AB (ii) area of ∆ABC.
Solution:
In ∆ABC and ∆EBD
∠1 = ∠2 (given)
∠B = ∠B (common)
⇒ ∆ABC ~ ∆EBD
Question 18.
In the given figure, ABC is a right-angled triangle with ∠BAC = 90°.
(i) Prove ∆ADB ~ ∆CDA.
(ii) If BD = 18 cm, CD = 8 cm, find AD.
(iii) Find the ratio of the area of ∆ADB is to area of ∆CDA.
Solution:
(i) In ∆ADB and ∆CDA :
∠ADB = ∠ADC [∵ each = 90°]
∠ABD = ∠CAD [∵ each = 90° – ∠BAD]
∴ ∆ADB ~ ∆CDA [by AA similarity axiom]
(ii) Since, ∆ADB ~ ∆CDA
Question 19.
In the given figure ∆ABC and ∆AMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove ∆ABC ~ ∆AMP
(ii) Find AB and BC.
Solution:
(i) In ∆ABC and ∆AMP,
∠A = ∠A (common)
∠ABC = ∠AMP (Each = 90°)
⇒ ∆ABC ~ ∆AMP (AA similarity)
(ii) From (i), \(\frac{\mathrm{AC}}{\mathrm{AP}}=\frac{\mathrm{BC}}{\mathrm{PM}}\)
(Sides are proportional)
⇒ \(\frac{10}{15}=\frac{B C}{12} \Rightarrow B C=\frac{12 \times 10}{15}\) = 8 cm
From right triangle ABC, we have
AC² = AB² + BC² (Pythagoras Theorem)
⇒ 10² = AB² + 8²
⇒ 100 = AB² + 64
⇒ AB² = 100 – 64 = 36
⇒ AB = 6 cm
Hence, AB = 6 cm, BC = 8 cm
Question 20.
In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm, DE = 4 cm and AC = 15 cm. Calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.
Solution:
(i) To prove : ∆ABC ~ ∆DEC
in ∆ABC and ∆DEC
∠ABC = ∠DEC = 90°
∠C = ∠C (common)
∴ ∆ABC ~ ∆DEC (by AA axion)
(ii) \(\frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{DE}}\)
\(\frac{15}{D C}=\frac{6}{4}\)
∴ CD = \(\frac{15 \times 4}{6}\)
CD = 10 cm
(iii) \(\frac{\text { area of } \triangle \mathrm{ABC}}{\text { area of } \triangle \mathrm{DEC}}=\frac{\mathrm{AB}^2}{\mathrm{DE}^2}\)
= \(\frac{6^2}{4^2}=\frac{36}{16}=\frac{9}{4}\) = 9 : 4
Question 21.
In ∆ABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA.
(ii) Find BC and CD.
(iii) Find area of ∆ACD : area of ∆ABC.
Solution:
In ∆ACD and ∆BCA
∠C = ∠C (common)
∠ABC = ∠CAD (Given)
∴ ∆ACD ≅ ∆BCA (by AA axiom)
Now, \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{CD}}{\mathrm{CA}}=\frac{\mathrm{AD}}{\mathrm{AB}}\)
\(\frac{4}{B C}=\frac{C D}{4}=\frac{5}{8}\)
∴ \(\frac{4}{\mathrm{BC}}=\frac{5}{8}\)
BC = \(\frac{4 \times 8}{5}=\frac{32}{5}\) = 6.4 cm
∴ \(\frac{\mathrm{CD}}{4}=\frac{5}{8} \Rightarrow \mathrm{CD}=\frac{5}{8} \times 4=\frac{5}{2}\) = 2.5 cm
(iii) ∵ ∆ACD ≅ ∆BCA
∴ \(\frac{\text { area of } \triangle \mathrm{ACB}}{\text { area of } \triangle \mathrm{BCA}}=\frac{\mathrm{AC}^2}{\mathrm{AB}^2}\)
= \(\frac{(4)^2}{(8)^2}=\frac{16}{64}=\frac{1}{4}\)
area (∆ACD): area (∆ABC) = 1 : 4
Question 22.
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:
(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of ∆ADE: area of quadrialteral BCED.
Solution:
(i) Consider ∆ADE and ∆ACB
∠A = ∠A (Common)
m∠B = m∠E = 90°
Thus by angle-angle similarity, triangles,
∆ACB ~ ∆ADE
(ii) Consider ∆ADE and ∆ACB
Since they are similar triangles, the sides are proportional
Thus, we have,
\(\frac{\mathrm{AE}}{\mathrm{AB}}=\frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}\) … (i)
Consider ∆ABC
By applying Pythagoras Theorem, we have,
AB² + BC² = AC²
⇒ AB² + 5² = 13²
AB² = (13)² – (5)²
AB² = 169 – 25
AB² = 144
AB = \(\sqrt{144}\)
⇒ AB = 12 cm
From equation (1), we have
\(\frac{4}{12}=\frac{\mathrm{AD}}{13}=\frac{\mathrm{DE}}{5}\)
\(\frac{1}{3}=\frac{\mathrm{AD}}{13}\)
⇒ AD = \(\frac { 13 }{ 3 }\) cm = 4.33 cm
Also \(\frac { 4 }{ 12 }\) = \(\frac { DE }{ 5 }\)
⇒ DE = \(\frac { 20 }{ 12 }\) = \(\frac { 5 }{ 3 }\) cm = 1.67 cm
(iii) We need to find the area of ∆ADE and quadrialteral BCED
Area of ∆ADE = \(\frac { 1 }{ 2 }\) x AE x DE
= \(\frac { 1 }{ 2 }\) x 4 x \(\frac { 5 }{ 3 }\)
= \(\frac { 10 }{ 3 }\) cm² = 3.34 cm²
Area of quad. BCED=Area of ∆ABC – Area of ∆ADE
= \(\frac { 1 }{ 2 }\) x BC x AB – \(\frac { 10 }{ 3 }\)
= \(\frac { 1 }{ 2 }\) x 5 x 12 – \(\frac { 10 }{ 3 }\)
= 30 – \(\frac { 10 }{ 3 }\)
= \(\frac{90-10}{3}=\frac{80}{3}\) cm² = 26.67 cm²
Thus ratio of areas of ∆ADE to quadrilateral
BCED = \(\frac{\frac{10}{3}}{\frac{80}{3}}=\frac{3.34}{26.67}\) = 3.34 : 26.67
Question 23.
In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.
(i) Prove ∆TPS ~ ∆TRQ.
(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.
(iii) Find area of quadrilateral PQRS if area of ∆PTS = 27 cm².
Solution:
(i) Since PQRS is a cyclic quadrilateral
∠RSP + ∠RQP = 180°
(Since sum of the opposite angles of a cyclic quadrilateral is 180°)
⇒ ∠RQP = 180° – ∠RSP … (i)
∠RQT + ∠RQP = 180°
(Since angles from a linear pair)
⇒ ∠RQP = 180° – ∠RQT … (ii)
From (i) and (ii),
180° – ∠RSP = 180° – ∠RQT
⇒ ∠RSP = ∠RQT …(iii)
In ∆TPS and ∆TRQ,
∠PTS = ∠RTQ (common angle)
∠RSP = ∠RQT [From (iii)]
∴ ∆TPS ~ ∆TRQ (AA similarity criterion)
(ii) Since ∆TPS ~ ∆TRQ, implies that corresponding sides are proportional that is,
\(\frac{\mathrm{SP}}{\mathrm{QR}}=\frac{\mathrm{TP}}{\mathrm{TR}}\)
\(\frac{S P}{4}=\frac{18}{6}\)
⇒ SP = \(\frac { 18×4 }{ 6 }\)
⇒ SP = 12 cm
(iii) Since ∆TPS ~ ∆TRQ
\(\frac { ar (∆TPS) }{ ar (∆TRQ) }\) = \(\frac { SP }{ RQ }\)
⇒ \(\frac { 27 }{ ar (∆TRQ) }\) = \(\frac { 12 }{ 4 }\)
⇒ ar(∆TRQ = \(\frac{27 \times 4}{12}\)
⇒ ar(∆TRQ) = 9 cm²
ar(PQRS) = ar(∆TPS) – ar(∆TRQ)
⇒ ar(PQRS) = 27 – 9
⇒ ar(PQRS) = 18 cm²