# Frank Solutions for Chapter 15 Similarity Class 10 ICSE Mathematics

1. In Î”ABC, DE is parallel to BC and AD/DB = 2/7. If AC = 5.6, find AE.

From the question it is given that, AD/DB = 2/7 and AC = 5.6

Consider Î”ABC,

DE is parallel to BC,

Let us assume AE = y

From Basic Proportionality Theorem (BPT) AD/DB = AE/EC

2/7 = y/(5.6 – y)

⇒ 2(5.6 – y) = 7y

⇒ 11.2 – 2y = 7y

⇒ 11.2 = 7y + 2y

⇒ 11.2 = 9y

⇒ y = 11.2/9

⇒ y = 1.24

Therefore, AE is 1.24

2. In Î”PQR, MN is drawn parallel to QR. If PM = x, MQ = (x – 2) and NR = (x – 1), find the value of x.

From the question it is given that, PM = x, MQ = (x – 2) and NR = (x – 1)

Consider Î”PQR,

MN is drawn parallel to QR,

From Basic Proportionality Theorem (BPT) PM/MQ = PN/NR

x/(x – 2) = (x + 2)/( x- 2)

⇒ x(x – 2) = (x + 2) (x – 2)

⇒ x2 – 2x = x2 – 4

⇒ x2 – x2 – 2x = – 4

⇒ -2x = – 4

⇒ x = – 4/-2

⇒ x = 2

Therefore, PM = x = 2

MQ = x – 2 = 2 – 2 =0

NR = x – 1 = 2 – 1 = 1

3. Î”ABC is similar to Î”PQR. If AB = 6 cm, BC = 9 cm, PQ = 9 cm and PR = 10.5 cm, find the lengths of AC and QR.

From the question it is given that, Î”ABC ~ Î”PQR

AB = 6 cm, BC = 9 cm, PQ = 9 cm and PR = 10.5 cm

Now, consider Î”ABC and Î”PQR

AB/PQ = BC/QR = AC/PR

6/9 = 9/M = L/10.5

Take first two fraction, 6/9 = 9/q

6M = 81

⇒ M = 81/6

⇒ M = 27/2

Then, QR = 13.5 cm

Now consider second and third fraction, 9/M = L/10.5

9/13.5 = L/10.5

⇒ 9(10.5) = 13.5L

⇒ 94.5 = 13.5L

⇒ L = 94.5/13.5

⇒ L = 7 cm

So, AC = 7 cm

4. ABCD and PQRS are similar figures. AB = 12 cm, BC = x cm, CD = 15 cm, AD = 10 cm, PQ = 8 cm, QR = 5 cm, RS = m cm and PS = n cm. Find the values of x, m and n.

From the question it is given, quadrilateral ABCD ~ quadrilateral PQRS. AB = 12 cm, BC = x cm, CD = 15 cm, AD = 10 cm, PQ = 8 cm, QR = 5 cm, RS = m cm and PS = n cm.

Then, AB/PQ = BC/QR = DC/SR = AD/SR

12/8 = x/5 = 15/m = 10/n

Consider, 12/8 = x/5

By cross multiplication we get,

12 × 5 = 8 × x

⇒ 60 = 8x

⇒ x = 60/8

⇒ x = 7.5 cm

Then, consider x/5 = 15/m

7.5/5 = 15/m

By cross multiplication we get,

7.5 × m = 15 × 5

⇒ 7.5m = 75

⇒ m = 75/7.5

⇒ m = 750/75

⇒ m = 10 cm

Now, consider 15/m = 10/n

15/10 = 10/n

By cross multiplication we get,

15 × n = 10 × 10

⇒ 15n = 100

⇒ n = 100/15

⇒ n = 6.67 cm

Therefore, the value of x is 7.5 cm, value of m is 10 cm and value of n is 6.67 cm.

5. AD and BC are two straight lines intersecting at O. CD and BA are perpendiculars from B and C on AD. If AB = 6 cm, CD = 9 cm, AD = 20 cm and BC = 25 cm, find the lengths of AO, BO, CO and DO.

From the question it is given that, AB = 6 cm, CD = 9 cm, AD = 20 cm and BC = 25 cm

We have to find the lengths of AO, BO, CO and DO.

Consider Î”AOB and Î”COD,

∠OAB = ∠ODC … [because both angles are equal to 90o]

∠AOB = ∠DOC … [because vertically opposite angles are equal]

Therefore, Î”AOB ~ Î”DOC [from AA corollary]

Then, AO/DO = OB/OC = AB/DC

x/(20 – x) = y/(25 – y) = 6/9

Consider, x/(20 – x) = 6/9

x/(20 – x) = 2/3

By cross multiplication we get,

3x = 2(20 – x)

⇒ 3x = 40 – 2x

⇒ 3x + 2x = 40

⇒ 5x = 40

⇒ x = 40/5

⇒ x = 8

Now, consider y/(25 – y) = 6/9

y/(25 – y) = 2/3

By cross multiplication we get,

3y = 2(25 – y)

⇒ 3y = 50 – 2y

⇒ 3y + 2y = 50

⇒ 5y = 50

⇒ y = 50/5

⇒ y = 10

Therefore, AO = x = 8 cm

OD = 20 – x = 20 – 8 = 12 cm

BO = y = 10 cm

OC = 25 – y = 25 – 10 = 15 cm

6. In Î”ABC, DE is drawn parallel to BC. If AD: DB = 2: 3, DE = 6 cm and AE = 3.6 cm find BC and AC.

From the question it is given that, AD: DB = 2: 3, DE = 6 cm and AE = 3.6 cm

Now consider the Î”ABC,

DE ∥ BC … [given]

From Basic Proportionality Theorem (BPT) AD/DB = AE/EC

2/3 = 3.6/m

⇒ 2m = 3 × 3.6

⇒ 2m = 10.8

⇒ m = 10.8/2

⇒ m = 5.4 cm

So, EC = 5.4 cm

Therefore, AC = 3.6 + x

= 3.6 + 5.4

= 9 cm

Now, consider the Î”ADE and Î”ABC,

∠ABC = ∠ADE … [because corresponding angles are equal]

∠ACB = ∠AED … [because corresponding angles are equal]

Therefore, Î”ABC ~ Î”ADE … [from AA corollary]

AE/AC = DE/BC

3.6/9 = 6/n

By cross multiplication we get,

3.6n = 9 × 6

⇒ 3.6n = 54

⇒ n = 54/3.6

⇒ n = 15

So, BC = 15 cm

7. D and E are points on the sides AB and AC respectively of Î”ABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, show that DE is parallel to BC.

From the question it is given that, AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

We have to show that, DE ∥ BC

Consider the Î”ABC,

= 14/42

AE/EC = 1.8/5.4

= 18/54

⇒ AE/EC = 1/3 …(ii)

From (i) and (ii)

Hence, it is proved that DE ∥ BC by converse BPT.

8. In Î”ABC, D and E are points on the sides AB and AC respectively. If AD = 4 cm, DB = 4.5 cm, AE = 6.4 cm and EC = 7.2 cm, find if DE is parallel to BC or not.

From the question it is given that, AD = 4 cm, DB = 4.5 cm, AE = 6.4 cm and EC = 7.2 cm

We have to show that, DE ∥ BC

Consider the Î”ABC,

= 40/45

⇒ AD/DB = 8/9 … (i)

AE/EC = 6.4/7.2

= 64/72

⇒ AE/EC = 8/9 … (ii)

From (i) and (ii)

Hence, it is proved that DE ∥ BC by converse BPT.

9. Î”ABC ~ Î”PQR such that AB = 1.8 cm and PQ = 2.1 cm. Find the ratio of areas of Î”ABC and Î”PQR.

From the question it is given that, Î”ABC ~ Î”PQR and AB = 1.8 cm, PQ = 2.1 cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

So, Area of Î”ABC/Area of Î”PQR = AB2/PQ2

= (1.8)2/(2.1)2

= (18/21)2

= (6/7)2

= 36/49

Therefore, the ratio of areas of Î”ABC and Î”PQR is 36: 49.

10. Î”ABC ~ Î”PQR, AD and PS are altitudes from A and P on sides BC and QR respectively. If AD: PS = 4: 9, find the ratio of the areas of Î”ABC and Î”PQR.

From the question it is given that, Î”ABC ~ Î”PQR and AD: PS = 4: 9

We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

So, Area of Î”ABC/Area of Î”PQR = AB2/PQ2 … [equation (i)]

Now, consider the Î”BAD and Î”QPS

∠ABD = ∠PQS … [because Î”ABC ~ Î”PQR]

∠AOB = ∠PSQ … [because both angles are equal to 90o]

Therefore, Î”BAD ~ Î”QPS … [from AA corollary]

So, AB/PQ = AD/PS … [equation (ii)]

From equation (i) and equation (ii),

Area of Î”ABC/Area of Î”PQR = AD2/PS2

= (4)2/(9)2

= 16/81

Therefore, the ratio of areas of Î”ABC and Î”PQR is 16: 81.

11. Î”ABC ~ Î”DEF. If BC = 3 cm, EF = 4 cm and area of Î”ABC = 54 cm2, find the area of Î”DEF.

From the question it is given that, Î”ABC ~ Î”DEF and BC = 3 cm, EF = 4 cm and area of Î”ABC = 54 cm2

We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

So, Area of Î”ABC/Area of Î”DEF = BC2/EF2

⇒ 54/Area of Î”DEF = (3)2/(4)2

⇒ 54/Area of Î”DEF = 9/16

⇒ Area of Î”DEF = (54 × 16)/9

⇒ Area of Î”DEF = 96 cm2

Therefore, area of Î”DEF is 96 cm2.

12. Î”ABC ~ Î”XYZ. If area of Î”ABC is 9 cm2and area of Î”XYZ is 16 cm2and if BC = 2.1 cm, find the length of YZ.

From the question it is given that, Î”ABC ~ Î”XYZ, area of Î”ABC is 9 cm2 and area of Î”XYZ is 16 cm2 and BC = 2.1 cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

So, Area of Î”ABC/Area of Î”XYZ = BC2/YZ2

9/16 = (2.1)2/(YZ)2

Taking square root on both sides we get,

3/4 = 2.1/YZ

⇒ YZ = (2.1 × 4)/3

⇒ YZ = 2.8

Therefore, YZ is 2.8 cm.

13. Prove that the area of Î”BCE described on one side BC of a square ABCD is one half the area of the similar Î”ACF described on the diagonal AC.

Consider the right angle triangle ABC,

We know that, Pythagoras theorem, AC2 = AB2 + BC2

AC2 = 2BC2 … [because AB = BC]

From the question it is given that, Î”BCE ~ Î”ACF

We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

So, Area of Î”BCE/Area of Î”ACF = BC2/AC2

= ½

Therefore, the ratio of areas of Î”BCE and Î”ACF is 1: 2.

14. In Î”ABC, MN BC

(a) If AN: AC = 5: 8, find the ratio of area of Î”AMN: area of Î”ABC.
(b) If AB/AM = 9/4, find area of trapezium MBCN/area of Î”ABC.
(c) If BC = 14 cm and MN = 6 cm, find area of Î”AMN/area of trapezium MBCN.

(a) From the it is given that, AN: AC = 5: 8

We have to find the ratio of area of Î”AMN: area of Î”AB

So, consider the Î”AMN and Î”ABC

∠AMN = ∠ABC … [because corresponding angles are equal]

∠ANM = ∠ACB … [because corresponding angles are equal]

Therefore, Î”AMN ~ Î”ABC … [from AA corollary]

We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

So, Area of Î”AMN/Area of Î”ABC = AN2/AC2

= (5/8)2

= 25/64

Therefore, the ratio of areas of Î”AMN and Î”ABC is 25: 64.

(b) From the it is given that, AB/AM = 9/4

In the above solution we prove that, Î”AMN ~ Î”ABC

So, Area of Î”AMN/Area of Î”ABC = AM2/AB2

= 42/92

= 16/81

(Area of Î”ABC – Area of trapezium MBCN)/area of Î”ABC = 16/81

By cross multiplication we get,

81(Area of Î”ABC – Area of trapezium MBCN) = area of Î”ABC × 16

⇒ (81 × Area of Î”ABC) – (81 × area of trapezium MBCN) = 16 × area of Î”ABC

⇒ 64 area of Î”ABC = 81 area of trapezium MBCN

Then, area of trapezium MBCN/area of Î”ABC = 65/81

Therefore, the ratio of areas of trapezium MBCN and Î”ABC is 65: 81.

(c) From the it is given that, BC = 14 cm and MN = 6 cm

In the above solution (a) we prove that, Î”AMN ~ Î”ABC

So, Area of Î”AMN/Area of Î”ABC = MN2/BC2

= 62/142

= 36/196

= 9/49

(Area of Î”AMN/Area of Î”AMN + Area of trapezium MBCN) = 9/49

By cross multiplication we get,

49(Area of Î”AMN) = 9 (Area of Î”AMN + Area of trapezium MBCN)

⇒ 49 Area of Î”AMN = 9 Area of Î”AMN + 9 Area of trapezium MBCN

⇒ 49 area of Î”AMN = 9 area of trapezium MBCN

Then, area of Î”ABC/area of trapezium MBCN = 9/40

Therefore, the ratio of area of Î”ABC and area of trapezium is 9: 40.

15. In Î”ABC, DE BC; DC and EB intersect at F, if DE/BC = 2/7, find area of Î”FDE/area of Î”FBC

From the question it is given that, DE/BC = 2/7, DE ∥ BC

Consider the Î”FDE and Î”FCB

∠FDE = ∠FCB … [because interior alternate angles are equal]

∠ANM = ∠ACB … [because interior alternate angles are equal]

Therefore, Î”FDE ~ Î”FCB … [from AA corollary]

We know that, the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

So, Area of Î”FDE/Area of Î”FBC = DE2/BC2

= (2/7)2

= 4/49

Therefore, the ratio of areas of Î”FDE and Î”FBC is 4: 49.

16. Figure shows Î”PQR in which ST || QR and SR and QT intersect each other at M. IF PT/TR = 5/3, find Ar.(Î”MTS)/Ar.(Î”MQR).

17. Figure shows Î”KLM, P and T on KL and KM respectively such that KLM = KTP. If KL/KT = 9/5, find (Ar. Î”KLM)/(Ar. Î”KTP).

18. In figure, DEF is a right-angled triangle with E = 90 ͦ. FE is produced to G and GH is drawn perpendicular to DF. If DE = 8 cm, DH = 8 cm, DH = 6 cm and HF = 4 cm, find Ar.(Î”DEF)/Ar.(Î”GHF).

19. In the figure, ABCD is a quadrilateral. F is a point on AD such that AF = 2. 1 cm and FD = 4.9 cm. E and G are points on AC and AB respectively such that EF || CD and GE || BC. Find Ar.(Î”BCD)/Ar.(Î”GEF).

### Exercise 15.2

1. The dimensions of a building are 50 m long, 40 m wide and 70 m high. A model of the same building is made with a scale factor of 1 : 500. Find the dimensions of the model.

2. A ship is 400 m long and 100 m wide. The length of its model is 20 cm. find the surface area of the deck of the model.

3. A model of a ship is made with a scale factor of 1 : 500. Find

(i) The length of the ship, if the model length is 60 cm.

(ii) The deck area of the model, if the deck area of the ship is 1500000 m2.

(iii) The volume of the ship, if the volume of its model is 200 cm2.

4. An aeroplane is 30 m long and its model is 15 cm long. If the total outer surface area of the model is 150 cm2, find the cost of painting the outer surface of the aerolplane at Rs 120 per m2, if 50 m2 is left out for windows.

5. The length of a river in a map is 54 cm. If 1 cm on the map represents 12500 m on land, find the length of the river.

6. 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 kilometers on the ground represented by 1 cm 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.

7. The actual area of an island is 1872 km2. On a map, this area is 117 cm2. If the length of the coastline is 44 cm on the map, find the length of its actual coastline.

8. On a map drawn to a scale of 1 : 25000, a triangular plot of a land is marked as ABC with AB = 6 cm, BC = 8 cm and ABC = 90 ͦ. Calculate the actual length of AB in km and the actual area of the plot in km2.

9. On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD is measured as AB = 12 cm and BC = 16 cm. Calculate the diagonal distance of the plot in km and the plot area in km2.