# 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.**

**Answer**

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.**

**Answer**

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)

⇒ x^{2} – 2x = x^{2} – 4

⇒ x^{2} – x^{2} – 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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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 90 ^{o}]**

∠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.**

**Answer**

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.**

**Answer**

We have to show that, DE ∥ BC

Consider the Î”ABC,

AD/DB = 1.4/4.2

= 14/42

⇒ AD/DB = 1/3 **…(i)**

AE/EC = 1.8/5.4

= 18/54

⇒ AE/EC = 1/3 **…(ii)**

From (i) and (ii)

AD/DB = AE/EC

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.**

**Answer**

We have to show that, DE ∥ BC

Consider the Î”ABC,

AD/DB = 4/4.5

= 40/45

⇒ AD/DB = 8/9 **… (i)**

AE/EC = 6.4/7.2

= 64/72

⇒ AE/EC = 8/9 **… (ii)**

From (i) and (ii)

AD/DB = AE/EC

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.**

**Answer**

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 = AB^{2}/PQ^{2}

= (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.**

**Answer**

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 = AB^{2}/PQ^{2} **… [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 90 ^{o}]**

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 = AD^{2}/PS^{2}

= (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 cm ^{2}, find the area of Î”DEF.**

**Answer**

^{2}

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 = BC^{2}/EF^{2}

⇒ 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 cm^{2}

Therefore, area of Î”DEF is 96 cm^{2}.

**12. ****Î”ABC ~ Î”XYZ. If area of Î”ABC is 9 cm ^{2}and area of Î”XYZ is 16 cm^{2}and if BC = 2.1 cm, find the length of YZ.**

**Answer**

^{2}and area of Î”XYZ is 16 cm

^{2}and BC = 2.1 cm

So, Area of Î”ABC/Area of Î”XYZ = BC^{2}/YZ^{2}

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.**

**Answer**

We know that, Pythagoras theorem, AC^{2} = AB^{2} + BC^{2}

AC^{2} = 2BC^{2} **… [because AB = BC]**

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

So, Area of Î”BCE/Area of Î”ACF = BC^{2}/AC^{2}

= 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.**

**Answer**

**(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]**

So, Area of Î”AMN/Area of Î”ABC = AN^{2}/AC^{2}

= (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 = AM^{2}/AB^{2}

= 4^{2}/9^{2}

= 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 = MN^{2}/BC^{2}

= 6^{2}/14^{2}

= 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**

**Answer**

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]**

So, Area of Î”FDE/Area of Î”FBC = DE^{2}/BC^{2}

= (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).**

**Answer:**

**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).**

**Answer:**

**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). **

**Answer:\**

**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). **

**Answer**

**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. **

**Answer**

**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. **

**Answer**

**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 m ^{2}. **

**(iii) The volume of the ship, if the volume of its model is 200 cm ^{2}. **

**Answer**

**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 cm ^{2}, find the cost of painting the outer surface of the aerolplane at Rs 120 per m^{2}, if 50 m^{2} is left out for windows. **

**Answer**

**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. **

**Answer**

**6. The scale of a map is 1 : 200000. A plot of land of area 20 km ^{2} 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 km ^{2} that can be represented by 1 cm^{2}. **

**(iii) The area on the map that represents the plot of land. **

**Answer**

**7. The actual area of an island is 1872 km ^{2}. On a map, this area is 117 cm^{2}. If the length of the coastline is 44 cm on the map, find the length of its actual coastline. **

**Answer**

**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 km ^{2}.**

**Answer**

**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 ****km ^{2}**

**.**

**Answer**