**Frank Solutions for Class 10 Maths Chapter 15 Similarity** are available here. At BYJUâ€™S, we have a team of expert faculty trying their best to provide exercise-wise solutions to students according to their level of understanding. Frank Solutions for Class 10 Maths Chapter 15 can be used as a mode of reference by the students to improve their conceptual knowledge and understand the different ways used to solve problems.

Chapter 15 – Similarity are figures having the same shape with corresponding sides proportional and corresponding angles equal. Students can learn different types of problems provided in this chapter on similarity by referring to Frank Solutions for Class 10 Maths. Click on the link given below to download the free PDF.

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

**Solution:-**

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 â€“ 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.**

**Solution:-**

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)

x^{2} â€“ 2x = x^{2} â€“ 4

x^{2} â€“ x^{2} â€“ 2x = – 4

-2x = – 4

x = – 4/-2

x = 2

Therefore, PM = x = 2

MQ = x â€“ 2 = 2 â€“ 2 =0

NR = x â€“ 1 = 2 â€“ 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.**

**Solution:-**

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.**

**Solution:-**

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 Ã— 5 = 8 Ã— 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 Ã— m = 15 Ã— 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 Ã— n = 10 Ã— 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.**

**Solution:-**

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,

âˆ OAB = âˆ ODC â€¦ [because both angles are equal to 90^{o}]

âˆ AOB = âˆ DOC â€¦ [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 â€“ x) = 6/9

x/(20 – x) = 2/3

By cross multiplication, we get,

3x = 2(20 – x)

3x = 40 â€“ 2x

3x + 2x = 40

5x = 40

x = 40/5

x = 8

Now, consider y/(25 â€“ y) = 6/9

y/(25 – y) = 2/3

By cross multiplication, we get,

3y = 2(25 – y)

3y = 50 â€“ 2y

3y + 2y = 50

5y = 50

y = 50/5

y = 10

Therefore, AO = x = 8 cm

OD = 20 â€“ x = 20 â€“ 8 = 12 cm

BO = y = 10 cm

OC = 25 â€“ y = 25 â€“ 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.**

**Solution:-**

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 âˆ¥ BC â€¦ [given]

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

2/3 = 3.6/m

2m = 3 Ã— 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,

âˆ ABC = âˆ ADE â€¦ [because corresponding angles are equal]

âˆ ACB = âˆ AED â€¦ [because corresponding angles are equal]

Therefore, Î”ABC ~ Î”ADE â€¦ [from AA corollary]

AE/AC = DE/BC

3.6/9 = 6/n

By cross multiplication, we get,

3.6n = 9 Ã— 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.**

**Solution:-**

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 âˆ¥ BC

Consider the Î”ABC,

AD/DB = 1.4/4.2

= 14/42

AD/DB = 1/3 â€¦ (i)

AE/EC = 1.8/5.4

= 18/54

AE/EC = 1/3 â€¦ (ii)

From (i) and (ii)

AD/DB = AE/EC

Hence, it is proved that DE âˆ¥ 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.**

**Solution:-**

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 âˆ¥ BC

Consider the Î”ABC,

AD/DB = 4/4.5

= 40/45

AD/DB = 8/9 â€¦ (i)

AE/EC = 6.4/7.2

= 64/72

AE/EC = 8/9 â€¦ (ii)

From (i) and (ii)

AD/DB = AE/EC

Hence, it is proved that DE âˆ¥ 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.**

**Solution:-**

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 = 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.**

**Solution:-**

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 = AB^{2}/PQ^{2} â€¦ [equation (i)]

Now, consider the Î”BAD and Î”QPS

âˆ ABD = âˆ PQS â€¦ [because Î”ABC ~ Î”PQR]

âˆ AOB = âˆ PSQ â€¦ [because both angles are equal to 90^{o}]

Therefore, Î”BAD ~ Î”QPS â€¦ [from AA corollary]

So, AB/PQ = AD/PS â€¦ [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.**

**Solution:-**

From the question it is given that, Î”ABC ~ Î”DEF and BC = 3 cm, EF = 4 cm and area of Î”ABC = 54 cm^{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 Ã— 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.**

**Solution:-**

From the question it is given that, Î”ABC ~ Î”XYZ, area of Î”ABC is 9 cm^{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 Ã— 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.**

**Solution:-**

Consider the right angle triangle ABC,

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

AC^{2} = 2BC^{2} â€¦ [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}

= Â½

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

**14. In Î”ABC, MN âˆ¥ BC**

**(a) If AN: AC = 5: 8, find the ratio of area of Î”AMN: area of Î”ABC**

**Solution:-**

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

âˆ AMN = âˆ ABC â€¦ [because corresponding angles are equal]

âˆ ANM = âˆ ACB â€¦ [because corresponding angles are equal]

Therefore, Î”AMN ~ Î”ABC â€¦ [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) If AB/AM = 9/4, find area of trapezium MBCN/area of Î”ABC.**

**Solution:-**

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 â€“ Area of trapezium MBCN)/area of Î”ABC = 16/81

By cross multiplication we get,

81(Area of Î”ABC â€“ Area of trapezium MBCN) = area of Î”ABC Ã— 16

(81 Ã— Area of Î”ABC) â€“ (81 Ã— area of trapezium MBCN) = 16 Ã— 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) If BC = 14 cm and MN = 6 cm, find area of Î”AMN/area of trapezium MBCN**

**Solution:-**

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 âˆ¥ BC; DC and EB intersect at F, if DE/BC = 2/7, find area of Î”FDE/area of Î”FBC**

**Solution:-**

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

Consider the Î”FDE and Î”FCB

âˆ FDE = âˆ FCB â€¦ [because interior alternate angles are equal]

âˆ ANM = âˆ ACB â€¦ [because interior alternate angles are equal]

Therefore, Î”FDE ~ Î”FCB â€¦ [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.

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