# Frank Solutions for Class 9 Maths Chapter 16 Similarity

Frank Solutions for Class 9 Maths Chapter 16 Similarity provide students with accurate answers to build a strong foundation for the topic. Students are suggested to refer to reliable resources in order to understand the concepts thoroughly. The main purpose of preparing solutions is to make learning fun and interesting among students. Regular practice of these solutions enhances confidence among students for examinations. For more conceptual knowledge, students can make use of the Frank Solutions for Class 9 Maths Chapter 16 Similarity PDF from the below-provided link.

Chapter 16 contains answers with simple step-by-step explanations, making solutions easy to follow. The solutions for each and every question are 100% accurate and are explained elaborately, clearing the confusion among students while solving textbook problems.

## Access Frank Solutions for Class 9 Maths Chapter 16 Similarity

1. In â–³ABC, D and E are the mid-points on AB and AC such that DE || BC

(i) If AD = 4, AE = 8, DB = x â€“ 4 and EC = 3x â€“ 19, find x.

(ii) If AD: BD = 4: 5 and EC = 2.5 cm, find AE.

(iii) If AD = 4x â€“ 3, AE = 8x â€“ 7, BD = 3x â€“ 1 and CE = 5x â€“ 3, find x.

(iv) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.

Solution:

âˆ D = âˆ B and âˆ C = âˆ E (DE || BC)

Hence,

AD / DB = AE / EC

âŸ¹ {4 / (x â€“ 4)} = {8 / (3x â€“ 19)}

âŸ¹ 4 (3x â€“ 19) = 8 (x â€“ 4)

âŸ¹ 12 x â€“ 76 = 8x â€“ 32

âŸ¹ 4x = 44

We get,

x = 11

âˆ D = âˆ B and âˆ C = âˆ E (DE || BC)

Hence,

AD / DB = AE / EC

âŸ¹ 4/5 = AE / 2.5

AE = (4 x 2.5) / 5

We get,

AE = 2 cm

âˆ D = âˆ B and âˆ C = âˆ E (DE || BC)

Hence,

AD / DB = AE / EC

âŸ¹ (4x â€“ 3) / (3x â€“ 1) = (8x â€“ 7) / (5x â€“ 3)

âŸ¹ (4x â€“ 3) (5x -3) = (8x â€“ 7) (3x â€“ 1)

âŸ¹ 20x2 â€“ 15x â€“ 12x + 9 = 24x2 â€“ 21x â€“ 8x + 7

âŸ¹ 4x2 â€“ 2x â€“ 2 = 0

By splitting middle term, we get

4x2 – 4x + 2x â€“ 2 = 0

4x (x -1) + 2 (x-1) = 0

x = -2/4 or x = 1

We know that, side of triangle can never be negative

Hence, x = 1

âˆ D = âˆ B and âˆ C = âˆ E (DE || BC)

Hence,

AD / DB = AE / EC

DB = 12 â€“ 8

DB = 4

âŸ¹ 8/4 = 12/ EC

âŸ¹ 8 x EC = 12 x 4

âŸ¹ EC = (12 x 4) / 8

We get,

âŸ¹ EC = 6 cm

2. In â–³ABC, D and E are points on AB and AC. Show that DE || BC for each of the following case or not:

(i) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

(ii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm

(iii) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm

Solution:

(i) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

AD / AB = (1.4) / (5.6)

AD / AB = (7 / 28)

AD / AB = (1 / 4)

AE / AC = (1.8) / (7.2)

AE / AC = (2 / 8)

AE / AC = (1/ 4)

AD / AB = AE / AC

Hence,

âŸ¹ âˆ D = âˆ B, âˆ E = âˆ C

But these are corresponding angles

Therefore,

DE || BC

(ii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm

AD / AB = (6.3) / (10.8)

AD / AB = (7/ 12)

AE / AC = (2.8) / (4.8)

AE / AC = (14 / 24)

AE / AC = (7/ 12)

âŸ¹ AD / AB = AE / AC

Hence,

âŸ¹ âˆ D = âˆ B, âˆ E = âˆ C

But these are corresponding angles

Therefore,

DE || BC

(iii) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm

AD / BD = (5.7) / (9.5)

AE / EC = (3.3) / (5.5)

AE / EC = (3/5)

AE / EC = 0.6

âŸ¹ AD / BD = AE / EC

Hence,

âŸ¹ âˆ D = âˆ B, âˆ E = âˆ C

But these are corresponding angles

Therefore,

DE || BC

3. In the figure, PQ is parallel to BC, AP: AB = 2: 7. If QC = 10 and BC = 21,

Find:

(i) AQ

(ii) PQ

Solution:

(i) Since PQ || BC

AP / PB = AQ / QC

âŸ¹ {AP / (AB â€“ AP)} = (AQ / QC)

âŸ¹ (2/5) = (AQ / 10)

âŸ¹ AQ = (2 x 10) / 5

We get,

AQ = 4

(ii) Since PQ || BC

(AP / AB) = (PQ / BC)

âŸ¹ (2/7) = (PQ / 21)

âŸ¹ PQ = (2 x 21) / 7

We get,

PQ = 6

4. In â–³ABC, DE is parallel to BC and DE: BC = 3: 8

Find:

(ii) AE, if AC = 16

Solution:

(i) Since DE || BC

(DE / BC) = (AD / AB)

âŸ¹ (3/8) = (AD / AB)

âŸ¹ (AD / AB) = (3/8)

Since DB = AB â€“ AD

âŸ¹ DB = 8 â€“ 3

âŸ¹ DB = 5

Hence,

(ii) DE: BC = 3: 8

Since DE || BC

(DE / BC) = (AE / AC)

âŸ¹ (3/8) = (AE / 16)

âŸ¹ AE = (3 x 16) / 8

We get,

AE = 6

5. In â–³ABC, point D divides AB in the ratio 5: 7. Find:

(i) AE / EC

(iii) AE / AC

(iv) BC, if DE = 2.5 cm

(v) DE, if BC = 4.8 cm

Solution:

Considering DE || BC

(i) AD / DB = AE / EC

âŸ¹ AE / EC = AD / DB

âŸ¹ AE / EC = (5/7)

(ii) AD / DB = (5/7)

Since, AB = AD + DB

âŸ¹ AB = 5 + 7

âŸ¹ AB = 12

Therefore,

AD / AB = (5 / 12)

(iii) AD / DB = AE /EC

âŸ¹ AE / EC = AD / DB

âŸ¹ AE / EC = (5/ 7)

Since, AC = AE + EC

âŸ¹ AC = 5 + 7

âŸ¹ AC = 12

Therefore,

AE / AC = (5/ 12)

(iv) Since, DE || BC

AD / AB = DE/ BC

âŸ¹ (5/ 12) = (2.5) / BC

âŸ¹ BC = (2.5 x 12) / 5

We get,

BC = 6 cm

(v) Since, DE || BC

AD / AB = DE / BC

âŸ¹ (5/ 12) = DE / (4.8)

âŸ¹ DE = (5 x 4.8) / 12

We get,

DE = 2 cm

6. In â–³PQR, AB is drawn parallel to QR. If PQ = 9 cm, PR = 6 cm and PB = 4.2 cm, find the length of AP.

Solution:

In â–³PQR

AB || QR

AP / PQ = PB / PR

âŸ¹ (AP / 9) = (4.2/ 6)

âŸ¹ AP = {(4.2 x 9)/ 6}

We get,

AP = 6.3 cm

7. In â–³ABC, MN is drawn parallel to BC. If AB = 3.5 cm, AM: AB = 5: 7 and NC = 2 cm, find:

(i) AM

(ii) AC

Solution:

(i) AM /AB = 5/ 7

AB = 3.5 cm (given)

Hence,

AM = {(5 x AB) / 7}

âŸ¹ AM = {(5 x 3.5) / 7}

On further calculation, we get,

AM = 2.5 cm

(ii) Since in â–³ABC, MN || BC and

AM / MB = AN / NC

Since AB = 3.5 cm and AM = 2.5 cm

Hence,

MB = AB â€“ AM

MB = 3.5 â€“ 2.5

MB = 1 cm

âŸ¹ AM/MB = AN/ NC

âŸ¹ (2.5 / 1) = (AN / 2)

âŸ¹ AN = (2.5 x 2) / 1

We get,

AN = 5 cm

We know,

AC = AN + NC

âŸ¹ AC = 5 + 2

We get,

AC = 7 cm

8. The sides PQ and PR of the â–³PQR are produced to S and T respectively. ST is drawn parallel to QR and PQ: PS = 3:4. If PT = 9.6 cm, find PR. If â€˜pâ€™ be the length of the perpendicular from P to QR, find the length of the perpendicular from P to ST in terms of â€˜pâ€™.

Solution:

Since QR is parallel to ST

By Basic Theorem of Proportionality

PQ / PS = PR / PT

âŸ¹ (3/4) = (PR / 9.6)

âŸ¹ PR = {(9.6 x 3) / 4}

On simplification, we get,

PR = 7.2 cm

Since QR is parallel to ST,

QM || SD

By Basic Theorem of Proportionality,

PQ / PS = PM / PD

âŸ¹ (3/4) = (p/ PD)

âŸ¹ PD = 4p/3

Therefore, the length of the perpendicular from P to ST in terms of p is 4p/3

9. In â–³ABC, DE || BC such that AD = 1.5 cm, DB = 3 cm and AE = 1 cm. Find AC

Solution:

Given in â–³ABC,

DE || BC

AD / AB = AE / AC

âŸ¹ (1.5/ 4.5) = (1/ AC)

On further calculation, we get,

AC = 3 cm

10. Given is a triangle with sides 3 cm, 5 cm and 6 cm. Find the sides of a triangle which is similar to the given triangle and its shortest side is 4.5 cm

Solution:

Since the two triangles are similar,

Hence, the ratio of the corresponding sides are equal

Consider x and y be the sides of the triangle where y is the longest side

(3/5) = (4.5/x)

On simplification, we get,

âŸ¹ x = 7.5 cm

(5/6) = (7.5/ y)

On further calculation, we get

âŸ¹ y = 9 cm

Hence, the sides of the triangles are 4.5 cm, 7.5 cm and 9 cm

11. Two figures are similar. If the ratio of their perimeters is 8: 16. What will be the ratio of the corresponding sides?

Solution:

We know that,

For two similar triangles, ratio of the corresponding sides is equal to the ratio of the perimeters of the triangles

Hence,

Ratio of the corresponding sides = (8/16) = (1/2)

That is, ratio of the corresponding sides is 1:2

12. Harmeet is 6 feet tall and casts a shadow of 3 feet long. What is the height of a nearby pole if it casts a shadow of 12 feet long at the same time?

Solution:

Harmeet and the pole will be perpendicular to the ground

So,

PQ || ST

In â–³PQR and â–³STR,

âˆ PQR = âˆ STR â€¦â€¦ (Both are right angles)

âˆ PRQ = âˆ SRTâ€¦â€¦.. (common angle)

â–³PQR âˆ¼ â–³STR â€¦â€¦. (AA criterion for similarity)

PQ / ST = QR / TR

âŸ¹ (h/6) = (12/ 3)

On simplification, we get,

h = 24 feet

Therefore, the height of the pole is 24 feet

13. The areas of two similar triangles are 16 cm2 and 9 cm2 respectively. If the altitude of the smaller triangle is 1.8 cm, find the length of the altitude corresponding to the larger triangle.

Solution:

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding altitudes.

Therefore,

{Area (â–³ABC) / Area (â–³DEF)} = (AL2 / DM2)

âŸ¹ (16 / 9) = (AL2 / 1.82)

âŸ¹ AL2 = {(16 x 3.24) / 9}

On further calculation, we get,

AL2 = 5.76

Hence,

AL = 2.4 cm

14. The areas of two similar triangles are 169 cm2 and 121 cm2 respectively. If one side of the larger triangle is 26 cm, find the length of the corresponding side of the smaller triangle.

Solution:

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides

Therefore,

{Area (â–³ABC) / Area (â–³DEF)} = (AB2 / DE2)

(169 / 121) = (262 / DE2)

DE2 = {(121x 262) / 169}

DE2 = {(121 x 676) / 169}

On simplification, we get,

DE2 = (81796 / 169)

DE2 = 484

Hence,

DE = 22 cm

15. In â–³ABC, DE is drawn parallel to BC cutting AB in the ratio 2: 3. Calculate:

(i) area (â–³ADE) / area (â–³ABC)

(ii) area (trapezium EDBC) / area (â–³ABC)

Solution:

Given

We know,

AB = 2 + 3

AB = 5