Similar triangles and their properties are the prime focus in this exercise. Basic results on proportionality and theorems are examined in the exercise questions. The RD Sharma Solutions Class 10 can be used by the students to clarify their doubts and as a preparation tool for exams. The RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.2 PDF provided below can be also be utilised by the students.

## RD Sharma Solutions for Class 10 Chapter 4 Triangles Exercise 4.2 Download PDF

### Access RD Sharma Solutions for Class 10 Chapter 4 Triangles Exercise 4.2

**1. In aÂ Î”Â ABC, D and E are points on the sides AB and AC respectively such that DE ||Â BC.**

**i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, Find AC. **

**Solution: **

Given:Â Î”Â ABC, DEÂ âˆ¥Â BC, AD = 6 cm, DB = 9 cm and AE = 8 cm.

Required to find AC.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

Let CE = x.

So then,

6/9 = 8/x

6x = 72 cm

x = 72/6 cm

x = 12 cm

âˆ´ AC = AE + CE = 12 + 8 = 20.

**ii) IfÂ AD/DB = 3/4Â and AC = 15 cm, Find AE.**

**Solution: **

Given:Â AD/BD = 3/4Â and AC = 15 cm [As DEÂ âˆ¥Â BC]

Required to find AE.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

Let, AE = x, then CE = 15-x.

â‡’ 3/4 = x/ (15â€“x)

45 â€“ 3x = 4x

-3x â€“ 4x = â€“ 45

7x = 45

x = 45/7

x = 6.43 cm

âˆ´ AE= 6.43cm

**iii) IfÂ AD/DB = 2/3Â and AC = 18 cm, Find AE.**

**Solution: **

Given:Â AD/BD = 2/3Â and AC = 18 cm

Required to find AE.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

Let, AE = x and CE = 18 â€“ x

â‡’ 23 = x/ (18â€“x)

3x = 36 â€“ 2x

5x = 36 cm

x = 36/5 cm

x = 7.2 cm

âˆ´ AE = 7.2 cm

**iv) If AD = 4 cm, AE = 8 cm, DB = x â€“ 4 cm and EC = 3x â€“ 19, find x.**

**Solution: **

Given: AD = 4 cm, AE = 8 cm, DB = x â€“ 4 and EC = 3x â€“ 19

Required to find x.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

Then,Â 4/ (x â€“ 4) = 8/ (3x â€“ 19)

4(3x â€“ 19) = 8(x â€“ 4)

12x â€“ 76 = 8(x â€“ 4)

12x â€“ 8x = â€“ 32 + 76

4x = 44 cm

x = 11 cm

**v) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.**

**Solution: **

Given: AD = 8 cm, AB = 12 cm, and AE = 12 cm.

Required to find CE,

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

8/4 = 12/CE

8 x CE = 4 x 12 cm

CE = (4 x 12)/8 cm

CE = 48/8 cm

âˆ´ CE = 6 cm

**vi) If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.**

**Solution: **

Given: AD = 4 cm, DB = 4.5 cm, AE = 8 cm

Required to find AC.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

4/4.5 = 8/AC

AC = (4.5 Ã— 8)/4Â cm

âˆ´AC = 9 cm

**vii) If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.**

**Solution: **

Given: AD = 2 cm, AB = 6 cm and AC = 9 cm

Required to find AE.

DB = AB â€“ AD = 6 â€“ 2 = 4 cm

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

2/4 = x/ (9â€“x)

4x = 18 â€“ 2x

6x = 18

x = 3 cm

âˆ´ AE= 3cm

**viii) IfÂ AD/BD = 4/5Â and EC = 2.5 cm, Find AE.**

**Solution: **

Given:Â AD/BD = 4/5Â and EC = 2.5 cm

Required to find AE.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

Then,Â 4/5 = AE/2.5

âˆ´ AE =Â 4 Ã— 2.55Â = 2 cm

**ix) If AD = x cm, DB = x â€“ 2 cm, AE = x + 2 cm, and EC = x â€“ 1 cm, find the value of x.**

**Solution: **

Given: AD = x, DB = x â€“ 2, AE = x + 2 and EC = x â€“ 1

Required to find the value of x.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

So, x/ (xâ€“2) = (x+2)/ (xâ€“1)

x(x â€“ 1) = (x â€“ 2)(x + 2)

x^{2}Â â€“ x â€“ x^{2}Â + 4 = 0

x = 4

**x) If AD = 8x â€“ 7 cm, DB = 5x â€“ 3 cm, AE = 4x â€“ 3 cm, and EC = (3x â€“ 1) cm, Find the value of x.**

**Solution: **

Given: AD = 8x â€“ 7, DB = 5x â€“ 3, AER = 4x â€“ 3 and EC = 3x -1

Required to find x.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

(8xâ€“7)/ (5xâ€“3) = (4xâ€“3)/ (3xâ€“1)

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

24x^{2}Â â€“ 29x + 7 = 20x^{2Â }â€“ 27x + 9

4x^{2}Â â€“ 2x â€“ 2 = 0

2(2x^{2}Â â€“ x â€“ 1) = 0

2x^{2}Â â€“ x â€“ 1 = 0

2x^{2}Â â€“ 2x + x â€“ 1 = 0

2x(x â€“ 1) + 1(x â€“ 1) = 0

(x â€“ 1)(2x + 1) = 0

â‡’ x = 1 or x = -1/2

We know that the side of triangle can never be negative. Therefore, we take the positive value.

âˆ´ x = 1.

**xi) If AD = 4x â€“ 3, AE = 8x â€“ 7, BD = 3x â€“ 1, and CE = 5x â€“ 3, find the value of x.**

**Solution: **

Given: AD = 4x â€“ 3, BD = 3x â€“ 1, AE = 8x â€“ 7 and EC = 5x â€“ 3

Required to find x.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

So,Â (4xâ€“3)/ (3xâ€“1) = (8xâ€“7)/ (5xâ€“3)

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

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

20x^{2}Â â€“ 12x â€“ 15x + 9 = 24x^{2}Â â€“ 29x + 7

20x^{2}Â -27x + 9 = 24^{2}Â -29x + 7

â‡’ -4x^{2}Â + 2x + 2 = 0

4x^{2}Â â€“ 2x â€“ 2 = 0

4x^{2}Â â€“ 4x + 2x â€“ 2 = 0

4x(x â€“ 1) + 2(x â€“ 1) = 0

(4x + 2)(x â€“ 1) = 0

â‡’ x = 1 or x = -2/4

We know that the side of triangle can never be negative. Therefore, we take the positive value.

âˆ´ x = 1

**xii) If AD = 2.5 cm, BD = 3.0 cm, and AE = 3.75 cm, find the length of AC.**

**Solution: **

Given: AD = 2.5 cm, AE = 3.75 cm and BD = 3 cm

Required to find AC.

By using Thales Theorem, [As DEÂ âˆ¥Â BC]

AD/BD = AE/CE

2.5/ 3 = 3.75/ CE

2.5 x CE = 3.75 x 3

CE = 3.75Ã—32.5

CE = 11.252.5

CE = 4.5

Now, AC = 3.75 + 4.5

âˆ´ AC = 8.25 cm.

**2. In aÂ **Î”**Â ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DEÂ **âˆ¥**Â BC:**

**i) AB = 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm.**

**Solution: **

Required to prove DEÂ âˆ¥Â BC.

We have,

AB = 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm**. **(Given)

So,

BD = AB â€“ AD = 12 â€“ 8 = 4 cm

And,

CE = AC â€“ AE = 18 â€“ 12 = 6 cm

Itâ€™s seen that,

AD/BD = 8/4 = 1/2

AE/CE = 12/6 = 1/2

Thus,

AD/BD = AE/CE

So, by the converse of Thaleâ€™s Theorem

We have,

DEÂ âˆ¥Â BC.

Hence Proved.

**ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm.**

**Solution: **

Required to prove DEÂ âˆ¥Â BC.

We have,

AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm**. **(Given)

So,

BD = AB â€“ AD = 5.6 â€“ 1.4 = 4.2 cm

And,

CE = AC â€“ AE = 7.2 â€“ 1.8 = 5.4 cm

Itâ€™s seen that,

AD/BD = 1.4/4.2 = 1/3

AE/CE = 1.8/5.4 =1/3

Thus,

AD/BD = AE/CE

So, by the converse of Thaleâ€™s Theorem

We have,

DEÂ âˆ¥Â BC.

Hence Proved.

**iii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.**

**Solution: **

Required to prove DEÂ âˆ¥Â BC.

We have

AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.

So,

AD = AB â€“ DB = 10.8 â€“ 4.5 = 6.3

And,

CE = AC â€“ AE = 4.8Â â€“ 2.8 = 2

Itâ€™s seen that,

AD/BD = 6.3/ 4.5 = 2.8/ 2.0 = AE/CE = 7/5

So, by the converse of Thaleâ€™s Theorem

We have,

DEÂ âˆ¥Â BC.

Hence Proved.

**iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, and EC = 5.5 cm.**

**Solution: **

Required to prove DEÂ âˆ¥Â BC.

We have

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

Now,

AD/BD = 5.7/9.5 =3/5

And,

AE/CE = 3.3/5.5 = 3/5

Thus,

AD/BD = AE/CE

So, by the converse of Thaleâ€™s Theorem

We have,

DEÂ âˆ¥Â BC.

Hence Proved.

**3. In aÂ **Î”**Â ABC, P and Q are the points on sides AB and AC respectively, such that PQÂ **âˆ¥**Â BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm. Find AB and PQ.**

**Solution: **

Given: Î”Â ABC, AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm. Also, PQÂ âˆ¥Â BC.

Required to find: AB and PQ.

By using Thales Theorem, we have [As itâ€™s given that PQÂ âˆ¥Â BC]

AP/PB = AQ/ QC

2.4/PB = 2/3

2 x PB = 2.4 x 3

PB = (2.4 Ã— 3)/2Â cm

â‡’ PB = 3.6 cm

Now finding, AB = AP + PB

AB = 2.4 + 3.6

â‡’ AB = 6 cm

Now, considering Î”Â APQ and Î”Â ABC

We have,

âˆ A = âˆ A

âˆ APQ = âˆ ABC (Corresponding angles are equal, PQ||BC and AB being a transversal)

Thus, Î”Â APQ and Î”Â ABC are similar to each other by AA criteria.

Now, we know that

Corresponding parts of similar triangles are propositional.

â‡’ AP/AB = PQ/ BC

â‡’ PQ = (AP/AB) x BC

= (2.4/6) x 6 = 2.4

âˆ´ PQ = 2.4 cm.

**4. In aÂ **Î”**Â ABC, D and E are points on AB and AC respectively, such that DEÂ **âˆ¥**Â BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm. Find BD and CE.**

**Solution: **

Given: Î”Â ABC such that AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm. Also DEÂ âˆ¥Â BC.

Required to find: BD and CE.

As DEÂ âˆ¥Â BC, AB is transversal,

âˆ APQ = âˆ ABC (corresponding angles)

As DEÂ âˆ¥Â BC, AC is transversal,

âˆ AED = âˆ ACB (corresponding angles)

InÂ Î”Â ADE andÂ Î”Â ABC,

âˆ ADE=âˆ ABC

âˆ AED=âˆ ACB

âˆ´Â Î”Â ADE =Â Î”Â ABC (AA similarity criteria)

Now, we know that

Corresponding parts of similar triangles are propositional.

â‡’ AD/AB = AE/AC = DE/BC

AD/AB = DE/BC

2.4/ (2.4 + DB) = 2/5 [Since, AB = AD + DB]

2.4 + DB = 6

DB = 6 â€“ 2.4

DB = 3.6 cm

In the same way,

â‡’ AE/AC = DE/BC

3.2/ (3.2 + EC) = 2/5 [Since AC = AE + EC]

3.2 + EC = 8

EC = 8 â€“ 3.2

EC = 4.8 cm

âˆ´ BD = 3.6 cm and CE = 4.8 cm.