# Class 10 RD Sharma Solutions – Chapter 4 Triangles – Exercise 4.2

**Question 1: In a **Δ**ABC, D and E are points on the sides AB and AC respectively such that DE || BC.**

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

**Solution: **

Given:Δ ABC whereLength of side AD = 6 cm, DB = 9 cm, AE = 8 cm

Also, DE ∥ BC,

To find : Length of side ACBy using Thales Theorem we get, (As DE ∥ BC)

AD/BD = AE/CE – equation 1

Let CE = x.

So then putting values in equation 1, we get

6/9 = 8/x

6x = 72 cm

x = 72/6 cm

x = 12 cm

∴ AC = AE + CE = 12 + 8 = 20.

Therefore, Length of side AC is 20 cm.

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

**Solution:**

Given:Length of side AD/BD = 3/4 and AC = 15 cm

To find: Length of side AEBy using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Let, AE = x

Then CE = 15 – x.

Now, putting values in equation 1,

⇒ 3/4 = x/ (15 – x)

45 – 3x = 4x

-3x – 4x = – 45

7x = 45

x = 45/7

x = 6.43 cm

∴ AE= 6.43cm

Therefore, Length of side AE is 6.43 cm

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

**Solution:**

Given:Length of side AD/BD = 2/3 and AC = 18 cm

To find: Length of side AE.By using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Let, AE = x

Then CE = 18 – x

Now, putting values in equation 1,

⇒ 23 = x/ (18 – x)

3x = 36 – 2x

5x = 36 cm

x = 36/5 cm

x = 7.2 cm

∴ AE = 7.2 cm

Therefore, Length of side AE is 7.2 cm

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

**Solution:**

Given:Length of side AD = 4 cm, AE = 8 cm, DB = x – 4 and EC = 3x – 19

To Find: length of side x.By using Thales Theorem, we get,

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

4/ (x – 4) = 8/ (3x – 19)

4(3x – 19) = 8(x – 4)

12x – 76 = 8(x – 4)

12x – 8x = – 32 + 76

4x = 44 cm

x = 11 cm

Therefore, Length of side x is 11 cm

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

**Solution:**

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

To find: Length of side CE.

By using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

8/4 = 12/CE

8 x CE = 4 × 12 cm

CE = (4 × 12)/8 cm

CE = 48/8 cm

∴ CE = 6 cm

Therefore, Length of side x is 6 cm

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

**Solution:**

Given:Length of side AD = 4 cm, DB = 4.5 cm, AE = 8 cm

To find: Length of side AC.

By using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

4/4.5 = 8/AC

AC = (4.5 × 8)/4 cm

∴AC = 9 cm

Therefore, Length of side AC is 9 cm

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

**Solution:**

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

To find: Length of side AC.Length of DB = AB – AD = 6 – 2 = 4 cm

By using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

2/4 = x/(9 – x)

4x = 18 – 2x

6x = 18

x = 3 cm

∴ AE = 3cm

Therefore, Length of side AE is 3 cm

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

**Solution:**

Given:Length of side AD/BD = 4/5 and EC = 2.5 cm

To find: Length of side AC.By using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

Then, 4/5 = AE/2.5

∴ AE = 4 × 2.55 = 2 cm

Therefore, Length of side AE is 2 cm

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

**Solution:**

Given:Length of side AD = x, DB = x – 2, AE = x + 2 and EC = x – 1

To find: Value of x.By using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

x/(x – 2) = (x + 2)/(x – 1)

x(x – 1) = (x – 2)(x + 2)

x

^{2}– x – x^{2}+ 4 = 0∴ x = 4

Therefore, the value of x is 4

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

**Solution:**

Given:Length of side AD = 8x – 7, DB = 5x – 3, AER = 4x – 3 and EC = 3x -1

To find : Value of xBy using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

(8x – 7)/(5x – 3) = (4x–3)/ (3x–1)

(8x – 7)(3x – 1) = (5x – 3)(4x – 3)

24x

^{2}– 29x + 7 = 20x^{2}– 27x + 94x

^{2}– 2x – 2 = 02(2x

^{2}– x – 1) = 02x

^{2}– x – 1 = 02x

^{2}– 2x + x – 1 = 02x(x – 1) + 1(x – 1) = 0

(x – 1)(2x + 1) = 0

⇒ x = 1 or x = -1/2

Since, we know that the side of triangle is always positive.

Therefore, we take the positive value.

∴ x = 1.

Therefore, the value of x is 1.

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

**Solution:**

Given:Length of side AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3

To find : Value of xBy using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

We get,

(4x – 3)/(3x – 1) = (8x – 7)/(5x – 3)

(4x – 3)(5x – 3) = (3x – 1)(8x – 7)

4x(5x – 3) -3(5x – 3) = 3x(8x – 7) -1(8x – 7)

20x

^{2}– 12x – 15x + 9 = 24x^{2}– 29x + 720x

^{2}-27x + 9 = 24x^{2}– 29x + 7⇒ -4x

^{2}+ 2x + 2 = 04x

^{2}– 2x – 2 = 04x

^{2}– 4x + 2x – 2 = 04x(x – 1) + 2(x – 1) = 0

(4x + 2)(x – 1) = 0

⇒ x = 1 or x = -2/4

We know that the side of triangle is always positive

Therefore, we only take the positive value.

∴ x = 1

Therefore, the value of x is 1.

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

**Solution:**

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

To find : Length of side ACBy using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

We get,

2.5/ 3 = 3.75/ CE

2.5 x CE = 3.75 x 3

CE = 3.75×32.5

CE = 11.252.5

CE = 4.5

Now, AC = 3.75 + 4.5

∴ AC = 8.25 cm.

Therefore, the value of AC is 8.25cm

**Problem 2. In a** Δ**ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE ∥ BC:**

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

**Solution:**

Given:Length of side AB = 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm

To show : Side DE is parallel to BC ( DE || BC )Now,

Length of side BD = AB – AD

⇒ 12 – 8 = 4 cm

And,

Length of side CE = AC – AE = 18 – 12 = 6 cm

Now,

AD/BD = 8/4 = 1/2 – equation 1

AE/CE = 12/6 = 1/2 – equation 2

Now from equation 1 and 2

AD/BD = AE/CE

So, by the converse of Thale’s Theorem

We get, DE ∥ BC.Hence, Proved.

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

**Solution:**

Given:Length of side AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm

To show : Side DE is parallel to BC ( DE || BC )Now,

Length of side BD = AB – AD = 5.6 – 1.4 = 4.2 cm

And,

Length of side CE = AC – AE = 7.2 – 1.8 = 5.4 cm

Now,

AD/BD = 1.4/4.2 = 1/3 – equation 1

AE/CE = 1.8/5.4 =1/3 – equation 2

Now from equation 1 and 2

AD/BD = AE/CE

So, by the converse of Thale’s Theorem

We get, DE ∥ BC.Hence, Proved.

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

**Solution:**

Given:Length of side AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.

To show: Side DE is parallel to BC ( DE || BC )Now,

Length of side AD = AB – DB = 10.8 – 4.5 = 6.3

And,

Length of side CE = AC – AE = 4.8 – 2.8 = 2

Now,

AD/BD = 6.3/ 4.5 = 2.8/ 2.0 = 7/5 – equation 1

AE/CE = 7/5 – equation 2

Now from equation 1 and 2

AD/BD = AE/CE

So, by the converse of Thale’s Theorem

We get, DE ∥ BC.Hence Proved

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

**Solution:**

Given:Length of side AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, and EC = 5.5 cm

To show: Side DE is parallel to BC (DE || BC)Now,

Length of side AD/BD = 5.7/9.5 = 3/5 – equation 1

And,

Length of side AE/CE = 3.3/5.5 = 3/5 – equation 2

Now from equation 1 and 2

AD/BD = AE/CE

So, by the converse of Thale’s Theorem

We get, DE ∥ BC.Hence, Proved

**Problem 3: In a **Δ**ABC, P and Q are the points on sides AB and AC respectively, such that PQ ∥ BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm. Find AB and PQ.**

**Solution:**

Given:In ΔABC,

Length of side AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm.

Also, PQ ∥ BC.

To find: Length of side AB and PQNow,

Since it’s given that PQ ∥ BC

So by using Thales Theorem, we get

AP/PB = AQ/ QC

2.4/PB = 2/3

2 x PB = 2.4 × 3

PB = (2.4 × 3)/2 cm

⇒ PB = 3.6 cm

Therefore, Length of PB is 3.6 cm

Now finding, AB = AP + PB

AB = 2.4 + 3.6

⇒ AB = 6 cm

Therefore, Length of AB is 6 cmNow, considering ΔAPQ and ΔABC

We have,

∠A = ∠A (Common angle)

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

Thus, ΔAPQ and ΔABC are similar to each other by AA criteria.

Now, we know that corresponding parts of similar triangles are propositional.

Therefore,

⇒ AP/AB = PQ/ BC

⇒ PQ = (AP/AB) x BC

= (2.4/6) x 6 = 2.4

∴ PQ = 2.4 cm.

Therefore, Length of PQ is 2.4 cm and AB is 6cm

**Problem 4: In a **Δ**ABC, D and E are points on AB and AC respectively, such that DE ∥ BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm. Find BD and CE.**

**Solution:**

Given:In Δ ABC,

Length of side AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm

Also, DE ∥ BC

To find: Length of side BD and CE.As DE ∥ BC, AB is transversal,

∠APQ = ∠ABC (corresponding angles) – equation 1

As DE ∥ BC, AC is transversal,

∠AED = ∠ACB (corresponding angles) – equation 2

In Δ ADE and Δ ABC,

Now from equation 1 and 2 we get,

∠ADE = ∠ABC

∠AED = ∠ACB

∴ ΔADE = ΔABC (By AA similarity criteria)

Now, we know that

Corresponding parts of similar triangles are proportional.

Therefore,

⇒ AD/AB = AE/AC = DE/BC

AD/AB = DE/BC

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

2.4 + DB = 6

DB = 6 – 2.4

DB = 3.6 cmLength of side DB is 3.6 cm

In the same way, we get

⇒ AE/AC = DE/BC

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

3.2 + EC = 8

EC = 8 – 3.2

EC = 4.8 cm∴ BD = 3.6 cm and CE = 4.8 cm.

Length of side BD is 3.6 cm and CE is 4.8 cm

**Problem 5: For **the **figure given, state PQ ∥ EF.**

**Solution:**

Given :Length of side EP = 3 cm, PG = 3.9 cm, FQ = 3.6 cm and QG = 2.4 cm

To find: PQ ∥ EF or not.According to Thales Theorem,

PG/GE = GQ/FQ

Now,

3.9/3 ≠ 3.6/ 2.4

As we can see it is not proportional.

So, PQ is not parallel to EF.

**Problem 6: M and N are the points on the sides PQ and PR respectively, of a **Δ**PQR. For each of the following cases, state whether MN ∥ QR.**

**(i) PM = 4 cm, QM = 4.5 cm, PN = 4 cm, NR = 4.5 cm.**

**(ii) PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, PN = 0.32 cm.**

(i)

Given:In the ∆PQR

M and N are points on PQ and PR respectively

PM = 4 cm, QM = 4.5 cm, PN = 4 cm, RN = 4.5 cm

To find: MN || QR or notAccording to Thales Theorem,

PM/QM = PN/NR

4/4.5 = 4/4.5

Hence, MN || QR

(ii)

Given :Length of side PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, and PN = 0.32 cm.

To find: that MN ∥ QR or not.MQ = PQ – PM = 1.28 – 0.16 = 1.12 cm

RN = PR – PN = 2.56 – 0.32 =2.24 cm

According to Thales Theorem,

PM/QM = PN/ RN

PM/QN = 0.16/1.12 = 1/7 – equation 1

PN/RN = 0.32/ 2.24 = 1/7 – equation 2

From equation 1 and 2

PM/QM = PN/ RN

Therefore, MN || QR

**Problem 7: In three-line segments OA, OB, and OC, points L, M, N respectively are so chosen that LM ∥ AB and MN ∥ BC but neither of L, M, and N nor A, B, C are collinear. Show that LN ∥ AC. **

**Solution: **

Given :OA, OB and OC, points are L, M, and N respectively

Such that LM || AB and MN || BC

To prove: LN ∥ ACNow,

In ΔOAB, Since, LM ∥ AB,

Then, OL/LA = OM/ MB (By using BPT) – equation 1

In ΔOBC, Since, MN ∥ BC

Then, OM/MB = ON/NC (By using BPT)

Therefore, ON/NC = OM/ MB – equation 2

From equation 1 and 2, we get

OL/LA = ON/NC

Therefore, In ΔOCA By converse BPT, we get

LN || AC

Hence proved

**Problem 8: If D and E are the points on sides AB and AC respectively of a **Δ**ABC such that DE ∥ BC and BD = CE. Prove that **ΔABC** is isosceles.**

**Solution: **

Given :In Δ ABC, DE ∥ BC and BD = CE.

To prove: that Δ ABC is isosceles.According to Thales Theorem,

AD/DB = AE/EC

But BD = CE (Given)

Therefore, we get,AD = AE

Now, BD = CE and AD = AE.

So, AD + BD = AE + CE.

Therefore, AB = AC.

Therefore, ΔABC is isosceles.Hence proved