# RD Sharma Solutions Class 10 Triangles Exercise 4.2

### RD Sharma Class 10 Solutions Chapter 4 Ex 4.2 PDF Free Download

#### Exercise 4.2

1. In a $\Delta$ ABC, D and E are points on the sides AB and AC respectively such that DE $\parallel$ BC.

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

ii) If $\frac{AD}{DB} = \frac{3}{4}$ and AC = 15 cm, Find AE.

iii) If $\frac{AD}{DB} = \frac{2}{3}$ and AC = 18 cm, Find AE.

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

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

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

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

viii) If $\frac{AD}{BD} = \frac{4}{5}$ and EC = 2.5 cm, Find AE.

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

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

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

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

Solution:

i) Given: $\Delta$ ABC AND DE $\parallel$ BC

To find AC,

Given that,

AD = 6 cm, DB = 9 cm and AE = 8 cm.

Hence, Using Thales Theorem

$\frac{AD}{BD} = \frac{AE}{CE}$

Let CE=x. then,

$\frac{6}{9} = \frac{8}{x}$

6x = 72 cm

x = 72/6 cm

x = 12 cm

∴, AC = 12 + 8 = 20.

ii) Given: $\frac{AD}{BD} = \frac{3}{4}$ and AC = 15 cm

To find: AE,

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

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

⇒$\frac{3}{4} = \frac{x}{15 – x}$

45 – 3x = 4x

-3x – 4x = – 45

7x = 45

x = 45/7

x = 6.43 cm

∴, AE= 6.43cm

iii) Given: $\frac{AD}{BD} = \frac{2}{3}$ and AC = 18 cm

To find: AE,

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

Let, AE = x and CE = 18 – x

⇒$\frac{2}{3} = \frac{x}{18 – x}$

3x = 36 – 2x

5x = 36 cm

x = 36/5 cm

x = 7.2 cm

∴, AE = 7.2 cm

iv) Given: AD = 4 cm, AE = 8 cm, DB = x – 4 and EC = 3x – 19

To find x,

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

Then, $\frac{4}{x – 4} = \frac{8}{3x – 19}$

4(3x – 19) = 8(x – 4)

12x – 76 = 8(x – 4)

12x – 8x = – 32 + 76

4x = 44 cm

x = 11 cm

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

To find CE,

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

$\frac{8}{4} = \frac{12}{CE}$

8CE = 4 x 12 cm

CE = (4 x 12)/8 cm

CE = 48/8 cm

CE = 6 cm

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

To find = AC

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

$\frac{4}{4.5} = \frac{8}{AC}$

$AC = \frac{4.5 \times 8 }{4}$ cm

AC = 9 cm

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

To find= AE

DB = 6 – 2 = 4 cm

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

$\frac{2}{4} = \frac{x}{9 – x}$

4x = 18 – 2x

6x = 18

x = 3 cm

∴, AE= 3cm

viii) Given: $\frac{AD}{BD} = \frac{4}{5}$ and EC = 2.5 cm

To find= AE

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

Then, $\frac{4}{5} = \frac{AE}{2.5}$

AE = $\frac{4 \times 2.5}{5}$ = 2 cm

ix) It is given that AD = x, DB = x – 2, AE = x + 2 and EC = x – 1

We have to find the value of x

So, $\frac{AD}{BD} = \frac{AE}{CE}$ (using Thales Theorem)

Then, $\frac{x}{x – 2} = \frac{x + 2}{x – 1}$

X(x – 1) = (x – 2)(x + 2)

x2 – x – x2 + 4 = 0

x =4

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

To find= x

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

$\frac{8x – 7}{5x – 3} = \frac{4x – 3}{3x – 1}$

(8x – 7)(3x – 1) = (5x – 3)(4x – 3)

24x2 – 29x + 7 = 20x2 – 27x + 9

4x2 – 2x – 2 = 0

2(2x2 – x – 1) = 0

2x2 – x – 1 = 0

2x2 – 2x + x – 1 = 0

2x(x – 1) + 1(x – 1) = 0

(x – 1)(2x + 1) = 0

x = 1 or x = -1/2

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

∴, x = 1.

xi)  Given: AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3

To find= x

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

Then, $\frac{4x – 3}{3x – 1} = \frac{8x – 7}{5x – 3}$

(4x – 3)(5x – 3) = (3x – 1)(8x – 7)

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

20x2 – 12x – 15x + 9 = 24x2 – 29x + 7

20x2 -27x + 9 = 242 -29x + 7

Then,

-4x2 + 2x + 2 = 0

4x2 – 2x – 2 = 0

4x2 – 4x + 2x – 2 = 0

4x(x – 1) + 2(x – 1) = 0

(4x + 2)(x – 1) = 0

X = 1 or x = -2/4

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

∴, x = 1

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

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{CE}$

$\frac{2.5}{3} = \frac{3.75}{CE}$

2.5CE = 3.75 x 3

$CE = \frac{3.75\times 3}{2.5}$

$CE = \frac{11.25}{2.5}$

CE = 4.5

Now, AC = 3.75 + 4.5

AC = 8.25 cm.

2. In a $\Delta$ ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE $\parallel$ BC.

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

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

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

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

Solution:

i) Given: D and R are the points on sides AB and AC.

To find= DE $\parallel$ BC.

Hence, Using Thales Theorem,

$\frac{AD}{DB} = \frac{AE}{CE}$

$\frac{8}{4} = \frac{12}{6}$

2 = 2  (LHS = RHS)

∴, DE $\parallel$ BC.

ii)  Given: D and E are the points on sides AB and AC

To prove= DE $\parallel$ BC

Hence, Using Thales Theorem,

$\frac{AD}{DB} = \frac{AE}{CE}$

$\frac{1.4}{4.2} = \frac{1.8}{5.4}$

$\frac{1}{3} = \frac{1}{3}$  (RHS)

∴, DE $\parallel$ BC.

iii)  Given: D and E are the points on sides AB and AC.

To prove= DE $\parallel$ BC.

Hence, Using Thales Theorem,

$\frac{AD}{DB} = \frac{AE}{CE}$

AD = AB – DB = 10.8 – 4.5 = 6.3

And,

EC = AC – AE = 4.8  – 2.8 = 2

Now,

$\frac{6.3}{4.5} = \frac{2.8}{2.0}$

∴, DE $\parallel$ BC.

iv) Given: D and E are the points on sides AB and Ac.

To prove: DE $\parallel$ BC.

Hence, Using Thales Theorem,

$\frac{AD}{DB} = \frac{AE}{CE}$

$\frac{5.7}{9.5} = \frac{3.3}{5.5}$

$\frac{3}{5} = \frac{3}{5}$   (LHS = RHS)

Hence, DE $\parallel$ BC.

3. In a $\Delta$ ABC, P and Q are the points on sides AB and AC respectively, such that PQ $\parallel$ BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm, Find AB and PQ.

Solution:

Given: AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm.

To find= AB and PQ.

Hence, Using Thales Theorem,

$\frac{AP}{PB} = \frac{AQ}{QC}$

$\frac{2.4}{PB} = \frac{2}{3}$

2PB = 2.4 x 3 cm

$PB = \frac{2.4 \times 3}{2}$ cm

PB = 3.6 cm

Now, AB = AP + PB

AB = 2.4 + 3.6

AB = 6 cm

As PQ $\parallel$ BC, AB is transversal,

$\Delta$ APQ  = $\Delta$ ABC    (by corresponding angles)

As PQ $\parallel$ BC, AC is transversal,

$\Delta$ APQ  = $\Delta$ ABC    (by corresponding angles)

In $\Delta$ ABQ and $\Delta$ ABC,

$\angle APQ = \angle ABC$

$\angle AQP = \angle ACB$

Hence, $\Delta$ APQ = $\Delta$ ABC (angle angle similarity)

We know that the corresponding sides of similar triangles are proportional,

⇒$\frac{AP}{AB} = \frac{PQ}{BC} = \frac{AQ}{AC}$

$\frac{AP}{AB} = \frac{PQ}{BC}$

$\frac{2.4}{6} = \frac{PQ}{6}$

∴, PQ = 2.4 cm.

4.  In a $\Delta$ ABC, D and E are points on AB and AC respectively, such that DE $\parallel$ BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm, and BC = 5 cm. Find BD and CE.

Solution:

Given: AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm.

To find= BD and CE.

As DE $\parallel$ BC, AB is transversal,

$\angle APQ = \angle ABC$

As DE $\parallel$ BC, AC is transversal,

$\angle AED = \angle ACB$

In $\Delta$ ADE and $\Delta$ ABC,

$\angle ADE = \angle ABC$

$\angle AED = \angle ACB$

∴, $\Delta$ ADE = $\Delta$ ABC (angle angle similarity)

We know that the corresponding sides of similar triangles are proportional,

⇒  $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$

$\frac{AD}{AB} = \frac{DE}{BC}$

$\frac{2.4}{2.4 + DB} = \frac{2}{5}$

2.4 + DB = 6

DB = 6 – 2.4

DB = 3.6 cm

Same way,

$\frac{AE}{AC} = \frac{DE}{BC}$

$\frac{3.2}{3.2 + EC} = \frac{2}{5}$

3.2 + EC = 8

EC = 8 – 3.2

EC = 4.8 cm

∴, BD = 3.6 cm and CE = 4.8 cm.

5.  In figure 4.35 given below, state PQ $\parallel$ EF.

Fig 4.35

Solution:

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

To check: PQ $\parallel$ EF or not.

Hence, Using Thales Theorem,

$\frac{PG}{GE} = \frac{GQ}{FQ}$

$\frac{3.9}{3} \neq \frac{3.6}{2.4}$

Hence, it is not proportional.

∴, PQ is not parallel to EF.

6. M and N are the points on the sides PQ and PR respectively, of a $\Delta$ PQR. For each of the following cases, state whether MN $\parallel$ 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.

Solution:

(i) Given: PM = 4 cm, QM = 4.5 cm, PN = 4 cm, and NR = 4.5 cm.

To check= MN $\parallel$ QR or not.

Hence, Using Thales Theorem,

$\frac{PM}{QM} = \frac{PN}{NR}$

$\frac{4}{4.5} = \frac{4}{4.5}$

∴, MN $\parallel$ QR.

(ii) Given: PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, and PN = 0.32 cm.

To check= MN $\parallel$ QR or not.

Hence, Using Thales Theorem,

$\frac{PM}{QM} = \frac{PN}{NR}$

$\frac{PM}{MQ} = \frac{0.16}{1.12}$ = 1/7

$\frac{PN}{NR} = \frac{0.32}{2.24}$ = 1/7

But, since,

$\frac{0.16}{1.12} = \frac{0.32}{2.24}$

∴, MN $\parallel$ QR.

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

Solution:

In $\Delta$ OAB, Given that, LM $\parallel$ AB,

Then, using BPT,

$\frac{OL}{LA} = \frac{OM}{MB}$

Now,

In $\Delta$ OBC, Given that, MN $\parallel$ BC,

Then, using BPT,

$\frac{OM}{MB} = \frac{ON}{NC}$

Hence,  $\frac{ON}{NC} = \frac{OM}{MB}$

We can infer from the above equations that,

$\frac{OL}{LA} = \frac{ON}{NC}$

Now, In $\Delta$ OCA,

$\frac{OL}{LA} = \frac{ON}{NC}$

By converse BPT

LN $\parallel$ AC

8. If D and E are the points on sides AB and AC respectively of a $\Delta$ ABC such that DE $\parallel$ BC and BD = CE. Prove that $\Delta$ ABC is isosceles.

Solution:

Given: $\Delta$ ABC, DE $\parallel$ BC and BD = CE.

To prove= $\Delta$ ABC is isosceles.

Hence, Using Thales Theorem,

$\frac{AD}{BD} = \frac{AE}{EC}$

∴, $\Delta$ ABC is isosceles.