# RD Sharma Solutions Class 10 Triangles Exercise 4.3

### RD Sharma Class 10 Solutions Chapter 4 Ex 4.3 PDF Free Download

1. In a $\Delta$ ABC, AD is the bisector of $\angle$ A , meeting side BC at D.

(i) if BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm, find DC.

(ii) if BD = 2 cm, AB = 5 cm, and DC = 3 cm, find AC.

(iii) if AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm, find BD.

(iv) if AB = 10 cm, AC = 14 cm, and BC = 6 cm, find BD and DC.

(v) if AC = 4.2 cm, DC = 6 cm, and BC = 10 cm, find AB.

(vi) if AB = 5.6 cm, BC = 6 cm, and DC = 3 cm, find BC.

(vii) if AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm, find AC.

(viii) if AB = 10 cm, AC = 6 cm, and BC = 12 cm, find BD and DC.

Solution:

(i) if BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm, find DC.

(ii) if BD = 2 cm, AB = 5 cm, and DC = 3 cm, find AC.

(iii) if AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm, find BD.

(iv) if AB = 10 cm, AC = 14 cm, and BC = 6 cm, find BD and DC.

(v) if AC = 4.2 cm, DC = 6 cm, and BC = 10 cm, find AB.

(vi) if AB = 5.6 cm, BC = 6 cm, and DC = 3 cm, find BC.

DC = 2.8 cm

And, BC = 2.8 + 3

BC = 5.8 cm

(vii) if AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm, find AC.

(viii) if AB = 10 cm, AC = 6 cm, and BC = 12 cm, find BD and DC.

x = 7.5

Now,

DC = 12 – BD

DC = 12-x = 12 – 7.5

DC = 4.5

2. In figure 4.57, AE is the bisector of the exterior $\angle CAD$ meeting BC produced in E. If AB = 10 cm, AC = 6 cm, and BC = 12 cm, find CE.

Solution:

Given: AE is the bisector of the exterior <CAD

AB = 10 cm, AC = 6 cm, and BC = 12 cm.

Since AE is the bisector of the exterior <CAD.

6x+72 = 10x

10x-6x = 72

4x = 72

x = 18

CE = 18 cm

3. In fig. 4.58, $\Delta ABC$ is a triangle such that $\frac{AB}{AC} = \frac{BD}{DC}$, $\angle B = 70$, $\angle C = 50$, find $\angle BAD$.

Solution:

Given: $\frac{AB}{AC} = \frac{BD}{DC}$, $\angle B = 70$ and $\angle C = 50$

To find= $\angle BAD$

From $\Delta ABC$,

$\angle A$ = 180 – (70 + 50)

= 180 – 120

= 60

Therefore, AD is the bisector of $\angle A$ since, $\frac{AB}{AC} = \frac{BD}{DC}$

4. In Fig. 4.59, check whether AD is the bisector of $\angle A$ of $\Delta ABC$ in each of the following:

(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm

(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm

(iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm

(iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm

(v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm

Solution:

(i)

Given: AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm

To check: If AD is bisector of $\angle A$

$\frac{AB}{AC} = \frac{5}{10} = \frac{1}{2}$

$\frac{BD}{CD} = \frac{1.5}{3.5} = \frac{3}{7}$

Since, $\frac{AB}{AC} \neq \frac{BD}{CD}$

Therefore, AD is not the bisector of $\angle A$.

(ii) Given: AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm.

To check: If AD is the bisector of $\angle A$

So, $\frac{AB}{AC} = \frac{BD}{DC}$

$\frac{4}{6} = \frac{1.6}{2.4}$

$\frac{2}{3} = \frac{2}{3}$ (proportional)

Therefore, AD is the bisector of $\angle A. (iii) Given: AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm. To check: if AD is the bisector of \(\angle A$

DC = BC – BD

DC = 24 – 6

DC = 18

So, $\frac{AB}{AC} = \frac{BD}{DC}$

$\frac{8}{24} = \frac{6}{18}$

$\frac{1}{3} = \frac{1}{3}$ (proportional)

Therefore, AD is the bisector of $\angle A.$

(iv) Given: AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm.

To check: If AD is the bisector of $\angle A$

$\frac{AB}{AC} = \frac{BD}{DC}$

$\frac{6}{8} = \frac{1.5}{2}$

$\frac{3}{4} = \frac{3}{4}$ (proportional)

Therefore, AD is the bisector of $\angle A$.

(v) Given: AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm.

To check: If AD is the bisector of $\angle A$

$\frac{AB}{AC} = \frac{5}{12}$

$\frac{BD}{CD} = \frac{2.5}{9} = \frac{5}{18}$

Since, $\frac{AB}{AC} \neq \frac{BD}{CD}$

Therefore, AD is not the bisector of $\angle A$.

5) In fig. 4.60, AD bisects $\angle A$, AB = 12 cm, AC = 20 cm, and BD = 5 cm, determine CD.

Solution:

Given: AD bisects $\angle A$

AB = 12 cm, AC = 20 cm, and BD = 5 cm.

To find =CD.

As AD is the bisector of $\angle A$

$\frac{AB}{AC} = \frac{BD}{DC}$

$\frac{12}{20} = \frac{5}{DC}$

12 x DC = 20 x 5

DC = 100/12

DC = 8.33 cm

∴ CD = 8.33 cm.

6. In $\Delta ABC$ (Fig. 4.60), if $\angle1 = \angle2$, prove that, $\frac{AB}{AC} = \frac{BD}{DC}$

Solution:

To prove: $\frac{AB}{AC} = \frac{BD}{DC}$

In $\Delta ABC$,

$\angle1 = \angle2$

So, AD is the angle bisector of $\angle A$

Hence,

$\frac{AB}{AC} = \frac{BD}{DC}$

7. D and E are the points on sides BC, CA and AB respectively. of a $\Delta ABC$ such that AD bisects $\angle A$, BE bisects $\angle B$ and CF bisects $\angle C$. If AB = 5 cm, BC = 8 cm, and CA = 4 cm, determine AF, CE, and BD.

Solution:

Given: AB = 5 cm, BC = 8 cm and CA = 4 cm.

To find= AF, CE and BD.

As AD is the bisector of $\angle A$

$\frac{AB}{AC} = \frac{BD}{CD}$

Then,

$\frac{5}{4} = \frac{BD}{BC – BD}$

$\frac{5}{4} = \frac{BD}{8 – BD}$

40 – 5BD = 4 BD

9BD = 40

So, BD = 40/9

As, BE is the bisector of $\angle B$

$\frac{AB}{BC} = \frac{AE}{EC}$

$\frac{AB}{BC} = \frac{AC – EC}{EC}$

$\frac{5}{8} = \frac{4 – CE}{CE}$

5CE = 32 – 8CE

5CE + 8CE = 32

13CE = 32

Hence, CE = $\frac{32}{13}$

Now, As CF is the bisector of $\angle C$

$\frac{BC}{CA} = \frac{BF}{AF}$

$\frac{8}{4} = \frac{AB – AF}{AF}$

$\frac{8}{4} = \frac{5 – AF}{AF}$

8AF = 20 – 4AF

12AF = 20

So, 3AF = 5

Therefore, AF = 5/3 cm, CE = 32/12 cm and BD = 40/9 cm