## Chapter 4 triangles R.D. Sharma Solutions for Class 10th Math Exercise 4.3

**Exercise 4.3**

1. In a ∆ABC, AD is the bisector of ∠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 = l0 cm, AC =14 cm and BC =6 cm, find BD and DC.

(v) If AC = 4.2 cm, DC = 6 cm and 10 cm, find AB

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

(vii) If AD = 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)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

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

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

**Solution**

3. In Fig. 4.58, △ABC is a triangle such that AB/AC = BD/DC, ∠B = 70°, ∠C = 50° . Find ∠BAD.

**Solution**

4. In fig. 4.59, check whether AD is the bisector of ∠A of ∆ABC in each of the following:

(i) AB = 5cm, AC = 10cm, 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 = 24cm

(iv) AB = 6 cm, AC = 8cm, BD = l.5 cm and CD = 2 cm

(v) AB = 5 cm, AC = l2 cm, BD = 2.5 cm and BC = 9cm

**Solution**

5. In Fig. 4.60, AD bisects ∠A, AB = 12 cm, AC = 20 cm and BD = 5 cm, determine CD.

**Solution**

It is given that AD bisects ∠A . Also, AB = 12 cm, AC = 20 cm and BD = 5 cm.

We have to find CD.

Since AD is the bisector of ∠A

Then AD/AC = BD/DC

12cm/20cm = 5cm/DC

12cm × Dc = 20cm × 5 cm

DC = 100/12 cm

We have to find CD.

Since AD is the bisector of ∠A

Then AD/AC = BD/DC

12cm/20cm = 5cm/DC

12cm × Dc = 20cm × 5 cm

DC = 100/12 cm

= 8.33 cm

Hence, CD = 8.33 cm

Hence, CD = 8.33 cm

6. In △ABC (Fig., 4.59), if ∠1 = ∠2, prove that AB/AC = BD/DC.

**Solution**

7. D, E and F are the points on sides BC, CA and AB respectively of ∆ABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C. If AB = 5 cm, BC = 8 cm and CA = 4 cm.

**Solution**