RD Sharma Solutions for Class 10 Chapter 4 Triangles Exercise 4.3

RD Sharma Solutions Chapter 4 Triangles Exercise 4.3 Class 10 Maths

Chapter Name	RD Sharma Chapter 4 Triangles
Book Name	RD Sharma Mathematics for Class 10
Other Exercises	Exercise 4.1 Exercise 4.2 Exercise 4.4 Exercise 4.5 Exercise 4.6 Exercise 4.7
Related Study	NCERT Solutions for Class 10 Maths

Exercise 4.3 Solutions

1. In a ΔABC, AD is the bisector of ∠A, meeting side BC at D. 
(i) If BD = 2.5cm, AB = 5cm  and AC = 4.2cm, find DC. 
(ii) If BD = 2cm, AB = 5cm and DC = 3cm, find AC. 
(iii) If AB = 3.5 cm , AC = 4.2cm  and DC = 2.8 cm, find BD. 
(iv) If AB = 10 cm, AC = 14cm and BC = 6cm, find BD and DC. 
(v)  If  AC = 4.2cm, DC = 6 cm and 10cm, find AB 
(vi) If AB = 5.6 cm, AC = 6cm and DC = 3 cm, find BC. 
(vii) If AD = 5.6 cm, BC = 6cm and BD = 3.2cm, find AC. 
(viii) If AB = 10 cm, Ac = 6 cm and BC = 12 cm, find BD and DC. 

Solution

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

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

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

= 7/3 = 2.33 cm 
∴ BD = 2.3 cm

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

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

We have, 
BC = 10cm , DC = 6 cm and AC = 4.2cm 
∴ BD = BC - DC = 10 - 6 = 4 cm 
⇒ BD = 4cm 
In ΔABC, AD is the bisector of ∠A. 
We know that, the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. 

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

(vii) If AD = 5.6 cm, BC = 6cm and BD = 3.2cm, find AC. 

In ∆ABC, AD is the bisector of ∠A. 
We know that, the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the containing the angle . 

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

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

Solution

In ΔABC, AD is the bisector of ∠A. 
We know that, the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. 

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

Solution

We have, if a line through one vertex of a triangle divides the opposite side in the ratio of the other two sides, then the line bisects the angle at the vertex. 

4. In ΔABC (fig., 4.59), if ∠1 = ∠2,  prove that AB/AC = BD/DC . 

Solution

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

Solution

6. In fig., 4.60, check whether AD is the bisector of ∠A of ΔABC in each of the following : 

(i) AB = 5cm, AC = 10cm, BD = 1.5cm and CD = 3.5cm 
(ii)  AB = 4cm, AC = 6 cm, BD = 1.6cm and CD = 2.4cm 
(iii) AB = 8 cm, AC = 24 cm, BD = 6cm and BC = 24 cm 
(iv) AB = 6cm, AC = 8cm , BD = 1.5cm and CD = 2cm. 
(v) AB = 5cm, AC = 12cm, BD = 2.5cm and BC = 9 cm

Solution