RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3: Students will learn about the internal and external bisectors of a triangle angle from this exercise. The chapter-by-chapter RD Sharma Class 10 Solutions created by Kopykitab experts is a very beneficial material for students to refer to and confidently prepare for their exams. Students can also use the links below to download RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.3 PDF.

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RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3

Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.3- Important Question with Answers

Question 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. (C.B.S.E. 1996)
(ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC. (C.B.S.E. 1992)
(iii) If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD. (C.B.S.E. 1992)
(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. (C.B.S.E. 1997C)
(vi) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, find BC. (C.B.S.E. 2001C)
(vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC. (C.B.S.E. 2001C)
(viii) If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC. (C.B.S.E. 2001)
Solution:
In ∆ABC, AD is the angle bisector of ∠A which meet BC at D
(i) BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm








⇒ 6x = 10 (12 – x) = 120 – 10x
⇒ 6x + 10x = 120
⇒ 16x = 120
x = 7.5
BD = 7.5 cm and DC = 12 – 7.5 = 4.5 cm

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

Solution:
In ∆ABC, AE is the bisector of exterior ∠A which meets BC produced at E.
AB = 10 cm, AC = 6 cm, BC = 12 cm Let CE = x, then BE = BC + CE = (12 + x)

Question 3.
In the figure, ∆ABC is a triangle such that ABAC = BDDC , ∠B = 70°, ∠C = 50°. Find ∠BAD.

Solution:

Question 4.
In the figure, check whether AD is the bisector of ∠A of ∆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) AB = 5 cm, AC = 10 cm, BD = 1.5 cm, CD = 3.5 cm

Question 5.
In figure, AD bisects ∠A, AB = 12 cm AC = 20 cm, and BD = 5 cm. Determine CD.

Solution:

Question 6.
In the figure, In ∆ABC, if ∠1 = ∠2, prove that ABAC = BDDC.

Solution:
Given: In ∆ABC,
AD is a line drawn from A meeting BC in D Such that ∠1 = ∠2

Question 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, determine AF, CE, and BD.
Solution:
In ∆ABC, AD, BE and CF are the bisector of ∠A, ∠B and ∠C respectively
AB = 5 cm, BC = 8 cm and CA = 4 cm

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