

RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.3: RD Sharma Solutions for Class 10 Maths Chapter 4 Exercise 4.3 provides detailed explanations and step-by-step solutions to help students understand the concepts in the chapter on Triangles.
Solving these questions helps students strengthen their understanding of triangle properties, such as the basic proportionality theorem, similarity of triangles and more. By practicing with these solutions, students gain confidence in applying theorems and solving problems independently, which is valuable for performing well in exams.RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.3 PDF
1. In a Δ 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.
Solution:
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. And BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm. Required to find: DC
Since AD is the bisector of ∠ A meeting side BC at D in Δ ABC
⇒ AB/ AC = BD/ DC
5/ 4.2 = 2.5/ DC
5DC = 2.5 x 4.2
∴ DC = 2.1 cm
(ii) if BD = 2 cm, AB = 5 cm, and DC = 3 cm, find AC.
Solution:
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. And BD = 2 cm, AB = 5 cm, and DC = 3 cm. Required to find: AC
Since AD is the bisector of ∠ A meeting side BC at D in Δ ABC
⇒ AB/ AC = BD/ DC
5/ AC = 2/ 3
2AC = 5 x 3
∴ AC = 7.5 cm
(iii) if AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm, find BD.
Solution:
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. And AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm. Required to find: BD
Since AD is the bisector of ∠ A meeting side BC at D in Δ ABC
⇒ AB/ AC = BD/ DC
3.5/ 4.2 = BD/ 2.8
4.2 x BD = 3.5 x 2.8
BD = 7/3
∴ BD = 2.3 cm
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(iv) if AB = 10 cm, AC = 14 cm, and BC = 6 cm, find BD and DC.
Solution:
Given: In Δ ABC, AD is the bisector of ∠A meeting side BC at D. And, AB = 10 cm, AC = 14 cm, and BC = 6 cm Required to find: BD and DC.
Since AD is the bisector of ∠A
We have,
AB/AC = BD/DC (AD is bisector of ∠ A and side BC)
Then, 10/ 14 = x/ (6 – x)
14x = 60 – 6x
20x = 60
x = 60/20
∴ BD = 3 cm and DC = (6 – 3) = 3 cm.
(v) if AC = 4.2 cm, DC = 6 cm, and BC = 10 cm, find AB.
Solution:
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. And AC = 4.2 cm, DC = 6 cm, and BC = 10 cm. Required to find: AB
Since AD is the bisector of ∠ A meeting side BC at D in Δ ABC
⇒ AB/ AC = BD/ DC
AB/ 4.2 = BD/ 6
We know that,
BD = BC – DC = 10 – 6 = 4 cm
⇒ AB/ 4.2 = 4/ 6
AB = (2 x 4.2)/ 3
∴ AB = 2.8 cm
(vi) if AB = 5.6 cm, AC = 6 cm, and DC = 3 cm, find BC.
Solution:
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. And AB = 5.6 cm, AC = 6 cm, and DC = 3 cm. Required to find: BC
Since AD is the bisector of ∠ A meeting side BC at D in Δ ABC
⇒ AB/ AC = BD/ DC
5.6/ 6 = BD/ 3
BD = 5.6/ 2 = 2.8cm
And we know that,
BD = BC – DC
2.8 = BC – 3
∴ BC = 5.8 cm
(vii) if AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm, find AC.
Solution:
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. And AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm. Required to find: AC
Since AD is the bisector of ∠ A meeting side BC at D in Δ ABC
⇒ AB/ AC = BD/ DC
5.6/ AC = 3.2/ DC
And we know that
BD = BC – DC
3.2 = 6 – DC
∴ DC = 2.8 cm
⇒ 5.6/ AC = 3.2/ 2.8
AC = (5.6 x 2.8)/ 3.2
∴ AC = 4.9 cm
(viii) if AB = 10 cm, AC = 6 cm, and BC = 12 cm, find BD and DC.
Solution:
Given: Δ ABC and AD bisects ∠A, meeting side BC at D. AB = 10 cm, AC = 6 cm, and BC = 12 cm. Required to find: DC
Since AD is the bisector of ∠ A meeting side BC at D in Δ ABC
⇒ AB/ AC = BD/ DC
10/ 6 = BD/ DC …….. (i)
And we know that
BD = BC – DC = 12 – DC
Let BD = x,
⇒ DC = 12 – x
Thus (i) becomes,
10/ 6 = x/ (12 – x)
5(12 – x) = 3x
60 -5x = 3x
∴ x = 60/8 = 7.5
Hence, DC = 12 – 7.5 = 4.5cm and BD = 7.5 cm
2. In figure 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, find CE.
Solution:
Given: AE is the bisector of the exterior ∠ CAD and AB = 10 cm, AC = 6 cm, and BC = 12 cm. Required to find: CE
Since AE is the bisector of the exterior
∠
CAD.
BE / CE = AB / AC
Let’s take CE as x.
So, we have
BE/ CE = AB/ AC
(12+x)/ x = 10/ 6
6x + 72 = 10x
10x – 6x = 72
4x = 72
∴ x = 18
Therefore, CE = 18 cm.
3. In fig. 4.58, Δ ABC is a triangle such that AB/AC = BD/DC, ∠B=70 o , ∠C = 50 o , find ∠BAD.
Solution:
Given: Δ ABC such that AB/AC = BD/DC, ∠B = 70 o and ∠C = 50 o Required to find: ∠BAD
We know that,
In ΔABC,
∠A = 180 – (70 + 50) [Angle sum property of a triangle]
= 180 – 120
= 60
o
Since,
AB/AC = BD/DC,
AD is the angle bisector of angle ∠A.
Thus,
∠BAD = ∠A/2 = 60/2 = 30
o
