RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 (Updated for 2024)

RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3

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.

Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 Free PDF

 


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

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

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 1

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

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 2

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

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 3

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

(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.

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 4

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

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 5

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

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 6

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

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 7

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

C:\Users\Tnluser\Downloads\RD-sharma-class-10-chapter-4.3---1.8.png

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

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 9

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=70o, ∠C = 50o, find ∠BAD.

Solution:

Given: Δ ABC such that AB/AC = BD/DC, ∠B = 70o and ∠C = 50o

Required to find: ∠BAD

R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 10

We know that,

In ΔABC,

∠A = 180 – (70 + 50) [Angle sum property of a triangle]

= 180 – 120

= 60o

Since,

AB/AC = BD/DC,

AD is the angle bisector of angle ∠A.

Thus,

∠BAD = ∠A/2 = 60/2 = 30o

We have provided complete details of RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

FAQs on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3

Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3?

Yes, learning all of the concepts included in RD Sharma Solutions for Class 10 Maths Chapter 4 Exercise 4.3 is required in order to achieve high scores on the Class 10 board exams. These solutions were created by subject matter specialists who compiled model questions that covered all of the textbook’s exercise questions.

Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 free PDF?

You can download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 free PDF from the above article.

What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3?

1. Correct answers according to the latest CBSE guidelines and syllabus.
2. The RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 is written in simple language to assist students in their board examination, & competitive examination preparation.

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