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
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
(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=70o, ∠C = 50o, find ∠BAD.
Solution:
Given: Δ ABC such that AB/AC = BD/DC, ∠B = 70o and ∠C = 50o
Required to find: ∠BAD
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
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