Ex-4.3 Triangles, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Q.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.

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

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

(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.

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

(vii) if AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm, find AC.

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

Sol: (i) It is given that BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm.

In Δ ABC,  AD is the bisector of ∠ A , meeting side BC at D.

We need to find DC,

Since, AD is ∠ A bisector,

Then,  AB/AC = 2.5/DC

5/4.2 = 2.5/DC

5DC = 4.2 x 2.5

DC = (4.2 x 2.5)/5

DC = 2.1

(ii) It is given that BD = 2 cm, AB = 5 cm, and DC = 3 cm

In Δ ABC, AD is the bisector of ∠ A, meeting side BC at D

We need to find AC.

Since, AD is ∠ A bisector.

Therefore, AB/AC = BD/DC  (since AD is the bisector of ∠ A and side BC)

Then, 5/AC = 2/3

2AC = 5 x 3

AC = 15/2

AC = 7.5 cm

(iii) It is given that AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm

In Δ ABC, AD is the bisector of ∠ A, meeting side BC at D

We need to find BD.

Since, AD is ∠ A bisector

Therefore,  AB/AC = BD/DC  (since, AD is the bisector of ∠ A and side  BC)

Then,  3.5/4.2 = BD/2.8

BD = (3.5 x 2.8)/4.2

BD = 7/3

BD = 2.3 cm

(iv) It is given that AB = 10 cm, AC = 14 cm, and BC = 6 cm

In Δ ABC, AD is the bisector of ∠ A meeting side BC at D

We need to find BD and DC.

Since, AD is bisector of ∠ A

Therefore, 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 = 3 cm.

 (v) It is given that AC = 4.2 cm, DC = 6 cm, and BC = 10 cm.

In Δ ABC, AD is the bisector of  ∠ A, meeting side BC at D.

We need to find out AB,

Since, AD is the bisector of ∠ A

Therefore, AC/AB = DC/BD

Then, 4.2/AB = 6/4

6AB = 4.2 x 4

AB = (4.2 x 4)/6

AB = 16.8/6

AB = 2.8 cm

(vi) It is given that AB = 5.6 cm, BC = 6 cm, and DC = 3 cm

In Δ ABC,  AD is the bisector of  ∠ A,  meeting side BC at D

We need to find BC,

Since, AD is the ∠ A bisector

Therefore, AC/AB = BD/DC

Then, 6/5.6 = 3/DC

DC = 2.8 cm

And, BC = 2.8 + 3

BC = 5.8 cm

(vii) It is given that AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm

In Δ ABC,  AD is the bisector of ∠ A , meeting side BC at D

Therefore, AB/AC = BD/DC

5.6/AC = 3.2/2.8   (DC = BC – BD)

AC = (5.6 x 2.8)/3.2

AC = 4.9 cm

(viii) It is given that AB = 10 cm, AC = 6 cm, and BC = 12 cm

In Δ ABC, AD is the ∠ A bisector, meeting side BC at D.

We need to find BD and DC

Since, AD is bisector of ∠ A

So, AC/AB = DC/BD

Let BD = x cm

Then,

6x = 120 – 10x

16x = 120

x = 120/16

x = 7.5

Now, DC = 12 – BD

DC = 12 – 7.5

DC = 4.5

BD = 7.5 cm and DC = 4.5 cm.

Q2.) 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.

Sol:

It is given that AE is the bisector of the exterior ∠CAD

Meeting BC produced E and AB = 10 cm, AC = 6 cm, and BC = 12 cm.

Since AE is the bisector of the exterior ∠CAD.

So, BE/CE = AB/AC

72 + 6x = 10x

4x = 72

x = 18

CE = 18 cm

Q.3) Δ ABC is a triangle such that AB/AC = BD/DC, ∠B = 70, ∠C = 50, find ∠BAD.

Sol:

It is given that in Δ ABC, AB/AC = BD/DC, ∠B = 70 and ∠C = 50

We need to find ∠ BAD

In Δ ABC,

∠A = 180 – (70 + 50)

= 180 – 120

= 60

Since, AB/AC = BD/DC

Therefore, AD is the bisector of ∠A

Hence, ∠BAD = 60/2 = 30

Q.4)  Check whether AD is the bisector of ∠A of ΔABC 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    

Sol: (i) It is given that AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm

We have to check whether AD is bisector of ∠ A

First we will check proportional ratio between sides.

Now,

AB/AC = 5/10 = 1/2

BD/CD = 1.5/3.5 = 3/7

Since, AB/AC ≠ BD/CD

Hence, AD is not the bisector of ∠ A.

(ii) It is given that AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm.

We have to check whether AD is the bisector of ∠ A

First we will check proportional ratio between sides.

So,  AB/AC = BD/DC

4/6 = 1.6/2.4

2/3 = 2/3     (it is proportional)

Hence, AD is the bisector of ∠ A.

(iii) It is given that AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm.

We have to check whether AD is the bisector of ∠ A

First we will check proportional ratio between sides.

DC = BC – BD

DC = 24 – 6

DC = 18

So,  AB/AC = BD/DC

8/24 = 6/18

1/3 = 1/3   (it is proportional)

Hence, AD is the bisector of ∠ A.

(iv) It is given that AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm.

We have to check whether AD is the bisector of ∠ A

First, we will check proportional ratio between sides.

So,  AB/AC = BD/DC

6/8 = 1.5/2

3/4 = 3/4     (it is proportional)

Hence, AD is the bisector of ∠ A.

(v) It is given that AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm.

We have to check whether AD is the bisector of ∠ A

First, we will check proportional ratio between sides.

So,  AB/AC = 5/12

BD/CD = 2.5/9 = 5/18

Since, AB/AC ≠ BD/CD

Hence, AD is not the bisector of ∠ A.

Q.5) In fig. AD bisects ∠A, AB = 12 cm, AC = 20 cm, and BD = 5 cm, determine CD.

Soln.: It is given that AD bisects ∠ A

AB = 12 cm, AC = 20 cm, and BD = 5 cm.

We need to find CD.

Since AD is the bisector of  ∠ A

then,  AB/AC = BD/DC

12/20 = 5/DC

12 x DC = 20 x 5

DC = 100/12

DC = 8.33 cm

∴ CD = 8.33 cm.

Q6.) In ΔABC, if ∠1 =  ∠2, prove that, AB/AC = BD/DC

 Sol:  We need to prove that, AB/AC = BD/DC

In  ΔABC,

∠1 =  ∠2

So, AD is the bisector of ∠A

Therefore,

ABAC = BDDC

Q.7)  D and E are the points on sides BC, CA and AB respectively. of a 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 AF, CE, and BD.

Sol: It is given that AB = 5 cm, BC = 8 cm and CA = 4 cm.

We need to find out, AF, CE and BD.

Since, AD is the bisector of ∠A

AB/AC = BD/CD

Then,

40 – 5BD = 4 BD

9BD = 40

So, BD = 40/9

Since, BE is the bisector of ∠ B

So, AB/BC = AE/EC

5CE = 32 – 8CE

5CE + 8CE = 32

13CE = 32

So, CE =  32/13

Now, since, CF is the bisector of ∠C

So, BC/CA = BF/AF

8AF = 20 – 4AF

12AF = 20

So, 3AF = 5

AF = 5/3 cm, CE = 32/12 cm

and BD = 40/9 cm