RD Sharma Solutions Class 10 Triangles Exercise 4.3

RD Sharma Solutions Class 10 Chapter 4 Exercise 4.3

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

Exercise 4.3

Q.1)  In a \(\Delta\) ABC, AD is the bisector of \(\angle\) 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 \(\Delta\) ABC,  AD is the bisector of \(\angle\) A , meeting side BC at D.

We need to find DC,

Since, AD is \(\angle\) A bisector,

Then,  \(\frac{AB}{AC} = \frac{2.5}{DC}\)

\(\frac{5}{4.2} = \frac{2.5}{DC}\)

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 \(\Delta\) ABC, AD is the bisector of \(\angle\) A, meeting side BC at D

We need to find AC.

Since, AD is \(\angle\) A bisector.

Therefore, \(\frac{AB}{AC} = \frac{BD}{DC}\)  (since AD is the bisector of \(\angle\) A and side BC)

Then, \(\frac{5}{AC} = \frac{2}{3}\)

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 \(\Delta\) ABC, AD is the bisector of \(\angle\) A, meeting side BC at D

We need to find BD.

Since, AD is \(\angle\) A bisector

Therefore,  \(\frac{AB}{AC} = \frac{BD}{DC}\)  (since, AD is the bisector of \(\angle\) A and side  BC)

Then,  \(\frac{3.5}{4.2} = \frac{BD}{2.8}\)

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 \(\Delta\) ABC, AD is the bisector of \(\angle\) A meeting side BC at D

We need to find BD and DC.

Since, AD is bisector of \(\angle\) A

Therefore, \(\frac{AB}{AC} = \frac{BD}{DC}\)  (AD is bisector of \(\angle\) A and side BC)

Then, \(\frac{10}{14} = \frac{x}{6 -x}\)

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 \(\Delta\) ABC, AD is the bisector of  \(\angle\) A, meeting side BC at D.

We need to find out AB,

Since, AD is the bisector of \(\angle\) A

Therefore, \(\frac{AC}{AB} = \frac{DC}{BD}\)

Then, \(\frac{4.2}{AB} = \frac{6}{4}\)

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 \(\Delta\) ABC,  AD is the bisector of  \(\angle\) A,  meeting side BC at D

We need to find BC,

Since, AD is the \(\angle\) A bisector

Therefore, \(\frac{AC}{AB} = \frac{BD}{DC}\)

Then, \(\frac{6}{5.6} = \frac{3}{DC}\)

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 \(\Delta\) ABC,  AD is the bisector of \(\angle\) A , meeting side BC at D

Therefore, \(\frac{AB}{AC} = \frac{BD}{DC}\)

\(\frac{5.6}{AC} = \frac{3.2}{2.8}\)   (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 \(\Delta\) ABC, AD is the \(\angle\) A bisector, meeting side BC at D.

We need to find BD and DC

Since, AD is bisector of \(\angle\) A

So, \(\frac{AC}{AB} = \frac{DC}{BD}\)

Let BD = x cm

Then,

\(\frac{6}{10} = \frac{12 – x}{x}\)

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 \(\angle\)CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm, and BC = 12 cm, Find CE.

1

Sol:

It is given that AE is the bisector of the exterior \(\angle\)CAD

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

Since AE is the bisector of the exterior \(\angle\)CAD.

So, \(\frac{BE}{CE} = \frac{AB}{AC}\)

\(\frac{12 + x}{x} = \frac{10}{x}\)

72 + 6x = 10x

4x = 72

x = 18

CE = 18 cm

Q.3) \(\Delta\) ABC is a triangle such that \(\frac{AB}{AC} = \frac{BD}{DC}\), \(\angle\)B = 70, \(\angle\)C = 50, find \(\angle\)BAD.

2

Sol:

It is given that in \(\Delta\) ABC, \(\frac{AB}{AC} = \frac{BD}{DC}\), \(\angle\)B = 70 and \(\angle\)C = 50

We need to find \(\angle\) BAD

In \(\Delta\) ABC,

\(\angle\)A = 180 – (70 + 50)

= 180 – 120

= 60

Since, \(\frac{AB}{AC} = \frac{BD}{DC}\)

Therefore, AD is the bisector of \(\angle\)A

Hence, \(\angle\)BAD = 60/2 = 30

Q.4)  Check whether AD is the bisector of \(\angle\)A of \(\Delta\)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    

3

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 \(\angle\) A

First we will check proportional ratio between sides.

Now,

\(\frac{AB}{AC} = \frac{5}{10} = \frac{1}{2} \)

\(\frac{BD}{CD} = \frac{1.5}{3.5} = \frac{3}{7} \)

Since, \(\frac{AB}{AC} \neq \frac{BD}{CD}\)

Hence, AD is not the bisector of \(\angle\) 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 \(\angle\) A

First we will check proportional ratio between sides.

So,  \(\frac{AB}{AC} = \frac{BD}{DC}\)

\(\frac{4}{6} = \frac{1.6}{2.4}\)

\(\frac{2}{3} = \frac{2}{3}\)     (it is proportional)

Hence, AD is the bisector of \(\angle\) 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 \(\angle\) A

First we will check proportional ratio between sides.

DC = BC – BD

DC = 24 – 6

DC = 18

So,  \(\frac{AB}{AC} = \frac{BD}{DC}\)

\(\frac{8}{24} = \frac{6}{18}\)

\(\frac{1}{3} = \frac{1}{3}\)   (it is proportional)

Hence, AD is the bisector of \(\angle\) 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 \(\angle\) A

First, we will check proportional ratio between sides.

So,  \(\frac{AB}{AC} = \frac{BD}{DC}\)

\(\frac{6}{8} = \frac{1.5}{2}\)

\(\frac{3}{4} = \frac{3}{4}\)     (it is proportional)

Hence, AD is the bisector of \(\angle\) 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 \(\angle\) A

First, we will check proportional ratio between sides.

So,  \(\frac{AB}{AC} = \frac{5}{12}\)

\(\frac{BD}{CD} = \frac{2.5}{9} = \frac{5}{18} \)

Since, \(\frac{AB}{AC} \neq \frac{BD}{CD}\)

Hence, AD is not the bisector of \(\angle\) A.

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

4

Soln.: It is given that AD bisects \(\angle\) A

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

We need to find CD.

Since AD is the bisector of  \(\angle\) A

then,  \(\frac{AB}{AC} = \frac{BD}{DC}\)

\(\frac{12}{20} = \frac{5}{DC}\)

12 x DC = 20 x 5

DC = 100/12

DC = 8.33 cm

∴ CD = 8.33 cm.

 

Q6.) In \(\Delta\)ABC, if \(\angle\)1 = \(\angle\)2, prove that, \(\frac{AB}{AC} = \frac{BD}{DC}\)

4

 Sol:  We need to prove that, \(\frac{AB}{AC} = \frac{BD}{DC}\)

In  \(\Delta\)ABC,

\(\angle\)1 = \(\angle\)2

So, AD is the bisector of \(\angle\)A

Therefore,

\(\frac{AB}{AC} = \frac{BD}{DC}\)

 

Q.7)  D and E are the points on sides BC, CA and AB respectively. of a \(\Delta\)ABC  such that AD bisects  \(\angle\)A, BE bisects \(\angle\)B and CF bisects \(\angle\)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 \(\angle\)A

\(\frac{AB}{AC} = \frac{BD}{CD}\)

Then,

\(\frac{5}{4} = \frac{BD}{BC – BD}\)

\(\frac{5}{4} = \frac{BD}{8 – BD}\)

40 – 5BD = 4 BD

9BD = 40

So, BD = 40/9

Since, BE is the bisector of \(\angle\) B

So, \(\frac{AB}{BC} = \frac{AE}{EC}\)

\(\frac{AB}{BC} = \frac{AC – EC}{EC}\)

\(\frac{5}{8} = \frac{4 – CE}{CE}\)

5CE = 32 – 8CE

5CE + 8CE = 32

13CE = 32

So, CE = \(\frac{32}{13}\)

Now, since, CF is the bisector of \(\angle\)C

So, \(\frac{BC}{CA} = \frac{BF}{AF}\)

\(\frac{8}{4} = \frac{AB – AF}{AF}\)

\(\frac{8}{4} = \frac{5 – AF}{AF}\)

8AF = 20 – 4AF

12AF = 20

So, 3AF = 5

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

and BD = 40/9 cm


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