Question

In Fig. 2, a circle is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA and 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF.

Answer 1

Given :

In ∆ABC ,

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The lengths of tangents drawn

from an external points to a

circle are equal .

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AF = AD = x ,

BD = BE = y ,

CE = CF = z ,


AB = 12 cm => x+y = 12 ---( 1 )

BC = 8 cm => y + z = 8 ---( 2 )

CA = 10 cm => z + x = 10 ---( 3 )

Add ( 1 ) , ( 2 ) , ( 3 ) , we get

2( x + y + z ) = 30

=> x + y + z = 15 ---- ( 4 )

Now ,

Subtract ( 1 ) from ( 4 ) , we get

z = 3 ,

Subtracting ( 2 ) from ( 4 ). we get

x = 7 ,

Subtracting ( 3 ) , from ( 4 ) we get

y = 5

Therefore ,

AD = x = 7 cm ,

BE = y = 5 cm ,

CF = z = 3 cm

••••

Answer 2

Let AD = x

Hence, BD = 12 - z

BE =x

Hence, EC = 8-x

CF = y

Hence, AF = 10 - y

BD = BE

CE = CF

AD = AF

That is,

12-z = x

= x + z = 12 ...eq(1)

8-x = y

=y+x = 8 ... eq(2)

10-y = z

z + y = 10 ... eq(3)

Add x , y and z

x + y + z =( 12 - z) + (8- x) + (10 - y)

= 2( x+ y + z) = 30

x + y + z = 15 ...eq (4)

Now,

Subtract eq(1) from eq (4)

x + y+ z - ( x + z)

= 15-12

y = CF = 3

Now,
Subtract eq(2) from eq(4)

x + y + z - (y+x)

= 15-8

z = AD = 7

Subtract eq (3) from eq (4)

x + y + z -(z +y)

= 15 - 10

x = BE = 5

Therefore,
We got the following values :-

AD = 7
BE = 5
CF = 3

Thanks!!