Question

.13. In Figure, a circle inscribed in triangle ABC touches its sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.

Answer 1

Solution:

We know that tangents from an external point are equal in length.

Given :

AB = 12 cm

BC= 8 cm

AC = 10 cm

Now let us take A as external point,thus

AD = AF

let AD = AF = x cm

So, DB = 12-x = BE [tangent from external point B]

CE = 8-12+x = -4+x = CF [tangent from external point C]

Since perimeter of ∆ABC does not change so

12+8+10= x+x+12-x+12-x-4+x-4+x

30=2x+16

2x=30-16

2x=14

x = 7 cm

So, AD = AE = 7 cm


BD=BE = 12-7=5 cm

CE=CF= -4+7=3 cm

Hope it helps you.

Answer 2

Answer:

Given :AB = 12 cm

            BC= 8 cm

            AC = 10 cm

let AD = AF = x cm [tangent from external point A]

DB = BE= 12-x  [tangent from external point B]

BE  = EC (O is perpendicular to BC and bisects BC at E )

EC = BC -BE  

     =8-12+x = -4+x = x-4

CE= CF [tangent from external point C]

AC = AF + FC

10 = x + x -4

14 = 2x

x= 14/2

x=7

so  AD =AF = x = 7 cm

FC = x- 4 =   7-4 =  3

so  FC = EC =  3 cm

BE = 12-x  =  12 - 7  = 5

so  BD = BE =  12- x = 5 cm

ASSUME IN THIS FIG . O  AS THE  MIDPOINT ..