A circle is inscribed in a …

Let AD be x cm, BE = y cm, CF = z cm

Then,           AD = AF = x cm          [Since tangents from an external point to the circle are equal]

                    BD = BE = y cm          [Since tangents from an external point to the circle are equal]

                    CE = CF = z cm          [Since tangents from an external point to the circle are equal]

According to question,

AD + BD = 12 cm          =>          x + y = 12          ..........(i)

BE + EC = 8 cm            =>          y + z = 8            ..........(ii)

CF + AC = 10 cm          =>          x + z = 10            ..........(iii)

Adding eq.(i), (ii) and (iii), we get

2(x + y + z) = 30           =>          x + y + z = 15         ..........(iv)

Solving eq. (iv) and (ii), we get x = 7 cm          =>          AD = 7 cm

Solving eq. (iv) and (iii), we get y = 5 cm          =>         BE = 5 cm

Solving eq. (iv) and (i), we get z = 3 cm          =>          CF = 3 cm

Ans. Given : A Circle inscribed in A triangle ABC

AB = 12cm, AC = 10cm, BC = 8cm

We Know that Tangent drawn from an external point are equal in length. 

Let AD = x 

Then AD = AE = x          [Tangent from Same point]

Similarly 

BD = BF = y 

CE = CF = z 

AB = AD + BD  

=> x + y  = 12         (1)

AC = AE + EC 

=> x + z = 10         (2)

BC = BF + FC

=> y + z = 8         (3)

Adding all three equations, we get

2(x+y+z) = 30
=> x + y + z = 15      (4)

from (1) x + y =  12

then z = 3 cm

Similarly  y = 5 cm and z = 7cm