Question

A circle inscribed in a triangle abc having ab = 12 cm, bc = 8 cm, ca = 10 cm touching ab at d,bc at e & ac at f . find the length of ad.

Answer 1

incircle touches the sides at mid point of the side. AB= 12, therefore ad= 6

Answer 2

$\bigstar{\pmb{\huge{\underline{\mathfrak{\red{Answer}}}}}}$

The tangents drawn from an external point to a circle are equal.

$\therefore$ $\sf{AD=AF=x(say)}$

$\sf{BD=BE=y(say)}$

$\sf{CE=CF=z(say)}$

Now,

$\sf{AB=x+y=12cm}$

$\sf{BC=y+z=8cm}$

$\sf{AC=z+x=10cm}$

On adding,

= $\sf{2(x+y+z)=30}$

= $\sf{x+y+z=15cm}$

Hence,

$\sf{AD=15-8=7cm}$ ,

$\sf{BE=15-10=5cm}$ ,

$\sf{CF=15-12=3cm}$