Question

A circle inscribed in triangle abc touches the sides ab bc and ac at point d e and f respectively if a b is 12 cm bc is 8 cm and ac is 10 cm then find the length of adb and cf

Answer 1

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Answer 2

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,

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

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

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