Question

A circle is inscribed in a triangle whose sides are $40cm,40cm$ and $48cm$ respectively. A smaller circle touching two equal sides of the triangle and to the first circle. Find the radius of the smaller circle.

• A. $2cm$
• B. $5cm$
• C. $4cm$
• D. None

Answer 1

The correct option is D None

$AB=AC=40cm$

$BC=48cm$

So, $BD=CD+24cm$

In, $△ABD$ applying Pythagoras theorem

$A{B}^2=B{D}^2=A{D}^2$

or, $AD=\sqrt{4}{0}^2-{24}^2=\sqrt{1}024=32cm$

$O$ centroid divide triangle in $2:1$ ratio,

$AO:OD=2:1$

So, $AO=\frac{2}{3}AD$ &$OD=\frac{AD}{3},AO=\frac{64}{3}$ & $OD=\frac{32}{3}$

Assume

$\angle AOZ=\angle AXY=90°$

$\frac{Xy}{OZ}=\frac{AX}{AO},AX=2r$

$AO-OX=2r$

$AO-OD-r=2r$

$3r=\frac{32}{3}r=\frac{32}{9}cm$