Question

In a triangle $ABC$, right angled at $B,BC=15cm$ and $AB=8cm$. A circle is inscribed in triangle $ABC$. Then the radius of the circle is

Answer 1

$AC=\sqrt{8^2+{15}^2}$

$=\sqrt{64+225}$

$=\sqrt{289}$

$=17$

Since tangents from same Limit to a circle are equal ........($\alpha$)

$CM=CN=x$

$BN=15-x$

$BN-BP=15-x$ (reasons ($\alpha$))

$AP=\left(8\right)-(15-x)=x-7$

$AP=AM=x-7$ (reasons ($\alpha$))

$AM+MC=AC$

$x-7+x=17$

$2x=24$

$x=12$

Now radius of circle , $r\Rightarrow$

$OP=ON=NB=BP=r=15-x$

($ONBP$ is a square)

$r=15-12$

$r=3$ units