Question

$ABC$ is a right angled triangle, right angled at $B$ such that $BC=6cm$ and $AB=8cm$. A circle with centre $O$ is inscribed in $△ABC$. The radius of the circle is

• A. $1cm$
• B. $2cm$
• C. $3cm$
• D. $4cm$

Answer 1

The correct option is B $2cm$

Here $AB=8cm,AC=6cm$

${AC}^2={AB}^2+{AC}^2$ by pythagoras theorem

$\Rightarrow {AC}^2=8^2+6^2=64+36=100cm$

$\therefore AC=10cm$

Now $ODBE$ is a square (as $\angle DBE={90}^{\circ }$ and by radius-tangent-perpendicularity theorem,$\angle ODB={90}^{\circ },$ and $DO=OE=radius$)

$\therefore DE=BE=EO=DO$ ..........$\left(1\right)$

Now, $AC=AF+FC$

$\Rightarrow AC=AD+EC$ since tangents drawn from external point of circle are equal

$\therefore 10=AE+EC$ .......$\left(2\right)$

Let $BD=BE=x$

$\therefore AB=AD+DB$

$\Rightarrow 8=AD+x$

$\therefore AD=8-x$ .....$\left(3\right)$

Again $EC=BC-BE=6-x$ .....$\left(4\right)$

From $\left(2\right),\left(3\right)$ and $\left(4\right)$

$10=8-x+6-x$

$\Rightarrow 2x=14-10=4$

$\therefore x=2cm$

Hence radius$=2cm$