Question

$ABC$ is an isosceles triangle inscribed in a circle.If $AB=AC=25$cm and $BC=14$cm find the radius of the circle.

Answer 1

Draw $AD\perp BC$

Hence $D$ is the midpoint of BC

That is $BD=DC=\left(\frac{1}{2}\right)BC$

$\therefore ,BD=DC=7$ cm

In right triangle ADB, by Pythagoras theorem we have

${AB}^2={AD}^2+{BD}^2$

$\Rightarrow {25}^2={AD}^2+7^2$

$\Rightarrow {AD}^2=625-49=576$

$\therefore AD=24$cm

Consider, right triangle $ODB$, by Pythagoras theorem we have

${OB}^2={OD}^2+{BD}^2$

Here $OD=AD–AO=24–r$, where $r$ is radius of the circle

$r^2={(24–r)}^2+7^2$

$\Rightarrow r^2=576+r^2-48r+49$

$\Rightarrow 48r=625$

$\Rightarrow r=\frac{625}{48}=13.02$cm