Question

ABC is an isosceles triangle inscribed in circle. If AB = AC = $12\sqrt{5}$ cm and BC = 24 cm, find the radius of the circle.

Answer 1

$AB=AC12\sqrt{5}$ and $BC=24cm$.
Join OB and OC and OA
Draw $AD\perp BC$ which will pass through
Centre O
OD bisect BC in D
$BD=CD=12cm$
In right $△ABD$
$A{B}^2=A{D}^2+B{D}^2$
$(12\sqrt{5}{)}^2=A{D}^2+B{D}^2$
$(12\sqrt{5}{)}^2=A{D}^2+(12{)}^2$
$144\times 5=A{D}^2+144$
$720-144=A{D}^2$
$A{D}^2=576(AD=\sqrt{576}=24$
Let radius of the circle $=OA=OB=OC=r$
$OD=AD-AO=24-r$
In right $△OBD$
$r^2=(12{)}^2+{(24-r)}^2$
$r^2=144+576+r^2-48r$
48r = 720
$r=720/48=48r$
$48r=720$
$r=720/48=15cm.$
Radius $=15cm.$