Question

In the adjoining figure, $\angle AOB={90}^o,AC=BC,OA=10cm$ and $OC=13cm$. The area of $△AOB$ is

Answer 1

Given $\angle AOB={90}^{\circ },AC=BC,OA=10cm,OC=13cm$

Since $AC=BC$, $C$ is the midpoint of $AB$

In right angled triangle the mid-point of the hypotenuse is equidistant from the vertices.

$\therefore OC=CA=CB$

$\Rightarrow AC=CB=13cm$

$\Rightarrow AB=AC+CB=13+13=26cm$

In right triangle $AOB$, $(AB{)}^2=(OA{)}^2+(OB{)}^2$

$\Rightarrow (OB{)}^2=(AB{)}^2-(OA{)}^2$

$=(26{)}^2-(10{)}^2$

$=576$

$\Rightarrow OB=\sqrt{576}=24$

$\therefore OB=24cm$

Hence area of triangle $AOB=\frac{1}{2}\times OA\times OB$

$=\frac{1}{2}\times 10\times 24=120$

Hence area of the triangle $AOB=120c{m}^2$