Question

In the above figure, $OCDE$ is a rectangle inscribed in a quadrant of a circle of radius $10cm$. If $OE=2\sqrt{5}$, find the area of the rectangle.

Answer 1

Given, $OCDE$ is a rectangle inscribed in a quadrant of a circle of radius $10cm$ and $OE=2\sqrt{5}cm$
In right triangle $OED$, we have
${\left(OD\right)}^2={\left(OE\right)}^2+{\left(ED\right)}^2\Rightarrow {\left(10\right)}^2={\left(2\sqrt{5}\right)}^2+{\left(ED\right)}^2\Rightarrow 100-20={\left(ED\right)}^2\Rightarrow ED=\sqrt{80}=4\sqrt{5}cm$
Now, area of rectangle $ODCE=OE\times ED=2\sqrt{5}\times 4\sqrt{5}=40{cm}^2$