Question

$ABCD$ is a rectangle inscribed in a quadrant of a circle having radius $10cm$. If $AD=2\sqrt{5}cm$, find the area of $ABCD$.

• A. $30{cm}^2$
• B. $50{cm}^2$
• C. $40{cm}^2$
• D. $35{cm}^2$

Answer 1

Answer: (C)

$40{cm}^2$

From the figure

$BC=AD=2\sqrt{5}$

Join $A$ to $D$ now $AD$ is the radius of the circle.

$\Rightarrow$AD=10$

Using Pythagoras theorem

${\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2{\left(AB\right)}^2+{\left(2\sqrt{5}\right)}^2={\left(10\right)}^2{\left(AB\right)}^2=100-20=80AB=\sqrt{80}=4\sqrt{5}$

Now area $=4\sqrt{5}\times 2\sqrt{5}=40{cm}^2$