# In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm.

Question:

In the given figure, $A B C D$ is a rectangle inscribed in a quadrant of a circle of radius $10 \mathrm{~cm}$. If $A D=2 \sqrt{5} \mathrm{~cm}$, then area of the rectangle is

(a) $32 \mathrm{~cm}^{2}$

(b) $40 \mathrm{~cm}^{2}$

(c) $44 \mathrm{~cm}^{2}$

(d) $48 \mathrm{~cm}^{2}$

Solution:

(b) $40 \mathrm{~cm}^{2}$

Radius of the circle, AC = 10 cm
Diagonal of the rectangle, AC = 10 cm

Now, $A B=\sqrt{A C^{2}-B C^{2}}=\sqrt{10^{2}-(2 \sqrt{5})^{2}}=\sqrt{80}=4 \sqrt{5} \mathrm{~cm}$

$\therefore$ Area of the rectangle $=A B \times A D=2 \sqrt{5} \times 4 \sqrt{5}=40 \mathrm{~cm}^{2}$