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 $=A B \times A D=2 \sqrt{5} \times 4 \sqrt{5}=40 \mathrm{~cm}^{2}$
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