Question

In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If $AD=2\sqrt{5}$ cm, then area of the rectangle is
(a) 32 cm²
(b) 40 cm²
(c) 44 cm²
(d) 48 cm²

Answer 1

(b) 40 cm²

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

$$\begin{gathered} \mathrm{Now},AB=\sqrt{A{C}^2-B{C}^2}=\sqrt{{10}^2-{\left(2\sqrt{5}\right)}^2}=\sqrt{80}=4\sqrt{5}\mathrm{cm} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{rectangle}=AB\times AD=2\sqrt{5}\times 4\sqrt{5}=40{\mathrm{cm}}^2 \end{gathered}$$