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) $32c{m}^2$

(b) $40c{m}^2$

(c) $44c{m}^2$

(d) $48c{m}^2$

Answer 1

(b) $40c{m}^2$
Radius of the circle, AC = 10 cm
Diagonal of the rectangle, AC = 10 cm
$Now,AB=A{C}^2-B{C}^2={10}^2-{25}^2=80=45cm\therefore Areaoftherectangle=AB\times AD=25\times 45=40c{m}^2$