Question

In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

• A. 200
• B. 228
• C. 245
• D. 230

Answer 1

The correct option is B

228

Area of shaded region= Area of quadrant OPBQ of circle - Area of square OABC (1)

It is given that OABC is a square which means that $\angle AOC={90}^{\circ }$ which is angle of sector of circle.

Radius of Quadrant of circle = r = Diagonal of square = OB

And, Diagonal of square = 0B = $\sqrt{(O{A}^2+O{B}^2)}$

= \(\sqrt {(20^2 + 20^2)}\) = $\sqrt{(400+400)}$ = $\sqrt{8}00$ = 20$\sqrt{2}0$ cm

Area of quadrant OPBQ = $\pi r^2\times \frac{v}{360}$ {where r is the radius and v is the angle of sector of circle.}

=$3.14\times 20\sqrt{2}0\times 20\sqrt{2}0\times \frac{90}{360}$ = 628 $c{m}^2$ (2)

Area of square OABC= side x side =20 x 20 = 400 $c{m}^2$

Putting (2) and (3) in (1), we get

Area of shaded portion = 628 – 400 = 228$c{m}^2$ (3)