Question

Find the area of shaded region, if in parallelogram ABCD, $\angle$PBQ=$\angle$SDR = ${60}^o$, BP=BQ=SD=RD= 21 cm, BC = 45 cm and MN = 20 cm.

• A. 231 $c{m}^2$
• B. 438 $c{m}^2$
• C. 669 $c{m}^2$
• D. 897 $c{m}^2$

Answer 1

The correct option is B 438 $c{m}^2$

As BP=BQ= radius, PQ is part of circle, region sector is sector with $\angle$PBQ is ${60}^o$
Similarly
As SD=RD= radius, SR is part of circle, region SDR is sector with $\angle$SDR is ${60}^o$

Area of shaded region
= area of parallelogram ABCD
- area of sector PBQ
- area of sector SDR
​
Area of parallelogram = base$\times$height
= BC$\times$MN = 45$\times$20 = 900 $c{m}^2$

Area of sector PBQ = $\frac{angle}{{360}^o}$$\times$$\pi$$r^2$
= $\frac{{60}^o}{{360}^o}$$\times$$\frac{22}{7}$$\times$${21}^2$ = $\frac{1}{6}$$\times$$\frac{22}{7}$$\times$441 = 11$\times$21
= 231 $c{m}^2$

As sector PBQ and SDR has same angle and radius, these two sector will have same area.
$⟹$Area of sector SDR = Area of sector PBQ = 231 $c{m}^2$

Area of shaded region = 900 - 231 - 231 = 438 $c{m}^2$