Question

Find the area of shaded region, if AB = 10 cm, CD = 6 cm and OP = 4 cm. (Use $\pi$ = 3.14)

• A. 14.13 $c{m}^2$
• B. 39.25 $c{m}^2$
• C. 65.38 $c{m}^2$
• D. 85.38 $c{m}^2$

Answer 1

The correct option is D 85.38 $c{m}^2$

radius = $\frac{diameter}{2}$
AO = $\frac{AB}{2}$ = $\frac{10}{2}$ = 5 cm
DP = $\frac{CD}{2}$ = $\frac{6}{2}$ = 3 cm

Area of shaded region
= Area of semicircle having radius AO + Area of semicircle having radius DP + Area of trapezium ABCD

Area of trapezium ABCD
= $\frac{1}{2}$$\times$(Sum of parallel sides)$\times$height
= $\frac{1}{2}$$\times$(AB+CD)$\times$OP
= $\frac{1}{2}$$\times$(10+6)$\times$4
= $\frac{1}{2}$$\times$16$\times$4 = 8$\times$4 = 32 $c{m}^2$

Area of semicircle = $\frac{1}{2}$$\times$$\pi$$r^2$
area of semicircle having radius AO
= $\frac{1}{2}$$\times$$\pi$$r^2$
= $\frac{1}{2}$$\times$3.14$\times$$5^2$
= $\frac{1}{2}$$\times$3.14$\times$25
= 39.25 $c{m}^2$

Area of semicircle having radius DP
= $\frac{1}{2}$$\times$$\pi$$r^2$
= $\frac{1}{2}$$\times$3.14$\times$$3^2$
= $\frac{1}{2}$$\times$3.14$\times$9
= 14.13 $c{m}^2$

Area of shaded region
= 39.25 + 14.13 + 32 = 85.38 $c{m}^2$