Question

In trapezium ABEF, BC = 4 cm, CD = 5 cm, DE=EF = 7 cm, then find the area of shaded region if FD represents arc of circle.

• A. 38.5 $c{m}^2$
• B. 59.5 $c{m}^2$
• C. 76 $c{m}^2$
• D. 97.5 $c{m}^2$

Answer 1

The correct option is B 59.5 $c{m}^2$

Given : AF = CE = CD+DE = 5+7 = 12 cm,
BE = BC+CD+DE = 4+5+7 = 16 cm.

As DE=EF= radius, FD is part of circle, region FED is sector with $\angle$FED is ${90}^o$

Area of shaded region
= Area of trapezium ABEF
- Area of sector DEF

Area of trapezium ABEF
= $\frac{1}{2}$$\times$(sum of parallel sides)$\times$height
= $\frac{1}{2}$$\times$(AF+BE)$\times$FE = $\frac{1}{2}$$\times$(12+16)$\times$7
= $\frac{1}{2}$$\times$28$\times$7 = 14$\times$7 = 98 $c{m}^2$

Area of sector DEF
= $\frac{angle}{{360}^o}$$\times$$\pi$$r^2$ = $\frac{{90}^o}{{360}^o}$$\times$$\frac{22}{7}$$\times$$7^2$
= $\frac{1}{4}$$\times$$\frac{22}{7}$$\times$$49$ = $\frac{22}{4}$$\times$$7$ = $\frac{11\times 7}{2}$ = 38.5 $c{m}^2$

Area of shaded region
= 98 - 38.5 = 59.5 $c{m}^2$