Question

Adjacent figure is consist of 4 semicircles and 2 quarter circles, if OA=OB=OC=OD = 28 cm, find the area of shaded region.

• A. 308 $c{m}^2$
• B. 616 $c{m}^2$
• C. 1232 $c{m}^2$
• D. 2462 $c{m}^2$

Answer 1

The correct option is C 1232 $c{m}^2$

Quarter circle has angle $\frac{{360}^o}{4}$ = ${90}^o$

Area of shaded region =
Area of semicircle having diameter OD
+Area of sector DOB having angle ${90}^o$
-Area of semicircle having diameter OB
+Area of semicircle having diameter OC
+Area of sector COA having angle ${90}^o$
-Area of semicircle having diameter OA

As OA=OB=OC=OD = 28 cm, hence
Area of semicircle having diameter OB = Area of semicircle having diameter OD
Area of semicircle having diameter OC = Area of semicircle having diameter OA
Hence
Area of shaded region
= Area of sector DOB having angle ${90}^o$
+ Area of sector COA having angle ${90}^o$

As both sectors has same radius and angle

Area of shaded region
= 2$\times$area of sector DOB having angle ${90}^o$
= 2$\times$($\frac{angle}{{360}^o}$$\times$$\pi$$r^2$) = 2$\times$($\frac{{90}^o}{{360}^o}$$\times$$\pi$$\times$${28}^2$)
= 2$\times$($\frac{1}{4}$$\times$$\frac{22}{7}$$\times$784) = $\frac{11\times 784}{7}$ = 1232 $c{m}^2$