Question

In the given figure, APB and CQD are semicircles of diameter 7 cm each, while ARC an BSD are semicircles of diameter 14 cm each. Find the (i) perimeter, (ii) area of the shaded region.

Answer 1

(i)

Given:
Diameter of semicircle APB & CQD = 7 cm
Diameter of semicircle ARC & BSD = 14 cm

From the given figure We have,

FOR SEMICIRCLE ARC:
Diameter AC =14 cm
Radius for semicircle ARC = $\frac{14}{2}$= 7 cm

FOR SEMICIRCLE APB
Diameter AB = 7 cm
Radius for semicircle AB = $\frac{7}{2}$ cm

FOR SEMICIRCLE BSD
Diameter BD =14 cm
Radius for semicircle BD = $\frac{14}{2}$ = 7 cm

FOR SEMICIRCLE CQD
Diameter CD=7 cm
Radius for semicircle CD =$\frac{7}{2}$ cm

PERIMETER OF CIRCLE = πr

Perimeter of the shaded region= Perimeters of semicircle (ARC + APB + BSD + CQD)

Perimeter of the shaded region= π×7 + π ×$\frac{7}{2}$ +π×7 + π ×$\frac{7}{2}$
= π(7+$\frac{7}{2}$+7+$\frac{7}{2}$)
= π(14+$\frac{7}{2}$+$\frac{7}{2}$) = π( 14 +7)
= $\frac{22}{7}$ × 21
= 22 × 3 = 66 cm

Hence, the Perimeter of the shaded region is 66 cm.

ii)

Area of APB and CDQ are equal because of r=$\frac{7}{2}cm$

Area of APB = $\frac{1}{2}\pi r^2$
= $\frac{1}{2}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}=\frac{77}{4}c{m}^2$

Area of CQD =$\frac{77}{4}c{m}^2$
Area of ARC & area of BSD are equal because r = 7cm
Area of ARC = $\frac{1}{2}\times \frac{22}{7}\times 7\times 7=77c{m}^2$
Area of BSD=$77c{m}^2$
By subtracting the area of ARC-area of APB= 77-$\frac{77}{4}c{m}^2$ = $\frac{231}{4}c{m}^2$
so, area of BSD-area of CQD =$\frac{231}{4}c{m}^2$
Area of shaded region =$\frac{231}{4}c{m}^2$+$\frac{231}{4}c{m}^2$=$\frac{462}{4}c{m}^2$