Question

In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region (in $c{m}^2$).

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• A. 98.01
• B. $95.01c{m}^2$
• C. $96.01c{m}^2$
• D. $90.01c{m}^2$

Answer 1

The correct option is A

98.01

Area of quadrant
$=\frac{90}{360}\times \pi \times 14\times 14=153.94c{m}^2$

Area of
$△ABC=0.5\times {14}^2=98c{m}^2$

The diameter
$=14\times \sqrt{2}=19.8cm.$

Area of semi-circle
$=0.5\pi \times {9.9}^2=153.95c{m}^2$

The area of segment
$=153.94-98=55.94c{m}^2$

The required area
$=153.95-55.94=98.01c{m}^2$.