Question

In the given figure, a semicircle is drawn on the hypotenuse of the right-angled triangle. Another arc of radius 30 cm is drawn passing through A and C with O as the center. What is the area of the shaded region? (Take $\pi$ as 3.14 and $\sqrt{3}$ = 1.732)

• A. $117.75sq.cm$
• B. $216sq.cm$
• C. $271.95sq.cm$
• D. $108sq.cm$

Answer 1

The correct option is C

$271.95sq.cm$

Area of the shaded region = Area of semicircle APC - Area of segment AQC
Area of semicircle = $\frac{\pi r^2}{2}=3.14\times 15\times \frac{15}{2}=353.25sq.cm$
Area of segment = Area of sector OAQC - Area of triangle OAC
Consider triangle OAC
Since, all sides are equal to 30 cm in the triangle, it is an equilateral triangle.
Hence, each angle is equal to 60 degrees.
Hence $\angle OAC=\angle AOC=\angle ACO={60}^{\circ }$.
Now, let the height be h.
$\sin {60}^{\circ }=\frac{h}{30}$
$\frac{\sqrt{3}}{2}=\frac{h}{30}$
$h=15\sqrt{3}$
Now, area of triangle OAC is $\frac{1}{2}\times \text{base}\times \text{h}$
$=\frac{1}{2}\times 30\times 15\sqrt{3}=225\sqrt{3}sq.cm=389.7sq.cm$
Area of the sector OAQC = $\frac{60}{360}\times \pi \times r^2=\frac{1}{6}\times 3.14\times 30\times 30=471sq.cm.$
Hence, area of segment AQC $=471-389.7=81.3sq.cm$
Hence, area of the shaded region = Area of semicircle - area of segment AQC
$=353.25-81.3=271.95sq.cm.$