Question

In the given 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 $c{m}^2$). (Take $\pi =\frac{22}{7}$)

• A. 98.01 $c{m}^2$
• B. 89.01 $c{m}^2$
• C. 87.01 $c{m}^2$
• D. 100.01 $c{m}^2$

Answer 1

The correct option is A 98.01 $c{m}^2$

Area of sector = $\frac{x^{\circ }}{{360}^{\circ }}\times \frac{22}{7}\times r^2$

$\because$ Quadrant subtends an angle of ${90}^{\circ }$ and its radius = 14 $cm$

$\therefore$ Area of quadrant
= $\frac{{90}^{\circ }}{{360}^{\circ }}\times \frac{22}{7}\times 14\times 14$ = 154 $c{m}^2$

Area of $△$ ABC = $\frac{1}{2}\times AB\times AC$

$\because AB=AC=r=14cm$

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

Applying pythagoras theorem in right $△$ABC,
$B{C}^2=A{B}^2+A{C}^2$
$B{C}^2={14}^2+{14}^2$
$BC=\sqrt{196+196}=14\sqrt{2}$ = 19.8 $cm$

Area of semi-circle = $\frac{1}{2}\times \pi \times r^2$

For semicircle with diameter BC,
r = $\frac{BC}{2}=9.9cm$

$\therefore$ Area of semi-circle with diameter BC
= 0.5 $\times$ $\frac{22}{7}\times {9.9}^2$ = 154.01 $c{m}^2$

Area of segment = [Area of quadrant - Area of $△$ ABC] = 154 - 98 = 56 $c{m}^2$

Now,
The area of shaded region = [Area of semicircle - Area of segment]
= 154.01 - 56
= 98.01 $c{m}^2$
$\therefore$ Area of shaded region = 98.01 $c{m}^2$