Question

Question 15
In Fig. 12.33, 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.

Answer 1

Radius of the the quadrant ABC of circle = 14 cm
AB = AC = 14 cm
BC is diameter of thesemicircle.
ABC is a right angled triangle.
By Pythagoras theorem in $\Delta ABC$,
$B{C}^2=A{B}^2+A{C}^2$
$\Rightarrow B{C}^2={14}^2+{14}^2$
$\Rightarrow BC=14\sqrt{2}cm$

Radius of semicircle
$=\frac{14\sqrt{2}}{2}cm=7\sqrt{2}cm$

Area of $\Delta ABC$
$=\frac{1}{2}\times \left(base\right)\times \left(height\right)$
$=\frac{1}{2}\times 14\times 14=98c{m}^2$

Area of quadrant
$=\frac{1}{4}\times \frac{22}{7}\times 14\times 14=154c{m}^2$
Area of the semicircle
$=\frac{1}{2}\times \frac{22}{7}\times 7\sqrt{2}\times 7\sqrt{2}=154c{m}^2$
Area of the shaded region = Area of the semicircle + Area of $\Delta ABC$ - Area of quadrant
$=154+98-154c{m}^2=98c{m}^2$