# On a circular table cover of radius 42 cm,

Question:

On a circular table cover of radius 42 cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design.[Use $\sqrt{3}=1.73, \pi=\frac{22}{7}$ ]

Solution:

Construction:  Join AO and extend it to D on BC.

Radius of the circle, r = 42 cm
∠OCD= 30°

$\cos 30^{\circ}=\frac{\mathrm{DC}}{\mathrm{OC}}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{DC}}{42}$

$\Rightarrow \mathrm{DC}=21 \sqrt{3}$

$\Rightarrow \mathrm{BC}=2 \times \mathrm{DC}=42 \sqrt{3}=72.66 \mathrm{~cm}$

$\sin 30^{\circ}=\frac{\mathrm{OD}}{\mathrm{OC}}$

$\Rightarrow \frac{1}{2}=\frac{\mathrm{OD}}{42}$

$\Rightarrow \mathrm{OD}=21 \mathrm{~cm}$

Now, $\mathrm{AD}=\mathrm{AO}+\mathrm{OD}=42+21=63 \mathrm{~cm}$

Area of shaded region = Area of circle − Area of triangle ABC

$=\pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{AD} \times \mathrm{AB}$

$=\frac{22}{7}(42)^{2}-\frac{1}{2} \times 63 \times 72.66$

$=5544-2288.79$

$=3255.21 \mathrm{~cm}^{2}$