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 $\sqrt{3}=1.73and\pi =\frac{22}{7}$.]

Answer 1

Join AO and extend it to D on BC.

Radius of the circle, r = 42 cm

$\angle OCD={30}^{o}cos{30}^o=\frac{DC}{OC}\Rightarrow \frac{\sqrt{3}}{2}=\frac{DC}{42}\Rightarrow DC=21\sqrt{3}\Rightarrow BC=2\times DC=2\times 21\sqrt{3}=72.66cmsin{30}^o=\frac{OD}{OC}\Rightarrow \frac{1}{2}=\frac{OD}{42}\Rightarrow OD=21cm$

Now, $AD=AO+OD=42+21=63cm$

Area of the shaded region = Area of circle – Area of triangle ABC

$=\pi (OA{)}^2–\frac{1}{2}\times AD\times AB=\frac{22}{7}\times (42{)}^2–12\times 63\times 72.66=5544–2288.79=3255.21c{m}^2$