Question

In a circular table cover of radius $32cm$, a design is formed leaving an equilateral triangle $ABC$ in the middle as shown in the given figure.Find the area of the design (shaded region)

Answer 1

Given : radius of circle = $32cm$

Area of design = Area of circle -Area of $\Delta ABC$

Firstly , we find area of a circle

Area of circle = $\pi r^2$

$=\frac{22}{7}\times (32{)}^2$

$=\frac{22}{7}\times 32\times 32$

$=\frac{22528}{7}c{m}^2$.....(a)

Now we will find the area of equilateral $\Delta ABC$

Construction :

Draw $OD\perp BC$

In $\Delta BOD$ and $\Delta COD$

$OB=OC$ (radii)

$OD=OD$ ( common)

$\angle ODB=\angle ODC\left({90}^0\right)$

$\Delta BOD\cong \Delta COD$ [ by RHS congruency ]

$\Rightarrow BD=DC$ [ by CPCT]

or $BC=2BD....\left(i\right)$

and,

$\angle BOD=\angle COD=\frac{1}{2}\angle BOC=\frac{120}{2}={60}^0$

Now , In $\Delta BOD$ we have

$sin{60}^0=\frac{BD}{OB}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{BD}{32}$

$\Rightarrow BD=16\surd 3cm$

From (i) $BC=2BD\Rightarrow BC=32\surd 3cm$

Now, Area of equilateral $\Delta ABC$

$=\frac{\sqrt{3}}{4}(side{)}^2$

$=\frac{\sqrt{3}}{4}\times \left(32\sqrt{3}\right)62$

$=768\surd 3c{m}^2...\left(b\right)$

therefore, Area of design = Area of circle -Area of $\Delta ABC$

$=\frac{22528}{7}-768\sqrt{3}c{m}^2$

[from (a) and (b) ]