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

"

Given:

In a circular table cover of the radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle.

To do:

We have to find the area of the design.

Solution:

Radius of the circle $r=32 \mathrm{~cm}$

Let $AD$ be the median of $ \triangle A B C$ and $O$ be the centre.

$AO=\frac{2}{3}AD$

$32=\frac{2}{3}AD$

This implies,

$A D=48 \mathrm{~cm}$

In $\triangle ABD$,

$A B^{2}=A D^{2}+BD^{2}$

$A B^{2}=(48)^{2}+(\frac{A B}{2})^{2}$

$(AB)^2-\frac{A B^{2}}{4}=(48)^{2}$

$\frac{3AB^2}{4}=48\times48$

$AB=\frac{48 \times 2}{\sqrt{3}}$

$=\frac{96}{\sqrt{3}}$

$=32 \sqrt{3}\ cm$

Area of equilateral triangle $\triangle ABC=\frac{\sqrt{3}}{4}(32 \sqrt{3})^{2}$

$=\frac{\sqrt{3}}{4} \times 32 \times 32 \times 3$

$=768 \sqrt{3} \mathrm{~cm}^{2}$

Area of the circle $=\pi r^{2}$

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

$=\frac{22}{7} \times 1024$

$=\frac{22528}{7} \mathrm{~cm}^{2}$

Area of the design $=$ Area of the circle $-$ Area of $\triangle A B C$

$=(\frac{22528}{7}-768 \sqrt{3}) \mathrm{cm}^{2}$

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