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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).
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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 $\triangle A B C$
$=(\frac{22528}{7}-768 \sqrt{3}) \mathrm{cm}^{2}$