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6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. Find the area of the design.

  

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Assume the centre of the circle to be point C and AD as the median of the equilateral triangle.

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

 32\ =\ \frac{2}{3}AD

AD\ =\ 48\ cm

Consider \Delta ABD,

AB^2\ =\ AD^2\ +\ BD^2

AB^2\ =\ 48^2\ +\ \left ( \frac{AB}{2} \right )^2

AB\ =\ 32\sqrt{3}\ cm

Thus the area of an equilateral triangle  =\ \frac{\sqrt{3}}{4}\times \left ( 32\sqrt{3} \right )^2

=\ 768\sqrt{3}\ cm^2

The area of the circle =\ \pi r^2\ =\ \pi\times 32^2

=\ \frac{22528}{7} cm^2

Hence the area of the design =\ \left ( \frac{22528}{7}\ -\ 768 \sqrt{3} \right )\ cm^2

Posted by

Devendra Khairwa

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