Find the area of a circle whose radius is equal to the side of an equilateral triangle of area $9 \sqrt{3} c m^{2}$

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Solution

Area of equilateral triangle $= \sqrt{3 / 4}$ $a^{2}$
$= 9 \sqrt{3}$
$\Rightarrow a = 6 c m$
$∴$ Area of circle $= π \times 6 \times 6 = 36 π c m^{2}$

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Find the area of a circle whose radius is equal to the side of an equilateral triangle of area 9√3cm2

Area of equilateral triangle =√3/4 a2 =9√3⇒a=6cm∴ Area of circle =π×6×6=36πcm2

Find the area of a circle whose radius is equal to the side of an equilateral triangle of area $9 \sqrt{3} c m^{2}$

Area of equilateral triangle $= \sqrt{3 / 4}$ $a^{2}$
$= 9 \sqrt{3}$
$\Rightarrow a = 6 c m$
$∴$ Area of circle $= π \times 6 \times 6 = 36 π c m^{2}$