Question

If ABC is an equilateral triangle having sides 24cm. If the centroid O is the centre of the circle, then radius of circle is *

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: ABC \: is \: an \: equilateral \: triangle \: of \: side \: 24.cm \: \: which \: is \: inscribed \: in \: circle \: with \: centre \: O$

$so \: we \: can \: say \: that \:the \: given \: circle \: circumscribes \: equilateral \: triangle \: ABC \\ so \: thus \: AO \: would \: be \: our \: circumradius \\ so \: we \: know \: that \: \\ circumradius \: of \: an \: equilateral = side \: \div \sqrt{3} \\ thus \: then \: accordingly \\ AO = AB \div \sqrt{3} \\ = 24 \div \sqrt{3} \\ rationalising \: the \: denominator \\ we \: get \: thus \\ AO = 24 \div \sqrt{3} \times \sqrt{3} \div \sqrt{3} \\ = 24 \sqrt{3} \times \sqrt{3} \div \sqrt{3} \\ = 24 \sqrt{3} \div \sqrt{3} \times \sqrt{3} \\ = 24 \sqrt{3} \div 3 \\ = 8 \sqrt{3} \\ = 8 \times 1.73 \\ = 13.84.cm$

$hence \: \: radius \: of \: equilateral \: triangle \: ABC \: of \: side \: 24.cm \: is \: 8 \sqrt{3} .cm \: ie \: approximately \: 13.84.cm$