Question

The side of equilateral triangle is 7 root 3 cm. a circle is drawn touching to side BC and producd side AB and BC . Find the area of circle.​

Answer 1

Answer:

hey here is your solution

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refer the reference diagram uploaded above

Step-by-step explanation:

$to \: find = \\ area \: of \: inscribed \: circle \: (incircle) \\ \\ so \: here \ \\ △ABC \: is \: an \: equilateral \: triangle \: which \\ is \: inscribed \: in \: adjoining \: circle \\ ie \: \: the \: given \: circle \: circumscribes \: \: △ABC \\ \\ hence \: the \: given \: circle \: is \: circumcircle \: of \: △ABC \\ and \: OB \: is \: its \: \: circumradius \\ \\ so \: we \: know \: that \\ circumradius \: of \: an \: equilateral \: triangle = \frac{side}{ \sqrt{3} } \\ \\ hence \: accordingly \\ \\ OB = \frac{AB }{ \sqrt{ 3} } \\ \\ = \frac{7 \sqrt{3} }{ \sqrt{3} } \\ \\ OB =r = 7.cm$

$thus \: then \\ area \: of \: circle = \pi.r {}^{2} \\ \\ = \frac{22}{7} \times (7) {}^{2} \\ \\ = \frac{22}{7} \times 7 \times 7 \\ \\ = 22 \times 7 \\ \\ area \: of \: circle = 154.cm {}^{2}$