Question

A circle is inscribed in an equilateral triangle whose side is 12 cm.Find the radius of the circle and the area of the shaded part

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ 1.radius \: (r) \: of \: incircle \: \\ 2.area \: of \: shaded \: part \\ \\ so \: here \: \\ △ABC \: is \: an \: equilateral \: triangle \\ and \: it \: circumscribes \: given \: circle \\ ie \: \: given \: circle \: is \: inscribed \: in \: △ABC \\ \\ so \: as \: it \: is \: an \: inscribed \: circle \: \\ it \: can \: be \: called \: as \: incircle \\ and \: its \: radius \: as \: inradius \: \\ \\ so \: we \: know \: that \\ area \: o f\: equilateral \: triangle \: = \frac{ \sqrt{3} }{4} \times side {}^{2} \\ \\A(△ABC) = \frac{ \sqrt{3} }{4} \times (12) {}^{2} \\ \\ = \frac{ \sqrt{3} }{4} \times 144 \\ \\ = 36 \times \sqrt{3} \\ \\ A(△ABC) = 36 \sqrt{3}.cm {}^{2} \: \: \: \: \: \: \: \: (1) \\ \\ similarly \: then \\ perimeter \: of \: △ABC = 4 \times side \\ = 4 \times 12 \\ = 48.cm \\ \\ inradius \: of \: \: triangle = \\ \frac{2 \times area \: of \: circumsribed \: triangle}{perimeter \:of \: that \: triangle \: } \\ \\ = \frac{2 \times A(△ABC) }{perimeter \: of \: △ABC} \\ \\ = \frac{2 \times 36 \sqrt{3} }{48} \\ \\ = \frac{72 \sqrt{3} }{48} \\ \\ r= \frac{3 \sqrt{3} }{2} . \: cm$

$so \: hence \: then \\ \\ area \: of \: incircle = \pi.r {}^{2} \\ \\ = \pi \times ( \: \frac{3 \sqrt{3} }{2} \: ) {}^{2} \\ \\ = \frac{27\pi}{4} \: .cm {}^{2} \: \: \: \: \: \: \: \: \: (2)$

$thus \: then \\ area \: of \: shaded \: region = \\ area \: of \: A(△ABC) \: - area \: of \: incircle \\ \\ = 36 \sqrt{3} - \frac{27\pi}{4} \\ \\ = \frac{(36 \sqrt{3} \times 4) - 27\pi }{4} \\ \\ = \frac{144 \sqrt{3} - (27 \times 3.14)}{4} \\ \\ = \frac{(144 \times 1.73) - 84.78}{4} \\ \\ = \frac{249.12 - 84.78}{4} \\ \\ = \frac{164 .34}{4} \\ \\ = 41.085 \: cm {}^{2}$