In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. Find area of remaining portion of the triangle (Take √3 = 1.732).
98.55 cm²
Area of the equilateral triangle of side 24 cm is \(\frac{\sqrt3}{4}(24)^2 = 144\sqrt3 = 144\times1.732 = 249.408\ \text{cm}^2\).
The inradius of an equilateral triangle of side \(a\) is \(r=\dfrac{a}{2\sqrt3}\); for \(a=24\), \(r=\dfrac{24}{2\sqrt3}=4\sqrt3\ \text{cm}\), so \(r^2=48\).
Area of the inscribed circle is \(\pi r^2 = 48\pi\); using \(\pi=22/7\), this equals \(48\times\frac{22}{7}=\frac{1056}{7}\approx150.857\ \text{cm}^2\).
Area of the remaining (shaded) portion = area of triangle − area of circle \(= 249.408-150.857 = 98.551\approx98.55\ \text{cm}^2\).
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