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An equilateral triangle of side 6 cm is inscribed in a circle. Then radius of the circle is
2√3 cm
3√2 cm
4√3 cm
√3 cm
Consider the above figure where ABC is an equilateral triangle, AM is a median and O is the point of intersection of the three medians of the triangle and also the center of the inscribing circle. In triangle BAM, angle BAM = a = (1/2)(60) = 30 degrees. In triangle BOM, angle BOM = 2a = 60 degrees BM = 1/2 BC = (1/2)(6) = 3 units radius r = OB = 3/(sin 60) = 2√3 units.
By: MIRZA SADDAM HUSSAIN ProfileResourcesReport error
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