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An equilateral triangle is inscribed in a circle of radius 6cm . find its sides ? |
| Answer» Let {tex}\\Delta ABC{/tex} is an equilateral triangle of each side \' 2a\' inscribed in a circle of radius 6 cm and centre O.Draw {tex}AD \\bot BC{/tex} .Then ,BD = DC [In a equilateral triangle, perpendicular bisects the base]Also , O lies on AD [ In a equilateral triangle circumcentre and centroid coincides]Thus , OA : OD = 2 : 1{tex}\\Rightarrow \\frac{6}{OD}=\\frac{2}{1}{/tex}{tex}\\Rightarrow OD=3 cm {/tex}, then AD = 9 cmNow , using pythagoras theorem in {tex}\\triangle ABD{/tex}BA2 = BD 2 + AD2(2a)2 = (a)2 + (9)24a2 = a2 + 813a2 = 81a2 = 27we get {tex}a=3\\sqrt3{/tex}Side of triangle = 2a = {tex}2(3\\sqrt3){/tex} = {tex}6\\sqrt3{/tex}cm | |