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Ncert ex 12.2 Q 6 with explanation anyone please help |
| Answer» Here, Area of equilateral triangle =\xa0{tex}{{\\sqrt 3 } \\over 4}{\\left( {{\\rm{Side}}} \\right)^2}{/tex}=\xa0{tex}{{\\sqrt 3 } \\over 4}{\\left( {{\\rm{15}}} \\right)^2}{/tex}\xa0= 97.425 sq. cmArea of circle =\xa0{tex}\\pi {r^2} = \\pi {\\left( {15} \\right)^2}{/tex}\xa0= 706.5 sq. cmArea of minor sector =\xa0{tex}{1 \\over 6} \\times 706.5{/tex}\xa0= 117.75 sq. cmTherefore, Area of minor segment = 117.75 - 97.425 = 20.325 sq. cmAnd, Area of major segment = 706.5 - 20.325 = 686.175 sq. cm | |