In a given figure ABC is …
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Sia ? 6 years, 5 months ago
In a right triangle BAC,
BC2 = AB2 + AC2 …….. By Pythagoras theorem
= (14)2 + (14)2 = 2(14)2
{tex}\Rightarrow{/tex} BC = {tex}14\sqrt2{/tex} cm
{tex}\Rightarrow{/tex} Radius of the semicircle
= {tex}\;\frac{14\sqrt2}2{/tex} cm = {tex}7\sqrt2{/tex} cm
Required area = Area BPCQB
= Area BCQB – Area BCPB
= Area BCQB – (Area BACPB - Area of {tex}\Rightarrow{/tex}BAC)
{tex}= \;\frac{180}{360}\mathrm\pi\left(7\sqrt2\right)^2\;-\left[\frac{90}{360}\mathrm\pi(14)^2\;-\frac{14\times\;14}2\right]{/tex}
{tex}\;\frac12\times\frac{22}7\times98\;-\left[\frac14\times\frac{22}7\times196\;-98\right]{/tex}
= 154 – (154 – 98) = 98 cm2
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