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Clearly, {tex}BD=\frac{BC}{2}=\frac{a}{2}{/tex} [In an equilateral triangle , altitude bisects the base]
In {tex}\triangle ABC{/tex}, {tex}\angle ADB=90^o{/tex}
Using pythgoras theorem,
AB2 = AD2 + BD2
{tex}\Rightarrow a^2=AD^2+(\frac{a}{2})^2{/tex}
{tex}\Rightarrow a^2=AD^2+(\frac{a^2}{4}){/tex}
{tex}\Rightarrow a^2 - (\frac{a^2}{4})=AD^2{/tex}
{tex}\Rightarrow (\frac{4a^2 - a^2}{4})=AD^2{/tex}
{tex}\Rightarrow AD=\frac{\sqrt3 a }{2}{/tex}
{tex}\therefore \triangle ABC{/tex} and {tex}\triangle ADE{/tex} are equilateral triangles and so equiangular.
{tex}\therefore \triangle ABC \sim \triangle ADE{/tex}
{tex}\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\frac{AD^2}{AB^2}{/tex}
{tex}\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\frac{\frac{\sqrt{3}a}{2}}{a^2}{/tex}
{tex}\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\frac{3}{4}{/tex}
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Yogita Ingle 6 years, 11 months ago
Area of semicircle = 1/2 πr2
= 1/2 × 3.14 × 10 × 10
= 3.14 × 5 × 10
=31.4 × 5
= 157 cm2
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