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Circumcircle

A triangle ha...

Question

A triangle has vertices $A, B$ and $C$ and the respective opposite sides have lengths $a, b$ and $c$ . This triangle is inscribed in a circle of radius $R$ . If $b = c = 1$ and the altitude from $A$ to side $B C$ has length $\sqrt{\frac{2}{3}}$ , then $R$ equals.

\frac{1}{\sqrt{3}}

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\frac{2}{\sqrt{3}}

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\frac{\sqrt{3}}{2}

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\frac{\sqrt{3}}{2 \sqrt{2}}

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Solution

The correct option is D $\frac{\sqrt{3}}{2 \sqrt{2}}$
Given $b = c = 1$
$l = \sqrt{\frac{2}{3}}$
Area of $△ l e = \frac{1}{2} . B c . d .$
$△ = \frac{1}{2} a . \sqrt{\frac{2}{3}}$
$a = \frac{2 △ \sqrt{3}}{\sqrt{2}} = \frac{2 \sqrt{3}}{\sqrt{2}} △$
we have to find $^{'} R^{'}$ which is circumradius
$R = \frac{a b c}{4 △} = \frac{(\frac{2 \sqrt{3}}{\sqrt{2}} △) .1 .1}{4 △}$
$\frac{\sqrt{3}}{2 \sqrt{2}}$

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