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  4. Let R be the point (3,7) and let P and Q be two points on the line x+y=5 such that P Q R is an equilateral triangle. Then the area of triangle PQR is :

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Q. Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $P Q R$ is an equilateral triangle. Then the area of $\triangle PQR$ is :

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Solution:

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$\sin 60^{\circ}=\frac{5 / \sqrt{2}}{ a } $
$a =\frac{5 \sqrt{2}}{3}$
Area of $\triangle PQR =\frac{\sqrt{3}}{4} a ^{2}=\frac{25}{2 \sqrt{3}}$