, 2016

Join $S$ to the midpoint $M$ of $QR$.

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Since $\triangle SQR$ is equilateral with side length 30, then $QM = MR = \frac{1}{2}QR = 15$.

Since $\triangle SQR$ is equilateral, then $SM$ is perpendicular to $QR$.

Since $\triangle PQR$ is isosceles with $PQ=PR$, then $PM$ is also perpendicular to $QR$.

Since $PM$ is perpendicular to $QR$ and $SM$ is perpendicular to $QR$, then $PM$ and $SM$ overlap, which means that $S$ lies on $PM$.

By the Pythagorean Theorem, $$PM = \sqrt{PQ^2 - QM^2} = \sqrt{39^2 - 15^2} = \sqrt{1521 - 225} = \sqrt{1296} = 36$$ By the Pythagorean Theorem, $$SM = \sqrt{SQ^2 - QM^2} = \sqrt{30^2 - 15^2} = \sqrt{900 - 225} = \sqrt{675} = 15\sqrt{3}$$ Therefore, $PS = PM - SM = 36 - 15\sqrt{3}$.

Since $QM$ is perpendicular to $PS$ extended, then the area of $\triangle PQS$ is equal to $\frac{1}{2}(PS)(QM)$.

(We can think of $PS$ as the base and $QM$ as the perpendicular height.)

Therefore, the area of $\triangle PQS$ equals $\frac{1}{2}(36-15\sqrt{3})(15) \approx 75.14$.

Of the given answers, this is closest to 75.