Red bold is not true... PS+SR = PR... so if you square (PS+SR)^2 = PR2... not what you have written

simplest way to solve this problem is sin60 = \(\sqrt{3}\)/2 = QS/PQ = 4\(\sqrt{3}\)/PQ
==> PQ = 8
so perimeter = 24
choice C.

Trigonometry is not tested on the GMAT, which means that EVERY GMAT geometry question can be solved without it.

In the figure shown, the length of line segment QS is \(4\sqrt{3}\). What is the perimeter of equilateral triangle PQR?
A. \(12\)
B. \(12 \sqrt{3}\)
C. \(24\)
D. \(24 \sqrt{3}\)
E. \(48\)

Since triangle PQR is equilateral then its all angles equal to 60°. So, right triangle QSR is a 30°-60°-90° right triangle (with angle R equal to 60°). Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\) and the leg opposite 60° (QS) corresponds with \(\sqrt{3}\), which makes QR equal to \(4\sqrt{3}*\frac{2}{\sqrt{3}}=8\) (\(\frac{QS}{QR}=\frac{\sqrt{3}}{2}\) --> \(QR=QS*\frac{2}{\sqrt{3}}=8\)).

Thus the perimeter is 3*8=24.

Answer: C.

For more on this subject check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
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Altitude of a equilateral triangle is \({\sqrt{3} * side}/2\)

\({\sqrt{3} * side}/2\) = \(4\sqrt{3}\).

So sides are 8

Perimeter of an equilateral triangle is 3* sides = 3*8 =>24

Hence answer is (C)
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