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agdimple333 is simple and easy. But how do I solve it without knowing sin (60)?

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. \(48\)

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Triangle.png [ 9.81 KiB | Viewed 5234 times ]

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 \(4\sqrt{3}*\frac{2}{\sqrt{3}}=8\) (\(\frac{QS}{QR}=\frac{\sqrt{3}}{2}\) --> \(QR=QS*\frac{2}{\sqrt{3}}=8\)).

In the figure shown, the length of line segment QS is [#permalink]

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19 Apr 2016, 06:31

Height in a an equilateral triangle equals \(\sqrt{3}\)/2 *a, where a is a side of such triangle. Hence a = \(\sqrt{3}\)/2 * a = 4\(\sqrt{3}\)
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