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

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Re: In the figure shown, the length of line segment QS is [#permalink]
agdimple333 is simple and easy. But how do I solve it without knowing sin (60)?
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Re: In the figure shown, the length of line segment QS is [#permalink]
Impenetrable wrote:
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. $$48$$
Attachment:

Triangle.png [ 9.81 KiB | Viewed 19561 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 $$4\sqrt{3}*\frac{2}{\sqrt{3}}=8$$ ($$\frac{QS}{QR}=\frac{\sqrt{3}}{2}$$ --> $$QR=QS*\frac{2}{\sqrt{3}}=8$$).

Thus the perimeter is 3*8=24.

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

Hope it helps.
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Re: In the figure shown, the length of line segment QS is [#permalink]
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Here is how I solved it.

An equilateral triangle has 3 equal sides, thus each angle is 60 degrees. Equilateral PQR has a line right down the middle, dividing the triangle PQR into two separate 30-60-90 triangles.

The properties of the 30-60-90 triangles are as follows:
Side with an angle of 30 degrees has a side length of $$x$$
Side with an angle of 60 degrees has a side length of $$x\sqrt{3}$$
Side with an angle of 90 degrees has a side length of $$2x$$

Knowing that $$4\sqrt{3}$$ = $$x\sqrt{3}$$ -----> $$4 = x$$ (the side length of the 60 degree angle), you can easily calculate the rest of the other sides of the triangle.
So, putting it all together the perimeter of Equilateral PQR (see attached figure) = $$8 + 8 + 4 + 4 = 24$$
Attachments

Capturegmat.JPG [ 19.37 KiB | Viewed 13244 times ]

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Re: In the figure shown, the length of line segment QS is [#permalink]
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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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Re: In the figure shown, the length of line segment QS is [#permalink]
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siddhans wrote:
Attachment:
Triangle.png
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. $$48$$

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