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

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Senior Manager
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In the figure shown, the length of line segment QS is [#permalink]  23 Jul 2011, 19:47
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Difficulty:

15% (low)

Question Stats:

81% (01:28) correct 19% (00:53) wrong based on 59 sessions
Attachment:

Triangle.png [ 9.81 KiB | Viewed 2811 times ]
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. $$48$$

[Reveal] Spoiler: What am i doing wrong here ?
What am i doing wrong here ?

PQ^2 = QS^2 + PS^2

PQ^2 = PS^2 + 48 ------ (1)

QR^2 = SR^2 + 48 ------(2)

PQ^2 + QR^2 = PS^2 + SR ^2 + 96 ----- 3

But, PS^2 + SR ^2 = PR ^2 ------- 4

Substitute 4 in 3....

PQ^2 + QR^2 = PR^2 + 96 -----5

PQ = PR = QR since equilateral triangle

Therefore PQ^2 = 96

PQ = sqrt (96)

But this doesnt yield the correct answer ....
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Oct 2013, 06:37, edited 3 times in total.
Edited the question and added the OA
Senior Manager
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Re: Perimeter [#permalink]  23 Jul 2011, 20:08
siddhans wrote:
http://img210.imageshack.us/img210/3589/gmatprepr.jpg

[Reveal] Spoiler: What am i doing wrong here ?
What am i doing wrong here ?

PQ^2 = QS^2 + PS^2

PQ^2 = PS^2 + 48 ------ (1)

QR^2 = SR^2 + 48 ------(2)

PQ^2 + QR^2 = PS^2 + SR ^2 + 96 ----- 3

But, PS^2 + SR ^2 = PR ^2 ------- 4

Substitute 4 in 3....

PQ^2 + QR^2 = PR^2 + 96 -----5

PQ = PR = QR since equilateral triangle

Therefore PQ^2 = 96

PQ = sqrt (96)

But this doesnt yield the correct answer ....

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.
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GMAT 2: 690 Q48 V37
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Kudos [?]: 14 [0], given: 24

Re: Perimeter [#permalink]  21 Mar 2012, 03:07
agdimple333 is simple and easy. But how do I solve it without knowing sin (60)?
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Re: Perimeter [#permalink]  21 Mar 2012, 03:34
Expert's post
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 2772 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.
_________________
Re: Perimeter   [#permalink] 21 Mar 2012, 03:34
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