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

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

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New post 23 Jul 2011, 20:47
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Attachment:
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Triangle.png [ 9.81 KiB | Viewed 4049 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)

Adding 1 and 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, 07:37, edited 3 times in total.
Edited the question and added the OA
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Re: Perimeter [#permalink]

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New post 23 Jul 2011, 21:08
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)

Adding 1 and 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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Re: Perimeter [#permalink]

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New post 21 Mar 2012, 04:07
agdimple333 is simple and easy. But how do I solve it without knowing sin (60)?
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Re: Perimeter [#permalink]

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New post 21 Mar 2012, 04:34
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
Triangle.png [ 9.81 KiB | Viewed 3813 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.

Answer: C.

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

Hope it helps.
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In the figure shown, the length of line segment QS is [#permalink]

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

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New post 19 Apr 2016, 08:30
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] 19 Apr 2016, 08:30
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