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

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Senior Manager
Joined: 29 Jan 2011
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In the figure shown, the length of line segment QS is  [#permalink]

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Updated on: 05 Oct 2013, 06:37
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00:00

Difficulty:

5% (low)

Question Stats:

80% (00:37) correct 20% (01:04) wrong based on 215 sessions

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Attachment:

Triangle.png [ 9.81 KiB | Viewed 8836 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. $$48$$

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 ....

Originally posted by siddhans on 23 Jul 2011, 19:47.
Last edited by Bunuel on 05 Oct 2013, 06:37, edited 3 times in total.
Edited the question and added the OA
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23 Jul 2011, 20:08
siddhans wrote:
http://img210.imageshack.us/img210/3589/gmatprepr.jpg

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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21 Mar 2012, 03:07
agdimple333 is simple and easy. But how do I solve it without knowing sin (60)?
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21 Mar 2012, 03:34
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 7913 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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In the figure shown, the length of line segment QS is  [#permalink]

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19 Apr 2016, 05:31
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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19 Apr 2016, 07:30
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

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

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19 Dec 2018, 08:39
Top Contributor
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. $$48$$

Let's first add the given information to the diagram.
If ∆PQR is an equilateral triangle, then the 3 angles are each 60°

At this point, we can see we have a special 30-60-90 right triangle, which we can compare with the BASE 30-60-90 right triangle

When we compare the sides that are opposite the 60° angle, we can see that 4√3 is 4 times the value of √3
So, the magnification factor is 4.
This means the other 2 lengths will be 4 times the sides of the BASE 30-60-90 right triangle

Now that we know that each side has length 8, the perimeter = 8 + 8 + 8 = 24

Cheers,
Brent
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Re: In the figure shown, the length of line segment QS is &nbs [#permalink] 19 Dec 2018, 08:39
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