## GMAT Prep (Pract2) Equilateral Triangle

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### GMAT Prep (Pract2) Equilateral Triangle

by dferm » Tue May 13, 2008 12:34 pm
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by aatech » Tue May 13, 2008 1:06 pm
Should be C.

Ht of eql triangle is (root3/2)* side = 4root3

solve to get side = 8

perimeter = 8*3= 24

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by dferm » Tue May 13, 2008 1:09 pm
Again not understanding ur logic..

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by aatech » Tue May 13, 2008 1:15 pm
QS is height of the triangle PQR...

Ht of equilateral traingle is calculated by formula

(root3)/2 * side

This is a std formula

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by kaf » Wed Apr 08, 2009 1:43 pm
aatech wrote:QS is height of the triangle PQR...

Ht of equilateral traingle is calculated by formula

(root3)/2 * side

This is a std formula
Can someone please confirm this if this is standard formula

thanks

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by moutar » Wed Apr 08, 2009 2:27 pm
Using Pythagoras (with side length a)

(a/2)^2 + (4root(3))^2 = a^2

(4root(3))^2 = a^2 - a^2/4 = 3a^2/4

Rooting both sides:

4root(3) = a*root(3)/2

Therefore a = 4root(3)*2/root(3) = 8

Perimeter = 3*8 = 24 = C.

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by fercho81 » Wed Apr 08, 2009 3:20 pm
A much easier way to solve it is the following. The problem says the triangle is equilateral, which as you know means all sides are equal. This also means that all sides have 60 degrees angles. QS bisects PR and forms a 90 degree angle, hence PS and SR are equal. Angle Q is cut in half so 60 is now 30.
At this moment you should recognize the 30:60:90 type of triangle being formed. This means that the side opposite the 30 degree angle or PS is x, the side opposite the 60 degree angle or QS will be x&#8730;3 and the side opposite the 90 degree angle or PQ is 2x. This is in the same way that a 45:45:90 triangle is x : x : x&#8730;2. These types of triangles are the most common so you can look them up in the Official Guide 11th edition page 130 or on the web.
Since you know the value of QS= 4&#8730;3 (looks familiar to the value of the side opposite to the 60 degree angle mentioned above 4&#8730;3 = x&#8730;3). Then from here you can determine that x = 4 and that 2x = 8. Since all sides are equal and you know the value for PQ = 2x = 8, then 8 * 3 = 24. Hope this helped.

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by fercho81 » Wed Apr 08, 2009 3:20 pm
A much easier way to solve it is the following. The problem says the triangle is equilateral, which as you know means all sides are equal. This also means that all sides have 60 degrees angles. QS bisects PR and forms a 90 degree angle, hence PS and SR are equal. Angle Q is cut in half so 60 is now 30.
At this moment you should recognize the 30:60:90 type of triangle being formed. This means that the side opposite the 30 degree angle or PS is x, the side opposite the 60 degree angle or QS will be x&#8730;3 and the side opposite the 90 degree angle or PQ is 2x. This is in the same way that a 45:45:90 triangle is x : x : x&#8730;2. These types of triangles are the most common so you can look them up in the Official Guide 11th edition page 130 or on the web.
Since you know the value of QS= 4&#8730;3 (looks familiar to the value of the side opposite to the 60 degree angle mentioned above 4&#8730;3 = x&#8730;3). Then from here you can determine that x = 4 and that 2x = 8. Since all sides are equal and you know the value for PQ = 2x = 8, then 8 * 3 = 24. Hope this helped.

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by krisraam » Wed Apr 08, 2009 4:35 pm
The height of the triangle = 4SQRT(3)

As this is a equilateral triangle

tan60 = 4sqrt(3)/(S/2)

==> S = 8

Perimeter = 24.

Thanks
Raama

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by Bhattu » Sat Apr 11, 2009 12:27 am
Fercho81's response is the best way to do it I think

Line segment QS divides the equilateral triangle 60-60-60 into 2 triangles with angles 30-60-90. The sides of such a triangle are in ratio of 1:sqroot3:2

Therefore 4:4sqroot3:8

Thus you have length of one side 8, multiply that into 3 as all sides are equal.