IMO C
h=a*(sqrt3)/2;a=side;h=height
4*sqrt(3)=a*sqrt(3)/2
a=8
perimeter=3*8=24
GMATprep: geometry
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xcusemeplz2009
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ershovici
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The answer is C
But the solution is much easier.
When we have a right triangle with one side equal x*3/2 the ather side is equal to x, and the hipotenuse 2x - so PS = 4, PQ = 8 - perimeter is 24
But the solution is much easier.
When we have a right triangle with one side equal x*3/2 the ather side is equal to x, and the hipotenuse 2x - so PS = 4, PQ = 8 - perimeter is 24
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briantime
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Since we have an equilateral triangle, the angles are all equal (60°).In the figure shown, the length of line segment QS is 4 sqrt(3). What is the perimeter of equilateral triangle PQR?
- 12
- 12 sqrt(3)
- 24
- 24 sqrt(3)
- 48
Angle QPS = 60
Angle QSP = 90
Therefore, PQS = 180 - 60 - 90 = 30 OR PQS = 1/2 * PQR = 1/2 * 60 = 30
In a 30° - 60° - 90° triangle, the sides have length ratios of 1 : sqrt(3) : 2.
Therefore, the ratios of our triangle are 4 : 4 sqrt(3) : 8.
2*4 + 2*8 = 24
Since this is an equilateral triangle we know that the length of PS = 1/2 the length of PQ. Let's call the length of PQ a.
By Pythagorean Theorem then we know that
PQ^2 = PS^2 + QS^2
Substituting in a for PQ, a/2 for PS and 4 sqrt 3 for QS we get
a^2 = (a/2)^2 + (4 sqrt 3)^2
or
a^2 = (a^2)/4 + 48
solving for a^2
a^2 = 64
or
a = 8
The perimeter of the triangle is thus 3a or 3 times the length of PQ or 24
By Pythagorean Theorem then we know that
PQ^2 = PS^2 + QS^2
Substituting in a for PQ, a/2 for PS and 4 sqrt 3 for QS we get
a^2 = (a/2)^2 + (4 sqrt 3)^2
or
a^2 = (a^2)/4 + 48
solving for a^2
a^2 = 64
or
a = 8
The perimeter of the triangle is thus 3a or 3 times the length of PQ or 24
jdd147 wrote:Since this is an equilateral triangle we know that the length of PS = 1/2 the length of PQ. Let's call the length of PQ a.
By Pythagorean Theorem then we know that
PQ^2 = PS^2 + QS^2
Substituting in a for PQ, a/2 for PS and 4 sqrt 3 for QS we get
a^2 = (a/2)^2 + (4 sqrt 3)^2
or
a^2 = (a^2)/4 + 48
solving for a^2
a^2 = 64
or
a = 8
The perimeter of the triangle is thus 3a or 3 times the length of PQ or 24
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 ....
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MBA.Aspirant
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