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by gmat_perfect » Wed Jul 07, 2010 12:09 am
The figure above shows a square inscribed within an equilateral triangle where the top of the square is parallel to the bottom of the triangle. If the side X of the square measures 12 inches, then what is the perimeter of the equilateral triangle (in inches)?

Please see the attached file for the figure.

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[/img]
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by kvcpk » Wed Jul 07, 2010 12:29 am
Let the triangle be PQR and the square be ABCD

Now, triangle PAB is equaliateral. Because, the angles are all 60 degrees.
Why?
angle(AQD) = 60
angle(BAD)=90
So Angle(PAB)=60

So PA=PB=12

Now, in triangle AQD, AD=12 and angle AQD = 60
So AQ = 12 / sin60 = 24/root(3)=8root(3)
QD = 12/tan60 = 12/root(3)=4root(3)

Similarly, BR=AQ and CR=QD

Therefore, perimeter = 2(12+8root(3)+4root(3)) +12
36+24root(3)

What is OA?
Last edited by kvcpk on Wed Jul 07, 2010 12:33 am, edited 1 time in total.
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by mj78ind » Wed Jul 07, 2010 12:32 am
3*(8sqrt3+12) .......

OR

12 * (2sqrt3 + 3)

OA pls
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by gmatmachoman » Wed Jul 07, 2010 1:13 am
kvcpk wrote:Let the triangle be PQR and the square be ABCD

Now, triangle PAB is equaliateral. Because, the angles are all 60 degrees.
Why?
angle(AQD) = 60
angle(BAD)=90
So Angle(PAB)=60

So PA=PB=12

Now, in triangle AQD, AD=12 and angle AQD = 60
So AQ = 12 / sin60 = 24/root(3)=8root(3)
QD = 12/tan60 = 12/root(3)=4root(3)

Similarly, BR=AQ and CR=QD

Therefore, perimeter = 2(12+8root(3)+4root(3)) +12
36+24root(3)

What is OA?
Let the triangle be ABC and square be PQRS

Tan 60 = PR/AP
sqrt(3) = 12/AP

AP = 12/V3




We know that AB= AP+PQ+QB

so AB = 12/V3 +12 +12/V3
= 12+24/V3
= 12 +8V3

Perimeter = AB+BC+AC
= 3 *AB
=3 *(12+8V3)
= 36 +24V3
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