The figure shows a square inscribed within an equilateral triangle. If the side X of the square measures 12 inches, then what is the perimeter of the equilateral triangle (in inches)?
A. 36
B. 36 x sqrt(3)
C. 36 x sqrt(3) + 24
D. 24 x sqrt(3) + 36
E. 72
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note: sqrt(x) stands for the square root of x
source: http://www.800score.com/gmat/
6 comments:
The main point here is not to "fell " in the trap of thinking that the side of the triangle might be 2 times X, leading to a (wrong) perimeter of 72.
A safer starting point would be to consider that an equilateral triangle has all internal angles equal to 60 degrees, whose tangent equals sqrt(3). Thus, the triangle base will measure X + 2*L, where X/L = sqrt(3), in another form L = X / sqrt(3).
Knowing this, the total perimeter P will be 3*(X + 2*L) = 3*[ X + 2*X/sqrt(3) ] = [2*sqrt(3) + 3]*X. For X = 12 (nches) we get P = 24*sqrt(3) + 36.
The correct answer is the option (D)
BUT I THOUGHT TRIGO WASN'T A PART OF THE GMAT SYLLABUS
Don't need trigo for that one...
The small triangle above the square is equilateral with each side equal to X.
It is equilateral as we can demonstrate all angles = 60 degrees.
Then it's just a matter of finding the hypotenuse (Y) of the 2 rectangle triangles on each side of the squares.
The smaller side of the rectangle triangle is Y/2 - this because all side of the big triangle = X + Y.
Pythagorus tells you that
X^2 + (Y/2)^2 = Y^2
do the maths and you'll find that Y= 8 x sqrt (3)
each side being Y + X
you have 24 x sqrt (3) + 36
You can also use the 30/60/90 triangle calc. if Hypotenuse equals 2Y. Then 12 = Y * sqrt(3), so 12*sqrt(3) = 3Y, so Y = 4*sqrt(3).
Then perimeter = 3*(12 + 8*sqrt(3) or 36 + 24*sqrt(3).
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