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25 Feb 2015, 03:00
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B
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Difficulty:
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82% (01:29) correct
18%
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wrong
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A pyramid has a square base of 6 cm, and the four lateral faces are four congruent equilateral triangles. What is the total surface area of the pyramid in square cm?
(A) \(36 + 18*\sqrt{3}\)
(B) \(36 + 36*\sqrt{3}\)
(C) 72
(D) \(72 + 36*\sqrt{3}\)
(E) \(72 + 72*\sqrt{3}\)
Kudos for a correct solution.
(A) \(36 + 18*\sqrt{3}\)
(B) \(36 + 36*\sqrt{3}\)
(C) 72
(D) \(72 + 36*\sqrt{3}\)
(E) \(72 + 72*\sqrt{3}\)
Kudos for a correct solution.
02 Mar 2015, 05:22
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Bunuel
MAGOOSH OFFICIAL SOLUTION:
First of all, of course, the base has an area of 36. For the lateral surfaces, it would be helpful to remember the formula for the area of an equilateral triangle. If that’s unfamiliar, take a look at this post: https://magoosh.com/gmat/2012/gmat-math- ... emorizing/
The area of one equilateral triangle is \(A = \frac{(s^2*\sqrt{3})}{4}\). We know the side of the equilateral triangle must be the same as the square: s = 6. Thus, one of these equilateral triangles has an area of \(A = \frac{(6^2*\sqrt{3})}{4} = 9*\sqrt{3}\). There are four identical triangles, so their combined area is \(A = 36*\sqrt{3}\). Now, add the square base, for a total surface area of \(A = 36 + 36*\sqrt{3}\).
Answer = B.
General Discussion
25 Feb 2015, 07:35
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Congruent equilateral triangles means all the triangles are equal. let it be x. Area of equilateral triangle = \(\frac{x^2 *sqrt(3)}{4}\)= \(\frac{36*sqrt(3)}{4}\)
we have 4 triangles and base so total surface area= 36 + 36*sqrt(3) answer B
we have 4 triangles and base so total surface area= 36 + 36*sqrt(3) answer B
