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Tetrahedron Lateral Surface Area

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by milanproda » Mon Dec 27, 2010 12:42 pm
The lateral surface area, A, of a regular tetrahedron is related to the length of its edge a by the following formula A= √3a^2. What is the difference in the length of the edge of the tetrahedron with the area of 625√3 and the length of the edge of the tetrahedron with the area 196√3?

A-3
B-9
C-11 ANSWER
D-90
E- 150
Milan Prodanovic
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by Tani » Mon Dec 27, 2010 3:01 pm
if √3a^2 = 625√3
then 625 = a^2
and a = 25

if √3a^2 = 196√3
the 196 = a^2
and a = 14

25-14 = 11

If you don't recognize the perfect squares up to 25, simply start factoring the numbers. Clearly 625 is divisible by 5 -- which gives you 125. Divide by 5 again and you get 25 = the square root.

Now, looking at the answers, if the larger is 25, the other side can only be 3, 9 or 11 less than 25. Squaring 22, 16, and 14 you find that 14^2 = 196 and 11 is your answer.
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by milanproda » Mon Dec 27, 2010 3:07 pm
I believe that the algebra confused me. I know from experience that 625 is 5^3, and eventually I figured out that 196 was 14^2. But I am not sure how I to apply those numbers into the formula. For example, I recognize that 196√3 is 14x14x3. However, since 625 is a cube, how can it be applied to the A= √3a^2 formula?
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by Tani » Mon Dec 27, 2010 3:37 pm
125 is 5^3; 625 is 5^4.
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