# SAT pyramid problem

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• May 30th 2012, 01:50 PM
Mat724
SAT pyramid problem
Before i start, there is a pyramid with vertex V and altitude h and edge of e and the edge of the base is m.

NOTE: FIGURE NOT DRAWN TO SCALE.
The pyramid shown above has an altitude h and a square base of side m. The four edges that meet at V, the vertex or the pyramid, each have length e. If e=m, what is the value of h in terms of m? Answer: M times [square root of 2]

Help please!
• May 30th 2012, 02:55 PM
HallsofIvy
Re: SAT pyramid problem
The base is a square with sides of length m so drawing a diagonal divides the square into two right triangles with leg lengths m. By the Pythagorean theorem, the diagonal has length d given by $d^2= m^2+ m^2= 2m^2$ so that $d= m\sqrt{2}$. If you drop a perpendicular from the top of the pyramid to the center of the base you have a right triangle with one leg of length h, one leg of length [tex]m\sqrt{2}/2[/itex] (half of a diagonal) and hypotenuse of length e. Put those into the Pythagorean theorem.