Question 239597
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Depends.  Is the height you gave the slant height, that is the measure from the apex of the pyramid down the face of one side to the midpoint of one edge of the base, or is it the pyramid height from the apex to the center of the square base?


If it is the slant height, then just use the triangle area formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{bh}{2}\ =\ \frac{214\,\cdot\,179}{2}]


If it is the actual height of the pyramid, then you need to incorporate Pythagoras to get the altitude of the triangle:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{bh}{2}\ =\ \frac{214\,\cdot\,\sqrt{179^2+107^2}}{2}]


Either way, you get to do your own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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