Originally Posted by

**davidlyness** Your problem could be interpreted one of two ways: either the slant height is doubled, or the perpendicular height is doubled.

Let's look our original pyramid, call it A. I'll assume you have a square-based pyramid. Then let each side of the base of the pyramid be b units, and the slant height (not the perpendicular height h units) be s units.

Your problem could be interpreted one of two ways: either the slant height is doubled, or the perpendicular height is doubled. If it is the perpendicular height, then pyramid A has volume $\displaystyle {V_A} = \frac{1}{3}bh$, and your 'doubled' pyramid B has volume $\displaystyle {V_B} = \frac{1}{3}\left( {2b} \right)\left( {2h} \right) = \frac{4}{3}bh = 4{V_A}$ - therefore, it is four times bigger in terms of volume, and so four times heavier.

The harder way is if the slant height is doubled. Then, you end up with your volumes as $\displaystyle {V_A} = \frac{1}{3}b{h_A} = \frac{b}{3}\sqrt {{s^2} - {{\left( {\frac{b}{2}} \right)}^2}} $ and $\displaystyle {V_B} = \frac{1}{3}(2b){h_B} = \frac{{2b}}{3}\sqrt {4{s^2} - {b^2}} $, which look a bit harder to work out, but the answer again comes out at 4. Therefore, I think your answer of 4 was correct.