1. ## Pyramid Volume Problem

If there are two identical pyramids except one is twice as large as the other (i.e. all linear dimension are twice those of the other pyramid), how much does more the larger one weigh? Assume they are both solid and made of the same stone. [hint: weight is proportional to volume.]
_______________________________________times as much

I've tried this problem a few time without getting it correct, since the pyramid is twice as large, doesn't that mean that the volume would increase by a factor of 4?

2. 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 ${V_A} = \frac{1}{3}bh$, and your 'doubled' pyramid B has volume ${V_B} = \frac{1}{3}\left( {2b} \right)\left( {2h} \right) = \frac{4}{3}bh = 4{V_A}$ - 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 ${V_A} = \frac{1}{3}b{h_A} = \frac{b}{3}\sqrt {{s^2} - {{\left( {\frac{b}{2}} \right)}^2}}$ and ${V_B} = \frac{1}{3}(2b){h_B} = \frac{{2b}}{3}\sqrt {4{s^2} - {b^2}}$, 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.

3. 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 ${V_A} = \frac{1}{3}bh$, and your 'doubled' pyramid B has volume ${V_B} = \frac{1}{3}\left( {2b} \right)\left( {2h} \right) = \frac{4}{3}bh = 4{V_A}$ - 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 ${V_A} = \frac{1}{3}b{h_A} = \frac{b}{3}\sqrt {{s^2} - {{\left( {\frac{b}{2}} \right)}^2}}$ and ${V_B} = \frac{1}{3}(2b){h_B} = \frac{{2b}}{3}\sqrt {4{s^2} - {b^2}}$, 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.
Unfortunately, I tried 4 and it seems that it is not the correct answer