Find a pyramid containing a given cuboid under some constraints.

Question

Can someone help out with this geometry problem plase?

I have a pyramid $P$ with square basis and each of its 4 triangles is equilateral. I also have cuboid $C$ with height 25 and width 50, something like this:

So the volume of $C$ is $25 . 50 . 50 = 62500$

I need to find out how big the square base of $P$ must be in order for $P$ to fully contain $C$ with the bottom of $C$ being at distance $26$ from the square base of $P$.Basically the cube must be fully inside the pyramid and to hand $26$ units above the pyramid's floor.

Can someone give advice on how to solve this?I have no other info on the pyramid.Is it even possible to solve this with so little information?

Answer 1

The suggested cross section is shown. Let $h$ be the height of the pyramid above the top of $C$and let $w$ be half the width of the base. By similar triangles, $$ \frac{w}{h + 51} = \frac{25}{h}, \quad\text{or}\quad w = 25\left(1 + \frac{51}{h}\right). $$

Answer 2

Solution

Let the pyramid $ABCDE$ whose sides have length $L$. See figure 1.

Figure 1

The right angled triangles $\triangle ENF$ and $\triangle EMB$ are isosceles. (Check it out).

Using Pythagoras we get: $$GF=50 \sqrt{2}$$ and $$DB=L \sqrt{2}$$.

So $NF=25 \sqrt{2}$ and $MB=\frac{L \sqrt{2}}{2}$.

As $EM=MB$ we get: $$L=50+51 \sqrt{2}$$