# A composite geometry is made up of cube and a square pyramid frustum. The slanted side AK stretch a side lengths of 5.62 forming an isosceles triangle. Calculate the volume, surface area and sketch the net of the surface area?

May 12, 2016

There are some assumptions that we have to make looking at the picture that are not explicitly specified in the problem. See below

#### Explanation:

1. The side length of a cube is $a = 5.84$
2. The pyramid frustum's top base is a square with a side $b = 3.15$
3. The pyramid before it was truncated was regular square pyramid
4. The plane of truncation (top base of a frustum) is parallel to its bottom base.
5. Side edge of the pyramid frustum is defined as $A K = c = 3.15$

Our major task is to define the height of the pyramid frustum $h$, all the other values are then easily obtainable. After we derive this height in terms of $a$, $b$ and $c$, we will leave all other calculations to the person who suggested this problem as they are just simple substitution of values into known formulas.

First of all, let's project the top base of the pyramid frustum onto the bottom base of this pyramid. On the bottom base we will see this projection as a square centrally located inside a bigger square - the bottom base. We know dimensions of both squares - $a$ outside and $b$ inside.

Using a Pythagorean Theorem, we can find a distance from the projection of point $A$ onto bottom base (let's call it $A '$ - left bottom vertex of an inner square) to point $K$ - left bottom vertex of the outer square:
$A ' {K}^{2} = {\left(\frac{a - b}{2}\right)}^{2} + {\left(\frac{a - b}{2}\right)}^{2} = {\left(a - b\right)}^{2} / 2$

Now, knowing $A K = c$ and $A ' K$, as calculated above, we can use again the Pythagorean Theorem to determine $A A ' = h$ - the height of the pyramid frustum:
$A A ' = h = \sqrt{{c}^{2} - {\left(a - b\right)}^{2} / 2}$

All the values required in the problem are now easily derived, knowing $a$, $b$ and $h$.