# A solid has a square base of side length $s.$ The upper edge is parallel to the base and has length $2s.$

I was studying for some quizzes when a wild question appears. It goes like this:

A solid has a square base of side length $s.$ The upper edge is parallel to the base and has length $2s.$ All other edges have lenght $s.$ Given that $s = 6\sqrt{2}$, what's the volume of solid?

My work:

It has a triangular prism, and I noticed that those red and green-lined solid is actually a part of a single triangular prism being cut diagonally.

Now I will let $V_1$ be the volume of a triangular prism , $V_2$ be the volume of green-lined solid, and $V_3$ be the volume of the red-lined solid. The total volume would be: $V_{total} = V_1 + V_2 + V_3$

Getting now the $V_1:$

$$V_1 = (area \space of \space the \space base)(height)$$

The area of the equilateral triangle base is $A = \frac{\sqrt{3}}{4} s^2.$ The height would be $s.$ Plugging it into $V_1:$

$$V_1 = \left ( \frac{\sqrt{3}}{4} s^2 \right)(s) = \frac{\sqrt{3}}{4} s^3$$

Getting now the $V_2:$

I did mention that the volume $V_2$ and $V_3$ is a a part of a single triangular prism being cut diagonally. With that in mind, let that the volume of that triangular prism be $V.$

Getting now the $V:$

$$V = (area \space of \space the \space base)(height)$$

The area of the equilateral triangle base is $A = \frac{\sqrt{3}}{4} s^2.$ The height would be $\frac{1}{2} s$. Plugging it into $V:$

$$V_1 = \left ( \frac{\sqrt{3}}{4} s^2 \right) \left( \frac{1}{2} s \right) =\left( \frac{\sqrt{3}}{8} \right) s^3$$

Then, $V_2 = \frac{V}{2} = \frac{\sqrt{3}}{16} s^3$

The value of $V_3$ is the same as $\frac{V}{2}.$

Then, the total volume of the solid would be: $$V_{total} = V_1 + V_2 + V_3$$ $$V_{total} = \frac{\sqrt{3}}{4} s^3 + \frac{\sqrt{3}}{16} s^3 + \frac{\sqrt{3}}{16} s^3$$ $$V_{total} = \frac{3 \sqrt{3}}{8} s^3$$

Since it was the given on the problem that $s = 6\sqrt{2},$ the volume would be $396.82$ cubic units.

But according to my notes, the volume would be 288 cubic units. Where did I messed up?

Update: Someone pointed out that the correct figure would be like this:

How did you get the volume of the colored solids (the green and the red ones)? I understand that the base having sides $s, s,$ and $s.$ would be the base of the colored solid.

• The problem is horribly flawed: There are multiple solids having different volumes that satisfy the description given. The problem can be repaired by saying that the solid has only the minimum of five faces. Aug 29, 2017 at 17:01
• @MarkFischler The problem has an illustration with it......so I think it just fine.....I just copied the figure..... Aug 29, 2017 at 17:10
• @anderstood It was my habit:-).......I sometimes forgot that the figure weren't mine...... Aug 29, 2017 at 17:13
• @anderstood It seems $2s$ does correspond to the edge of volume. I just copied the figure, so its fine........The figure is basically like a Toblerone box and its ends were cut diagonally that is described in the problem above....XD Aug 29, 2017 at 17:23
• Do the red and green solids placed side to side form a regular tetrahedron of side $s$? Aug 29, 2017 at 17:29

The bases of prism are not equilateral triangles: notice that only one of their sides is an edge of the solid. The other two sides have length $x$, with: $$x^2+\left({s\over2}\right)^2=s^2.$$
As a matter of fact, computing volumes as explained above I get $$V={\sqrt2\over3}s^3=288.$$
• That $x$ and $s$ are in the middle part of the figure, that black-lined solid? Aug 29, 2017 at 17:17
• "Edges of a polyhedron" is a quite well defined thing, I presume. If the problem says that "all other edges have length $s$" I come then to the conclusion that the diagram is wrong, because it marks with length $s$ a red segment which is not an edge. That red segment should have instead length $x$. Aug 29, 2017 at 17:38
• That's right. I was confused because, based on the drawing I had $A(s+s/3)$ which was not $288$. Indeed the drawing is wrong. Aug 29, 2017 at 17:53