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?

enter image description here

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:

enter image description here

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.

  • 1
    $\begingroup$ 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. $\endgroup$ Aug 29, 2017 at 17:01
  • $\begingroup$ @MarkFischler The problem has an illustration with it......so I think it just fine.....I just copied the figure..... $\endgroup$ Aug 29, 2017 at 17:10
  • $\begingroup$ @anderstood It was my habit:-).......I sometimes forgot that the figure weren't mine...... $\endgroup$ Aug 29, 2017 at 17:13
  • $\begingroup$ @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 $\endgroup$ Aug 29, 2017 at 17:23
  • $\begingroup$ Do the red and green solids placed side to side form a regular tetrahedron of side $s$? $\endgroup$ Aug 29, 2017 at 17:29

1 Answer 1


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: $$ x^2+\left({s\over2}\right)^2=s^2. $$

In addition, green and red solids should be viewed as two equal pyramids with triangular base, because they are not half of some prism.

As a matter of fact, computing volumes as explained above I get $$ V={\sqrt2\over3}s^3=288. $$

  • $\begingroup$ That $x$ and $s$ are in the middle part of the figure, that black-lined solid? $\endgroup$ Aug 29, 2017 at 17:17
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    $\begingroup$ "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 $x$. $\endgroup$ Aug 29, 2017 at 17:38
  • $\begingroup$ 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. $\endgroup$
    – anderstood
    Aug 29, 2017 at 17:53
  • $\begingroup$ Maybe make a nice figure based from your insight:-) and thanks for helping me out, man! $\endgroup$ Aug 30, 2017 at 2:51
  • $\begingroup$ I've updated my post, adding a correct figure to the problem above. I was confused on how you get the volume of the colored solids above......How'd you do that? $\endgroup$ Sep 5, 2017 at 2:46

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