# Way to find volume of the solid

A solid has a square base of side $s$ . The upper edge is parallel to the base and has length $2s$. All other edges have length $s$ . What is the volume of the solid ?

NB : The volume of the tetrahedron with all sides length l is $V = \dfrac{\sqrt2}{12}l^3$

• Obviously all the edges are straight lines... right? Jan 23, 2014 at 8:47
• Yes . You are right . Jan 23, 2014 at 9:05
• is that a section of Square pyramid Jan 23, 2014 at 9:15
• Especially given your "NB", it seems like you're describing a solid formed by slicing a regular tetrahedron by a plane parallel to (and half-way between) two opposite edges. In that case, the volume of the solid is half the volume of a regular tetrahedron of side $2s$.
– Blue
Jan 23, 2014 at 9:31
• Can you explain it ? I have been asked this question by my younger sister . Jan 23, 2014 at 9:34

We can divide the solid in two tetrahedra and a square pyramid. See fig.1. Fig.1

The pyramid height $GH$ has measure $x$, and the measure of $DB$ is the double of $x$. (Check it out).

Using pythagoras in $\triangle ABD$ we get: $$x=\frac{s \sqrt{2}}{2}$$ We know that the volume of the solid $V$ is: $$V=2V_{tetra}+V_{pyr}$$ where $V_{tetra}$ is the volume of each tetrahedron and $V_{pyr}$ is the volume of the square pyramid. Calculating each term we have: $$V=2 \frac{\sqrt{2}}{12}s^3+\frac{1}{3}s^2 \frac{s \sqrt{2}}{2} \Rightarrow$$ $$V= \frac{\sqrt{2}}{3}s^3$$

Especially given your "NB", it seems like you're describing a solid formed by slicing a regular tetrahedron by a plane parallel to (and half-way between) two opposite edges. In that case, the volume of the solid is half the volume of a regular tetrahedron of side $2s$. 