Consider the triangle drawn on a sheet of paper, along with the coordinate axes, whose vertices are the points $(0,0)$, $(34,0)$ and $(16,24)$.
The vertices of your middle triangle are the midpoints of their sides. A triangular pyramid isformed by folding the triangle along the sides of its middle triangle. Determine the volume of this pyramid.(S:$408$)
I calculated the area of the base but how to find the height of the pyramid?
$$S_{base}=\dfrac{DE\cdot h}{2}=\dfrac{12\cdot 17}{12}=102.$$
Best Answer
We claim that the upper vertex of the pyramid is $V=(16,12,12)$ (and the face $CDE$ is orthogonal to the base $DEF$). It suffices to verify that the distance between $V$ and the vertices of the base, i. e. $D=(8,12,0)$, $E=(25,12,0)$ and $F=(17,0,0)$: $$\begin{align} |VD|^2&=8^2+0^2+12^2=8^2+12^2=|AD|^2=|CD|^2\\ |VE|^2&=9^2+0^2+12^2=9^2+12^2=|CE|^2=|BE|^2\\ |VF|^2&=1+12^2+12^2=289=17^2=|BF|^2=|AF|^2. \end{align}$$ It follows that $h=12$ and $$V=\frac{S_{base}\cdot h}{3}=\frac{102\cdot 12}{3}=408.$$
More generally we may use use the Cayley-Menger Formula with $n=3$: $$\begin{align} V^2&=\frac{1}{(3!)^2\cdot 2^3}\det\begin{bmatrix}0 & d_{01}^2 & d_{02}^2 & d_{03}^2 & 1\\ d_{01}^2 & 0 & d_{12}^2 & d_{13}^2 & 1\\ d_{02}^2 & d_{12}^2 & 0 & d_{23}^2 & 1\\ d_{03}^2 & d_{13}^2 & d_{23}^2 & 0 & 1\\ 1 & 1& 1& 1 &0 \end{bmatrix}\\ &=\frac{1}{288}\det\begin{bmatrix}0 & 9^2+12^2 & 17^2 & 8^2+12^2 & 1\\ 9^2+12^2 & 0 & 8^2+12^2 & 17^2 & 1\\ 17^2 & 8^2+12^2 & 0 & 9^2+12^2 & 1\\ 8^2+12^2 & 17^2 & 9^2+12^2 & 0 & 1\\ 1 & 1& 1& 1 &0 \end{bmatrix}=408^2 \end{align}$$ where $d_{i j}$ is the Euclidean distances between vertices $V_i$ and $V_{j}$.