+44 Let $ABCD$ be a regular tetrahedron. Let $E$, $F$, $G,$ $H$ be the centers of faces $BCD$, $ACD$, $ABD$, $ABC$, respectively. The volume of pyramid $DEFG$ is $18.$ Find the volume of pyramid $EFGH$
+897 This isn't too hard of a question once you realize how to do it!
\(\Delta EFG \cdot \frac{2h}{3}=18\\ \Delta EFG =\frac{3\cdot 18}{2h}\\ [{EFGH}]=\Delta EFG\cdot \frac{1}{3}h\\ [{EFGH}]=\frac{3\cdot 18}{2h}\cdot \frac{1}{3}h\\ [{EFGH}]=9\)
The key is realizing that the pyramid is just 1/2 of EFG times the height!
Thanks! :)
