Pyramid with equilateral triangle as a base, length of side of pyramid is $s=3$(not a base side). Plane goes through pyramid, and contains base edge, and is normal to a side of pyramid. If surface area of that cutting through with plane through pyramid is $14$, what is the volume of the pyramid.
My attempt at solution: I know that figure formed by cutting through with plane is a isosceles triangle, and I tried to connect height of that pyramid with a side of the pyramid. But having trouble to find connection, can't really represent it with formula.
Best Answer
Note that by that plane, the pyramid is cut into two tetrahedrons with base area 14 and sum of their heights equal to $s=3$. Since the volume of tetrahedron is given by $\displaystyle V=\frac {Sh} 3$ where $S$ is a base area and $h$ is a height from the base to apex, the volume of the pyramid is $$ \frac{14(h_1+h_2)}{3}=\frac{14\times 3}{3}=14. $$