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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.

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1 Answer 1

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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 $$ \frac{14(h_1+h_2)}{3}=\frac{14\times 3}{3}=14. $$

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