# Triangular Pyramid Volume with Triangle Faces Given

In the triangular pyramid $$WXYZ$$, edge $$WX$$ has length $$3\,\mathrm{cm}$$. The area of face $$WXY$$ is $$15\,\mathrm{cm}^2$$ and the area of face $$WXZ$$ is $$12\,\mathrm{cm}^2$$. These two faces meet each other at $$30^\circ$$ angle. Find the volume of the tetrahedron in $$\mathrm {cm}^3$$.

## 1 Answer

Given that the area $$A_{WXY} = 15$$ and $$WX = 3$$, The corresponding height from the vertex $$Y$$ to the base line $$WX$$ is $$h=10$$. Let $$H$$ be the height of the pyramid from the vertex $$Y$$ to the base surface $$WXZ$$. Then, $$h$$ is at the angle $$30^\circ$$ with the surface $$WXZ$$ and we have

$$H = h\sin 30^\circ = 5$$

Thus, the volume of the pyramid is

$$\frac13 A_{WXZ}\cdot H = \frac13\cdot 12\cdot 5 = 20\>cm^3$$