# The base of a triangular pyramid is a triangle with corners at (6 ,7 ), (4 ,5 ), and (8 ,7 ). If the pyramid has a height of 6 , what is the pyramid's volume?

Jan 5, 2018

$4$

#### Explanation:

The area of a triangle with vertices $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$, $\left({x}_{3} , {y}_{3}\right)$ is given by the formula:

$A = \frac{1}{2} \left\mid {x}_{1} {y}_{2} + {x}_{2} {y}_{3} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} - {x}_{2} {y}_{1} - {x}_{3} {y}_{2} \right\mid$

Letting $\left({x}_{1} , {y}_{1}\right) = \left(6 , 7\right)$, $\left({x}_{2} , {y}_{2}\right) = \left(4 , 5\right)$ and $\left({x}_{3} , {y}_{3}\right) = \left(8 , 7\right)$ we find that the area of the base of the given pyramid is:

$\frac{1}{2} \left\mid \textcolor{b l u e}{6} \cdot \textcolor{b l u e}{5} + \textcolor{b l u e}{4} \cdot \textcolor{b l u e}{7} + \textcolor{b l u e}{8} \cdot \textcolor{b l u e}{7} - \textcolor{b l u e}{6} \cdot \textcolor{b l u e}{7} - \textcolor{b l u e}{4} \cdot \textcolor{b l u e}{7} - \textcolor{b l u e}{8} \cdot \textcolor{b l u e}{5} \right\mid$

$= \frac{1}{2} \left\mid 30 + 28 + 56 - 42 - 28 - 40 \right\mid = 2$

Another way of seeing this is by considering the points $\left(6 , 7\right)$ and $\left(8 , 7\right)$ as the base of the triangle, which is of length $2$. Then the apex of the triangle at $\left(4 , 5\right)$ is at height $2$ above the base. Then the area of the triangle is:

$\frac{1}{2} \cdot \text{base" * "height} = \frac{1}{2} \cdot \textcolor{b l u e}{2} \cdot \textcolor{b l u e}{2} = 2$

Then the volume of a pyramid is:

$\frac{1}{3} \cdot \text{base" * "height} = \frac{1}{3} \cdot \textcolor{b l u e}{2} \cdot \textcolor{b l u e}{6} = 4$