# The base of a triangular pyramid is a triangle with corners at (5 ,8 ), (3 ,4 ), and (4 ,8 ). If the pyramid has a height of 5 , what is the pyramid's volume?

Oct 3, 2016

$V = \frac{10}{3} u n i t {s}^{3}$

#### Explanation:

Let $\left({A}_{x} , {A}_{y}\right) = \left(5 , 8\right)$
Let $\left({B}_{x} , {B}_{y}\right) = \left(3 , 4\right)$
Let $\left({C}_{x} , {C}_{y}\right) = \left(4 , 8\right)$

According to Area of a triangle given 3 points the area, $\Delta$, of the base is:

$\Delta = | \frac{{A}_{x} \left({B}_{y} - {C}_{y}\right) + {B}_{x} \left({C}_{y} - {A}_{y}\right) + {C}_{x} \left({A}_{y} - {B}_{y}\right)}{2} |$

$\Delta = | \frac{5 \left(4 - 8\right) + 3 \left(8 - 8\right) + 4 \left(8 - 4\right)}{2} |$

$\Delta = | \frac{5 \left(- 4\right) + 3 \left(0\right) + 4 \left(4\right)}{2} |$

$\Delta = | \frac{- 20 + 16}{2} |$

$\Delta = | \frac{- 4}{2} |$

$\Delta = 2$

The volume of the pyramid

$V = \frac{1}{3} \Delta h$ where h is the height

$V = \frac{1}{3} \left(2\right) \left(5\right)$

$V = \frac{10}{3} u n i t {s}^{3}$