# The base of a triangular pyramid is a triangle with corners at (6 ,2 ), (3 ,1 ), and (4 ,2 ). If the pyramid has a height of 8 , what is the pyramid's volume?

Volume $V = \frac{1}{3} \cdot A h = \frac{1}{3} \cdot 1 \cdot 8 = \frac{8}{3} = 2 \frac{2}{3}$

#### Explanation:

Let ${P}_{1} \left(6 , 2\right)$, and ${P}_{2} \left(4 , 2\right)$,and ${P}_{3} \left(3 , 1\right)$

Compute the area of the base of the pyramid
$A = \frac{1}{2} \left[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{matrix}\right]$

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

$A = \frac{1}{2} \left[\begin{matrix}6 & 4 & 3 & 6 \\ 2 & 2 & 1 & 2\end{matrix}\right]$

$A = \frac{1}{2} \left(6 \cdot 2 + 4 \cdot 1 + 3 \cdot 2 - 2 \cdot 4 - 2 \cdot 3 - 1 \cdot 6\right)$

$A = \frac{1}{2} \left(12 + 4 + 6 - 8 - 6 - 6\right)$

$A = 1$

Volume $V = \frac{1}{3} \cdot A h = \frac{1}{3} \cdot 1 \cdot 8 = \frac{8}{3} = 2 \frac{2}{3}$

God bless....I hope the explanation is useful.