# The base of a triangular pyramid is a triangle with corners at (6 ,5 ), (2 ,2 ), and (4 ,7 ). If the pyramid has a height of 8 , what is the pyramid's volume?

Volume $V = 18 \frac{2}{3} = 18.67 \text{ }$cubic units

#### Explanation:

To solve for the volume of the pyramid which is
$V = \frac{1}{3} \cdot a r e a \cdot h e i g h t$

Solve for the area of the triangle first with ${P}_{1} \left(2 , 2\right) , {P}_{2} \left(6 , 5\right) , {P}_{3} \left(4 , 7\right)$

the area A matrix

$A = \frac{1}{2} \cdot \left[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{matrix}\right]$

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

$A = \frac{1}{2} \cdot \left[\begin{matrix}2 & 6 & 4 & 2 \\ 2 & 5 & 7 & 2\end{matrix}\right]$

$A = \frac{1}{2} \cdot \left[2 \left(5\right) + 6 \left(7\right) + 4 \left(2\right) - 6 \left(2\right) - 4 \left(5\right) - 2 \left(7\right)\right]$

$A = \frac{1}{2} \cdot \left(10 + 42 + 8 - 12 - 20 - 14\right)$

$A = \frac{1}{2} \cdot \left(60 - 46\right)$

$A = 7$

Now the volume V

$V = \frac{1}{3} \cdot a r e a \cdot h e i g h t$

$V = \frac{1}{3} \cdot 7 \cdot 8 = \frac{56}{3}$

$V = 18 \frac{2}{3}$