# The base of a triangular pyramid is a triangle with corners at (6 ,2 ), (3 ,5 ), and (4 ,2 ). If the pyramid has a height of 8 , what is the pyramid's volume?

$V = 8 \text{ }$cubic units

#### Explanation:

the volume $V$ of a triangular pyramid has the formula

Volume = 1/3 of Area of base * Height of pyramid

Let us compute the area $A$ of the base with
${P}_{1} \left({x}_{1} , {y}_{1}\right) = \left(6 , 2\right)$
${P}_{2} \left({x}_{2} , {y}_{2}\right) = \left(3 , 5\right)$
${P}_{3} \left({x}_{3} , {y}_{3}\right) = \left(4 , 2\right)$

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

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

$A = \frac{1}{2} \left[\begin{matrix}6 & 3 & 4 & 6 \\ 2 & 5 & 2 & 2\end{matrix}\right]$

$A = \frac{1}{2} \cdot \left[6 \cdot 5 + 3 \cdot 2 + 4 \cdot 2 - 3 \cdot 2 - 4 \cdot 5 - 6 \cdot 2\right]$

$A = \frac{1}{30 + 6 + 8 - 6 - 20 - 12}$

$A = \frac{1}{2} \left(44 - 38\right)$

$A = 3 \text{ }$square units

Let us now compute the volume V of the pyramid

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

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

$V = 8 \text{ }$cubic units

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