# The base of a triangular pyramid is a triangle with corners at (6 ,2 ), (1 ,7 ), and (5 ,4 ). If the pyramid has a height of 8 , what is the pyramid's volume?

Volume $V = \frac{20}{3} \text{ }$cubic units

#### Explanation:

Compute the area of the triangular base:

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

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

Let
${P}_{1} \left(6 , 2\right)$
${P}_{2} \left(5 , 4\right)$
${P}_{3} \left(1 , 7\right)$

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

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

$A = \frac{1}{2} \left(24 + 35 + 2 - 10 - 4 - 42\right)$

$A = \frac{1}{2} \left(61 - 56\right)$
$A = \frac{5}{2}$

Now compute the volume of the Pyramid

$V = \frac{1}{3} \cdot A \cdot h = \frac{1}{3} \cdot \frac{5}{2} \cdot 8$

$V = \frac{20}{3} \text{ }$cubic units

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