# 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 6 , what is the pyramid's volume?

$V = 4 \text{ }$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 ${P}_{1} \left(3 , 4\right) , {P}_{2} \left(5 , 8\right) , {P}_{3} \left(4 , 8\right)$

the area A matrix

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

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

$A = \frac{1}{2} \cdot \left[\begin{matrix}3 & 5 & 4 & 3 \\ 4 & 8 & 8 & 4\end{matrix}\right]$

$A = \frac{1}{2} \cdot \left[3 \left(8\right) + 5 \left(8\right) + 4 \left(4\right) - 5 \left(4\right) - 4 \left(8\right) - 3 \left(8\right)\right]$

$A = \frac{1}{2} \cdot \left(24 + 40 + 16 - 20 - 32 - 24\right)$

$A = \frac{1}{2} \cdot \left(80 - 76\right)$

$A = 2$

Now the volume V

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

$V = \frac{1}{3} \cdot 2 \cdot 6 = 4$

$V = 4 \text{ }$cubic units

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