# The base of a triangular pyramid is a triangle with corners at (6 ,7 ), (5 ,3 ), and (8 ,4 ). If the pyramid has a height of 6 , what is the pyramid's volume?

Volume $V = 11 \text{ }$cubic units

#### Explanation:

Compute the area of the triangular base first

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

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

The given points are ${P}_{1} \left(6 , 7\right)$, ${P}_{2} \left(5 , 3\right)$, ${P}_{3} \left(8 , 4\right)$

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

Area $A = \frac{1}{2} \left[\begin{matrix}6 & 5 & 8 & 6 \\ 7 & 3 & 4 & 7\end{matrix}\right]$

Area $A = \frac{1}{2} \cdot \left(6 \cdot 3 + 5 \cdot 4 + 8 \cdot 7 - 5 \cdot 7 - 8 \cdot 3 - 6 \cdot 4\right)$

Area $A = \frac{1}{2} \cdot \left(18 + 20 + 56 - 35 - 24 - 24\right)$

Area $A = \frac{1}{2} \cdot \left(94 - 83\right)$

Area $A = \frac{1}{2} \cdot \left(11\right) = 5.5$

Compute the volume of the triangular pyramid

$V = \frac{1}{3} \cdot A \cdot h = \frac{1}{3} \cdot \frac{11}{2} \cdot 6$

$V = 11 \text{ }$cubic units

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