# The base of a triangular pyramid is a triangle with corners at (5 ,1 ), (2 ,7 ), and (3 ,4 ). If the pyramid has a height of 4 , what is the pyramid's volume?

Jan 19, 2017

2 unit^3

#### Explanation:

Volume of pyramid #=1/3 x base area x height

Let ${x}_{1} = 5 , {x}_{2} = 2 , {x}_{3} = 3 , {y}_{1} = 1 , {y}_{2} = 7 \mathmr{and} {y}_{3} = 4$

Base area $= \frac{1}{2} \left[\left[\left({x}_{1} \cdot {y}_{2}\right) + \left({x}_{2} \cdot {y}_{3}\right) + \left({x}_{3} \cdot {y}_{1}\right)\right] - \left[\left({y}_{1} \cdot {x}_{2}\right) + \left({y}_{2} \cdot {x}_{3}\right) + \left({y}_{3} \cdot {x}_{1}\right)\right]\right]$

$= \frac{1}{2} \left[\left[\left(5 \cdot 7\right) + \left(2 \cdot 4\right) + \left(3 \cdot 1\right)\right] - \left[\left(1 \cdot 2\right) + \left(7 \cdot 3\right) + \left(4 \cdot 5\right)\right]\right]$
$= \frac{1}{2} \left[\left(35 + 8 + 3\right) - \left(2 + 21 + 20\right)\right]$
$= \frac{1}{2} \left(46 - 43\right)$
$= \frac{3}{2} u n i {t}^{2}$

The volume $= \frac{1}{3} \cdot \frac{3}{2} \cdot 4 = 2 u n i {t}^{3}$