# The base of a triangular pyramid is a triangle with corners at (3 ,5 ), (4 ,2 ), and (2 ,6 ). If the pyramid has a height of 7 , what is the pyramid's volume?

Jun 29, 2017

The volume is $= 2.33 {u}^{3}$

#### Explanation:

The area of the base triangle is calculated with the determinant

Therefore, the area of the base is

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

$A = \frac{1}{2} | \left(3 , 5 , 1\right) , \left(4 , 2 , 1\right) , \left(2 , 6 , 1\right) |$

$= \frac{1}{2} \left(3 \cdot | \left(2 , 1\right) , \left(6 , 1\right) | - 5 \cdot | \left(4 , 1\right) , \left(2 , 1\right) | + 1 \cdot | \left(4 , 2\right) , \left(2 , 6\right) |\right)$

$= \frac{1}{2} \left(3 \left(2 - 6\right) - 5 \left(4 - 2\right) + 1 \left(24 - 4\right)\right)$

$= \frac{1}{2} \left(- 12 - 10 + 20\right)$

$= \frac{1}{2} | - 2 | = 1$

The volume of the pyramid is

$= \frac{1}{3} \cdot h \cdot A = \frac{1}{3} \cdot 7 \cdot 1 = \frac{7}{3} = 2.33 {u}^{3}$