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

Oct 17, 2017

$13 {\text{ units}}^{3}$

#### Explanation:

the volume of a pyramid

$V = \frac{1}{3} \times \text{base area"xx "perpendicular height}$

in the question we have the height and the coordinates of the triangle's vertices.

To find the area of a triangle with coordinates

$\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) , \left({x}_{3} , {y}_{3}\right)$

we evaluate the determinant

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

for this question:

${A}_{\Delta} = \frac{1}{2} | \left(1 , 1 , 1\right) , \left(7 , 5 , 8\right) , \left(8 , 3 , 4\right) |$

expanding by ${R}_{1}$

${A}_{\Delta} = \frac{1}{2} \left[| \left(5 , 8\right) , \left(3 , 4\right) | - | \left(7 , 8\right) , \left(8 , 4\right) | + | \left(7 , 5\right) , \left(8 , 3\right) |\right]$

${A}_{\Delta} = \frac{1}{2} \left[- 4 + 36 - 19\right]$

${A}_{\Delta} = \frac{1}{2} \times 13 = \frac{13}{2}$

the volume of teh pyramid is therefore

$V = \frac{1}{\cancel{3}} \times \frac{13}{\cancel{2}} \times \cancel{6}$

$13 {\text{ units}}^{3}$