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

May 5, 2018

$7 {\text{ units}}^{3}$

#### Explanation:

the volume of a pyramid is found by the formula

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

so the real problem is finding the area of the base

we are given that the base is a triangle and its vertices are given as co-ordinates

for a triangle with co-cordinates

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

the area can be calculated by the determinant

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

we ahve therefore

$A = \pm \frac{1}{2} | \left(6 , 7 , 1\right) , \left(5 , 5 , 1\right) , \left(8 , 4 , 1\right) |$

$R {'}_{1} = {R}_{1} - {R}_{2}$

$A = \pm \frac{1}{2} | \left(1 , 2 , 0\right) , \left(5 , 5 , 1\right) , \left(8 , 4 , 1\right) |$

expand by Row 1

$A = \pm \frac{1}{2} \left[| \left(5 , 1\right) , \left(4 , 1\right) | - 2 | \left(5 , 1\right) , \left(8 , 1\right) | + 0\right]$

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

$A = \pm \frac{1}{2} \left[1 + 6\right]$

$A = \frac{7}{2} \text{ }$( taking the positive value)

$\therefore V = \frac{1}{3} \times 6 \times \frac{7}{2}$

$= 7 {\text{ units}}^{3}$