# The base of a triangular pyramid is a triangle with corners at (2 ,1 ), (3 ,6 ), and (4 ,8 ). If the pyramid has a height of 5 , what is the pyramid's volume?

Aug 22, 2017

$\frac{5}{2} {\text{ units}}^{3}$

#### Explanation:

The volume of a pyramid is given by the formula

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

We are given the height $= 5$

so the problem is essentially finding the area of the triangular base.

The most direct way of finding the area of a triangle from its coordinates

$\left(a , b\right) , \left(b , c\right) , \left(d , e\right) \text{ }$is to find the absolute value of the determinate

$\Delta = \frac{1}{2} | \left(a , b , 1\right) , \left(b , c , 1\right) , \left(d , e , 1\right) |$

using the coordinates given:

$\Delta = \frac{1}{2} | \left(2 , 1 , 1\right) , \left(3 , 6 , 1\right) , \left(4 , 8 , 1\right) |$

making the determinant simpler by row operations

$R {'}_{3} = {R}_{3} - {R}_{1}$

$\Delta = \frac{1}{2} | \left(2 , 1 , 1\right) , \left(3 , 6 , 1\right) , \left(2 , 7 , 0\right) |$

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

$\Delta = \frac{1}{2} | \left(2 , 1 , 1\right) , \left(1 , 5 , 0\right) , \left(2 , 7 , 0\right) |$

expanding by$\text{ } {C}_{3}$

$\Delta = \frac{1}{2} \left[1 | \left(1 , 5\right) , \left(2 , 7\right) | - 0 + 0\right]$

$\Delta = \frac{1}{2} \left(1 \times 7 - 5 \times 2\right) = \frac{1}{27} - 10 = - \frac{3}{2}$

So the area of the triangle is${\text{ " 3/2 " units}}^{2}$

$\therefore {V}_{P} = \frac{1}{3} \times \frac{3}{2} \times 5$

${V}_{p} = \frac{5}{2} {\text{ units}}^{3}$