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

Mar 2, 2016

$V \approx 3.35$

#### Explanation:

The volume of a pyramid is given by $V = \frac{1}{3} \cdot B a s e \cdot H e i g h t$
so the first step is to find the area of the base.

The simplest way to do that, given the three vertices, is to find the lengths of the three sides and use the area formula $A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where $s$ is the semi-perimeter and $a , b \mathmr{and} c$ are the three sides of the base.

${a}^{2} = {\left(4 - 2\right)}^{2} + {\left(7 - 4\right)}^{2} 4 + 9 = 13$
$a = \sqrt{13} \approx 3.6$

${b}^{2} = {\left(6 - 4\right)}^{2} + {\left(7 - 5\right)}^{2} = 4 4 = 8$
$b = 2 \sqrt{2} \approx 2.83$

$c = {\left(6 - 2\right)}^{2} + {\left(5 - 4\right)}^{2} = 16 + 1 = 17$
$c = \sqrt{17} \approx 4.12$

$s = \frac{\sqrt{13} + 2 \sqrt{2} + \sqrt{17}}{2} \approx 5.28$

$A = \sqrt{5.28 \left(5.28 - 3.6\right) \left(5.28 - 2.83\right) \left(5.28 - 4.12\right)} \approx 5.02$

Then the volume is $V = \frac{5.02 \cdot 2}{3} \approx 3.35$