# The base of a triangular pyramid is a triangle with corners at (6 ,8 ), (2 ,4 ), and (7 ,3 ). If the pyramid has a height of 2 , what is the pyramid's volume?

Mar 19, 2016

${V}_{{\Delta}_{p y r}} = B A \cdot H = 24 \cdot 2 = 48$

#### Explanation:

Given: Base Triangle Vertices: $A \left(2 , 4 , 0\right) , B \left(7 , 3 , 0\right) , C \left(6 , 8 , 0\right)$
Using Distance formula find the length of the sides of the base triangle, $\overline{A B} , \overline{B C} , \overline{C A}$:
$\overline{A B} = \sqrt{{\left(2 - 7\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{26}$
$\overline{B C} = \sqrt{{\left(7 - 6\right)}^{2} + {\left(3 - 8\right)}^{2}} = \sqrt{26}$
$\overline{C A} = \sqrt{{\left(6 - 2\right)}^{2} + {\left(8 - 4\right)}^{2}} = 4 \sqrt{2}$

Now you can use the Pyramid volume formula to calculate the Volume. $V = \frac{1}{3} B A \cdot H$
where $B A$, and $H$ are Base Area and Height of the pyramid respectively.
We are going to use Heron Formula to calculate the area of the triangle:
A_(Delta) = sqrt(p(p-bar(AB))(p-bar(BC))(p-bar(CA))
Where $p = \frac{P}{2} = \frac{\overline{A B} + \overline{B C} + \overline{C A}}{2}$
$p = \frac{\sqrt{26} + \sqrt{26} + 4 \sqrt{2}}{2}$
A_(Delta) = sqrt(((sqrt(26)+sqrt(26)+4sqrt(2))/2)((sqrt(26)+sqrt(26)+4sqrt(2))/2-sqrt(26))((sqrt(26)+sqrt(26)+4sqrt(2))/2-sqrt(26))((sqrt(26)+sqrt(26)+4sqrt(2))/2-4sqrt(2))
A_Delta = sqrt(((sqrt(26)+sqrt(26)+4sqrt(2)))/2(4sqrt(2))(4sqrt(2))((sqrt(26)+sqrt(26)-4sqrt(2))/2 )
${A}_{\Delta} = \sqrt{\left(2 \sqrt{26} + 4 \sqrt{2}\right) \left(2 \sqrt{26} - 4 \sqrt{2}\right) \left(4 \cdot 2\right)}$
${A}_{\Delta} = \sqrt{\left(4 \cdot 26 - 16 \cdot 2\right) \left(4 \cdot 2\right)} = 24$
and the Volume is:
${V}_{{\Delta}_{p y r}} = B A \cdot H = 24 \cdot 2 = 48$