# The base of a triangular pyramid is a triangle with corners at (2 ,2 ), (3 ,1 ), and (7 ,5 ). If the pyramid has a height of 6 , what is the pyramid's volume?

Mar 2, 2016

$8$

#### Explanation:

To find volume of a triangular pyramid of height $6$ and base a triangle with corners at $A \left(2 , 2\right)$, $B \left(3 , 1\right)$, and $C \left(7 , 5\right)$ we must find the area of the base triangle.

The sides of triangle can be found as follows.

$A B = \sqrt{{\left(3 - 2\right)}^{2} + {\left(1 - 2\right)}^{2}} = \sqrt{2} = 1.4142$

$B C = \sqrt{{\left(7 - 3\right)}^{2} + {\left(5 - 1\right)}^{2}} = \sqrt{16 + 16} = \sqrt{32} = 5.6568$

$C A = \sqrt{{\left(7 - 2\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{25 + 9} = \sqrt{34} = 5.831$

Using Heron's formula $s = \frac{1.4142 + 5.6568 + 5.831}{2} = 6.451$

and area of triangle is sqrt(6.451xx(6.451-1.4142)xx(6.451-5.6568)xx(6.451-5.831)
i.e. $\sqrt{6.451 \times 5.0368 \times 0.7942 \times 0.62} = 4$ (approx.)

As volume of pyramid $\frac{1}{3} \times h e i g h t \times a r e a o f b a s e$, it is $\frac{1}{3} \times 4 \times 6 = 8$