# The base of a triangular pyramid is a triangle with corners at (2 ,5 ), (6 ,9 ), and (3 ,8 ). If the pyramid has a height of 15 , what is the pyramid's volume?

Mar 2, 2016

$20.001$

#### Explanation:

To find volume of a triangular pyramid of height $15$ and base a triangle with corners at $A \left(2 , 5\right)$, $B \left(6 , 9\right)$, and $C \left(3 , 8\right)$ we must find the area of the base triangle.

The sides of triangle can be found as follows.

$A B = \sqrt{{\left(6 - 2\right)}^{2} + {\left(9 - 5\right)}^{2}} = \sqrt{32} = 5.6568$

$B C = \sqrt{{\left(6 - 3\right)}^{2} + {\left(9 - 8\right)}^{2}} = \sqrt{9 + 1} = \sqrt{10} = 3.1623$

$C A = \sqrt{{\left(3 - 2\right)}^{2} + {\left(8 - 5\right)}^{2}} = \sqrt{1 + 9} = \sqrt{10} = 3.1623$

Using Heron's formula $s = \frac{5.6568 + 3.1623 + 3.1623}{2} = 5.9907$

and area of triangle is sqrt(5.9907xx(5.9907-5.6568)xx(5.9907-3.1623)xx(5.9907-3.1623)
i.e. $\sqrt{5.9907 \times 0.3339 \times 2.8284 \times 2.8284} = \sqrt{16.0021} = 4.0002$ (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 $\frac{1}{3} \times 4.0002 \times 15 = 20.001$