# The base of a triangular pyramid is a triangle with corners at (8 ,5 ), (7 ,2 ), and (4 ,6 ). If the pyramid has a height of 7 , what is the pyramid's volume?

Jul 30, 2018

color(green)(V = 1/3 A_b h = (1/3) * 7.1171 * 7 = 16.6065, " cubic units"

#### Explanation:

Given : $A \left(8 , 5\right) , B \left(7 , 2\right) , C \left(4 , 6\right)$, h = 7

Using distance formula we can calculate the lengths of sides a, b, c.

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$a = \sqrt{{\left(7 - 4\right)}^{2} + {\left(2 - 6\right)}^{2}} = 5$

$b = \sqrt{{\left(4 - 8\right)}^{2} + {\left(6 - 5\right)}^{2}} = \sqrt{17}$

$c = \sqrt{{\left(8 - 7\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{10}$

Semi perimeter of base triangle $s = \frac{a + b + c}{2}$

$s = \frac{5 + \sqrt{17} + \sqrt{10}}{2} = \frac{12.2854}{2} \approx 6.2427$

Area of triangle ${A}_{t} = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

${A}_{t} = \sqrt{6.2427 \cdot \left(6.2427 - 5\right) \cdot \left(6.2427 - \sqrt{17}\right) \cdot \left(6.2427 - \sqrt{10}\right)}$

A_t ~~ 7.1171

Volume of pyramid $V = \left(\frac{1}{3}\right) \cdot {A}_{t} \cdot h$

color(green)(V = (1/3) * 7.1171 * 7 = 16.6065, " cubic units"