# The base of a triangular pyramid is a triangle with corners at (2 ,6 ), (5 ,3 ), and (8 ,7 ). If the pyramid has a height of 18 , what is the pyramid's volume?

Mar 24, 2016

${V}_{p y r} = \frac{1}{3} \left(10.5\right) \cdot 18 = 10.5 \cdot 6 = 63 \text{ cubic units}$

#### Explanation:

Given the vertices of the base triangle of a pyramid and altitude:
$A \left(2 , 6\right) , B \left(5 , 3\right) \mathmr{and} C \left(8 , 7\right)$
Altitude, $H = 18$ units
Required: The volume?
This problem can be solved as follows:
${V}_{p y r} = \frac{1}{3} \left(B A\right) \cdot H$ where $B A =$Base Area and $H =$Altitude
So the strategy is:
a) Use distance formula to determine the length of sides of the triangle: $\overline{A B} , \overline{B C} , \overline{C A} ,$
Distance formula of, ${P}_{1} \left({x}_{1} , {y}_{1} , {z}_{1}\right) , {P}_{2} \left({x}_{2} , {y}_{2} , {z}_{2}\right)$
$\overline{{P}_{1} {P}_{2}} = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
Since the base is on the x-y plane ${z}_{1} = {z}_{2} = 0$ Thus,
$\overline{A B} = \sqrt{{\left(2 - 5\right)}^{2} + {\left(6 - 3\right)}^{2}} = 3 \sqrt{2}$
$\overline{B C} = \sqrt{{\left(5 - 8\right)}^{2} + {\left(3 - 7\right)}^{2}} = 5$
$\overline{C A} = \sqrt{{\left(8 - 2\right)}^{2} + {\left(7 - 6\right)}^{2}} = \sqrt{37}$
b) Now calculate the area. Since you know all sides you can use Heron's formula:
$A r e {a}_{\Delta} = {A}_{\Delta} = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
where $a , b , \mathmr{and} d$ are sides of base triangle and s is half the perimeter $s = \frac{1}{2} \left(a + b + c\right) = \frac{1}{2} \left(5 + 3 \sqrt{2} + \sqrt{37}\right)$
${A}_{\Delta} = \sqrt{\frac{1}{2} \left(5 + 3 \sqrt{2} + \sqrt{37}\right) \left[\frac{1}{2} \left(5 + 3 \sqrt{2} + \sqrt{37}\right) - 5\right] \left[\frac{1}{2} \left(5 + 3 \sqrt{2} + \sqrt{37}\right) - 3 \sqrt{2}\right] \left[\frac{1}{2} \left(5 + 3 \sqrt{2} + \sqrt{37}\right) - \sqrt{37}\right]} \approx 10.5$ ${A}_{\Delta} = 10.5 \text{ square units}$
b) Now calculate the volume. : ${V}_{p y r} = \frac{1}{3} \left(B A\right) \cdot H$
${V}_{p y r} = \frac{1}{3} \left(10.5\right) \cdot 18 = 10.5 \cdot 6 = 63 \text{ cubic units}$