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

Nov 26, 2017

Volume of the pyramid with triangular base is $18$

#### Explanation:

Volume of triangular pyramid = $\left(\frac{1}{3}\right) A H$ where A is the area of the triangular base and H is the height of the pyramid.

Area of triangular base = $\left(\frac{1}{2}\right) b h$ where b is the base and h is the height of the triangle.

Base $= \sqrt{{\left(5 - 2\right)}^{2} + {\left(3 - 6\right)}^{2}} = \sqrt{{3}^{2} + {3}^{2}} = \textcolor{b l u e}{3 \sqrt{2}}$

Eqn of Base =is
$\left(\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}}\right) = \left(\frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}\right)$

$\left(\frac{y - 6}{3 - 6}\right) = \left(\frac{x - 2}{5 - 2}\right)$

$\left(y - 6\right) = \left(- x + 2\right)$
$y + x = 8$ Eqn. (1)
Slope of base $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
Slope $m = \frac{3 - 6}{5 - 2} = - 1$

Slope of altitude ${m}_{1} = - \left(\frac{1}{m}\right) = - \left(\frac{1}{- 1}\right) = 1$

Eqn of Altitude is
$\left(y - {y}_{3}\right) = {m}_{1} \left(x - {x}_{3}\right)$

$y - 2 = 1 \left(x - 8\right)$
$y - x = - 6$. Eqn (2)

Solving Eqns (1) & (2), we get coordinates of the base of the altitude.
Coordinates of base of altitude are (7,1)

Height of altitude$= \sqrt{{\left(8 - 7\right)}^{2} + {\left(2 - 1\right)}^{2}} = \textcolor{red}{\sqrt{2}}$

Area of triangular base $= \left(\frac{1}{2}\right) \textcolor{b l u e}{3 \sqrt{2}} \textcolor{red}{\sqrt{2}}$

Area $= 3$

Volume of pyramid = $\left(\frac{1}{3}\right) \cdot 3 \cdot 18 = 18$