# The base of a triangular pyramid is a triangle with corners at (8 ,5 ), (6 ,7 ), and (5 ,1 ). If the pyramid has a height of 8 , what is the pyramid's volume?

Jan 8, 2017

Pyramid's volume is $\frac{56}{3}$

#### Explanation:

Let they be

A(8;5);B(6;7);C(5;1);h=8

First, you would calculate the area of the triangle ABC, by getting the lenght of a side:

$A C = \sqrt{{\left({x}_{C} - {x}_{A}\right)}^{2} + {\left({y}_{C} - {y}_{A}\right)}^{2}} = \sqrt{{\left(5 - 8\right)}^{2} + {\left(1 - 5\right)}^{2}} = \sqrt{9 + 16} = 5$

and then the height relative to AC, by finding the distance between the point B and the line AC:

${m}_{A C} = \frac{{y}_{C} - {y}_{A}}{{x}_{C} - {x}_{A}} = \frac{1 - 5}{5 - 8} = \frac{4}{3}$

Then the equation of the line AC is

$y - {y}_{A} = {m}_{A C} \left(x - {x}_{A}\right)$

that's

$y - 5 = \frac{4}{3} \left(x - 8\right)$

$y - 5 = \frac{4}{3} x - \frac{32}{3}$

$3 y - 15 = 4 x - 32$

$4 x - 3 y - 17 = 0$ in the form $a x + b y + c = 0$

Then let's calculate the distance:

$d = | a {x}_{B} + b {y}_{B} + c \frac{|}{\sqrt{{a}^{2} + {b}^{2}}}$

$d = | 4 \cdot 6 - 3 \cdot 7 - 17 \frac{|}{\sqrt{{4}^{2} + {3}^{2}}} = | 24 - 21 - 17 \frac{|}{\sqrt{25}} = \frac{14}{5}$

Now let's calculate the area of the triangle:

$A r e a = \left(s i \mathrm{de} A C\right) \cdot \left(d\right) \cdot \frac{1}{2} = \cancel{5} \cdot {\cancel{14}}^{7} / \cancel{5} \cdot \frac{1}{\cancel{2}} = 7$

The pyramid's volume is obtained by:

$V = \frac{1}{3} A r e {a}_{b a s e} \cdot h = \frac{1}{3} \cdot 7 \cdot 8 = \frac{56}{3}$