# The base of a triangular pyramid is a triangle with corners at (9 ,5 ), (6 ,3 ), and (7 ,8 ). If the pyramid has a height of 15 , what is the pyramid's volume?

Feb 17, 2018

Volume is $32.5$units

#### Explanation:

,Area of the base is given by:

$A = \frac{1}{2} \left({x}_{2} - {x}_{1}\right) \left({y}_{1} + {y}_{2}\right) + \frac{1}{2} \left({x}_{3} - {x}_{2}\right) \left({y}_{2} + {y}_{3}\right) + \frac{1}{2} \left({x}_{1} - {x}_{3}\right) \left({y}_{3} + {y}_{1}\right)$

We have the coordinates

$\left({x}_{1} , {y}_{1}\right) \equiv \left(9 , 5\right)$
$\left({x}_{2} , {y}_{2}\right) \equiv \left(6 , 3\right)$
$\left({x}_{3} , {y}_{3}\right) \equiv \left(7 , 8\right)$

Thus,
$A = \frac{1}{2} \left(6 - 9\right) \left(5 + 3\right) + \frac{1}{2} \left(7 - 6\right) \left(3 + 8\right) + \frac{1}{2} \left(9 - 7\right) \left(8 + 5\right)$
$A = \frac{- 3 \times 8}{2} + \frac{1 \times 11}{2} + \frac{2 \times 13}{2}$

$A = - 12 + 5.5 + 13 = 1 + 5.5 = 6.5$

Area of the base is $A = 6.5$
height is $15$

Volume V is given by
$V = \frac{1}{3} A h$
where
A is the area of the base
h is the height of the pyramid

Substituting

$V = \frac{1}{3} \times 6.5 \times 15 = 6.5 \times 5 = 32.5$

Volume is $32.5$units

Feb 17, 2018

$\text{volume } = \frac{65}{2}$

#### Explanation:

$\text{the volume (V) of a pyramid is calculated using the formula}$

â€¢color(white)(x)V=1/3xx"area of base "xx" height"

$\text{the area (A) of the base can be found using}$

A=1/2|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2|

$\text{let } \left({x}_{1} , {y}_{1}\right) = \left(9 , 5\right) , \left({x}_{2} , {y}_{2}\right) = \left(6 , 3\right) , \left({x}_{3} , {y}_{3}\right) = \left(7 , 8\right)$

$A = \frac{1}{2} | 9 \left(3 - 8\right) + 6 \left(8 - 5\right) + 7 \left(5 - 3\right) |$

$\textcolor{w h i t e}{A} = \frac{1}{2} | - 45 + 18 + 14 | = \frac{13}{2}$

$\Rightarrow V = \frac{1}{3} \times \frac{13}{2} \times 15 = \frac{65}{2}$