Question

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?

Answer 1

Volume is `32.5`units

Explanation:

,Area of the base is given by:

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

We have the coordinates

`(x_1,y_1)-=(9,5)`
`(x_2,y_2)-=(6,3)`
`(x_3,y_3)-=(7,8)`

Thus,
`A=1/2(6-9)(5+3)+1/2(7-6)(3+8)+1/2(9-7)(8+5)`
`A=(-3xx8)/2+(1xx11)/2+(2xx13)/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=1/3Ah`
where
A is the area of the base
h is the height of the pyramid

Substituting

`V=1/3xx6.5xx15=6.5xx5=32.5`

Volume is `32.5`units

Answer 2

`"volume "=65/2`

Explanation:

`"the volume (V) of a pyramid is calculated using the formula"`

`•color(white)(x)V=1/3xx"area of base "xx" height"`

`"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|`

`"let "(x_1,y_1)=(9,5),(x_2,y_2)=(6,3),(x_3,y_3)=(7,8)`

`A=1/2|9(3-8)+6(8-5)+7(5-3)|`

`color(white)(A)=1/2|-45+18+14|=13/2`

`rArrV=1/3xx13/2xx15=65/2`