Question

The base of a triangular pyramid is a triangle with corners at `(5 ,2 )`, `(3 ,7 )`, and `(4 ,8 )`. If the pyramid has a height of `6`, what is the pyramid's volume?

Answer 1

`V=7`

Explanation:

The formula for the volume of a cone is

`V=1/3bh`

Here, the base is formed by the three specified points. We can use Heron's formula to determine the area of this triangle, but first we need to find the length of each side of the triangle using the distance formula.

Between `(5,2)` and `(3,7)`:

`d=sqrt((5-3)^2+(2-7)^2)=sqrt(4+25)=sqrt29`

Between `(3,7)` and `(4,8)`:

`d=sqrt((3-4)^2+(7-8)^2)=sqrt(1+1)=sqrt2`

Between `(4,8)` and `(5,2)`:

`d=sqrt((4-5)^2+(8-2)^2)=sqrt(1+36)=sqrt37`

These are the three side lengths of the triangle. To find the area, we should first find the semiperimeter `s`, which is `"1/2"` the triangle's perimeter.

`s=(sqrt29+sqrt2+sqrt37)/2`

Heron's formula states that a triangle with sides `a,b,c` and semiperimeter `s` has an area equal to

`A=sqrt(s(s-a)(s-b)(s-c))`

Hence:

`A=sqrt((sqrt29+sqrt2+sqrt37)/2((sqrt29+sqrt2+sqrt37)/2-sqrt29)((sqrt29+sqrt2+sqrt37)/2-sqrt2)((sqrt29+sqrt2+sqrt37)/2-sqrt37))`

You could plug this into a calculator to see that this equals `7/2`, but I'll walk through the algebra.

`A=sqrt((sqrt29+sqrt2+sqrt37)/2((sqrt29+sqrt2+sqrt37)/2-(2sqrt29)/2)((sqrt29+sqrt2+sqrt37)/2-(2sqrt2)/2)((sqrt29+sqrt2+sqrt37)/2-(2sqrt37)/2))`

`A=sqrt((sqrt29+sqrt2+sqrt37)/2((-sqrt29+sqrt2+sqrt37)/2)((sqrt29-sqrt2+sqrt37)/2)((sqrt29+sqrt2-sqrt37)/2))`

`A=1/4sqrt((sqrt29+sqrt2+sqrt37)(-sqrt29+sqrt2+sqrt37)(sqrt29-sqrt2+sqrt37)(sqrt29+sqrt2-sqrt37))`

The portion in the square root is in the form (distribute this if you don't believe me):

`(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)`

Within the square root, we have

`{:(a=sqrt29,=>,a^2=29,=>,a^4=841),(b=sqrt2,=>,b^2=2,=>,b^4=4),(c=sqrt37,=>,c^2=37,=>,c^4=1369):}`

Hence:

`A=1/4sqrt(2(29(2)+2(37)+37(29))-(841+4+1369))`

`A=1/4sqrt196=7/2`

So, to find the volume, we plug in

`V=1/3(7/2)(6)=7`