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

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Answer 1 · 2016-09-07T07:20:30

Volume of pyramid is $\frac{1}{3} \times 6 \times 8.0004 = 16.0008$ $1/3xx6xx8.0004=16.0008$

Find the distance between corners should give us three sides and using Heron's formula we can then find area of base triangle.

$Delta=sqrt(s(s-a)(s-b)(s-c))$ , where $s=1/2(a+b+c)$ , where sides of a triangle are $a$ , $b$ and $c$ .

Then area can be multiplied by height and divided by $3$ , which will give us volume of pyramid.

The sides of triangle formed by $(6,2)$ , $(4,5)$ and $(8,7)$ are

$a=sqrt((4-6)^2+(5-2)^2)=sqrt(4+9)=sqrt13=3.6056$

$b=sqrt((8-4)^2+(7-5)^2)=sqrt(16+4)=sqrt20=4.4721$ and

$c=sqrt((8-6)^2+(7-2)^2)=sqrt(4+25)=sqrt29=5.3852$

Hence $s=1/2(3.6056+4.4721+5.3852)=1/2xx13.4629=6.7315$

and $Delta=sqrt(6.7315xx(6.7315-3.6056)xx(6.7315-4.4721)xx(6.7315-5.3852)$

= $sqrt(6.7315xx3.1259xx2.2594xx1.3463)=sqrt64.0062=8.0004$

Hence volume of pyramid is $1/3xx6xx8.0004=16.0008$

The base of a triangular pyramid is a triangle with corners at (6 ,2 )(6,2)(6 ,2 ), (4 ,5 )(4,5)(4 ,5 ), and (8 ,7 )(8,7)(8 ,7 ). If the pyramid has a height of 6 66 , what is the pyramid's volume?