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

Sep 7, 2016

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

#### Explanation:

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

$\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{1}{2} \left(a + b + c\right)$, 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 $\left(6 , 2\right)$, $\left(4 , 5\right)$ and $\left(8 , 7\right)$ are

$a = \sqrt{{\left(4 - 6\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{4 + 9} = \sqrt{13} = 3.6056$

$b = \sqrt{{\left(8 - 4\right)}^{2} + {\left(7 - 5\right)}^{2}} = \sqrt{16 + 4} = \sqrt{20} = 4.4721$ and

$c = \sqrt{{\left(8 - 6\right)}^{2} + {\left(7 - 2\right)}^{2}} = \sqrt{4 + 25} = \sqrt{29} = 5.3852$

Hence $s = \frac{1}{2} \left(3.6056 + 4.4721 + 5.3852\right) = \frac{1}{2} \times 13.4629 = 6.7315$

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

= $\sqrt{6.7315 \times 3.1259 \times 2.2594 \times 1.3463} = \sqrt{64.0062} = 8.0004$

Hence volume of pyramid is $\frac{1}{3} \times 6 \times 8.0004 = 16.0008$