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

Mar 17, 2016

Volume of pyramid is $21.992$

#### Explanation:

Volume of a pyramid is given by $\frac{1}{3} \times \text{area of base} \times h e i g h t$

Hence, we have to first find area of base triangle and as all the three sides can be found between three points, we would be using Heron's formula i.e. area of triangle is given by $\sqrt{s \times \left(c - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{1}{2} \left(a + b + c\right)$.

Let the points be $A \left(7 , 6\right)$, $B \left(4 , 2\right)$ and $C \left(3 , 8\right)$. Hence

$a = \sqrt{{\left(3 - 4\right)}^{2} + {\left(8 - 2\right)}^{2}} = \sqrt{{\left(- 1\right)}^{2} + {6}^{2}} = \sqrt{1 + 36} = \sqrt{37} = 6.083$

$b = \sqrt{{\left(3 - 7\right)}^{2} + {\left(8 - 6\right)}^{2}} = \sqrt{{\left(- 4\right)}^{2} + {2}^{2}} = \sqrt{16 + 4} = \sqrt{20} = 4.472$

$c = \sqrt{{\left(7 - 4\right)}^{2} + {\left(6 - 2\right)}^{2}} = \sqrt{{3}^{2} + {4}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

Hence $s = \frac{1}{2} \left(6.083 + 4.472 + 5\right) = \frac{15.555}{2} = 7.777$ and

Area of triangle is sqrt(7.777(7.777-6.083)(7.777-4.472)(7.777-5) or

$\sqrt{7.777 \times 1.694 \times 3.305 \times 2.777} = \sqrt{120.913} = 10.996$

Hence volume of pyramid is $\frac{1}{3} \times 10.996 \times 6 = 2 \times 10.996 = 21.992$