# The base of a triangular pyramid is a triangle with corners at (6 ,2 ), (5 ,1 ), and (7 ,4 ). If the pyramid has a height of 12 , what is the pyramid's volume?

##### 1 Answer
Jul 3, 2016

Volume of pyramid is $2.0016$ cubic units.

#### Explanation:

As volume of pyramid one-third of base area multiplied by height, one should first find the area of base triangle.

Here the sides of a triangle are $a$, $b$ and $c$, then the area of the triangle $\Delta$ is given by the formula

$\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{1}{2} \left(a + b + c\right)$

and radius of circumscribed circle is $\frac{a b c}{4 \Delta}$

Hence let us find the sides of triangle formed by $\left(6 , 2\right)$, $\left(5 , 1\right)$ and $\left(7 , 4\right)$. This will be surely distance between pair of points, which is

$a = \sqrt{{\left(5 - 6\right)}^{2} + {\left(1 - 2\right)}^{2}} = \sqrt{1 + 1} = \sqrt{2} = 1.4142$

$b = \sqrt{{\left(7 - 5\right)}^{2} + {\left(4 - 1\right)}^{2}} = \sqrt{4 + 9} = \sqrt{13} = 3.6056$ and

$c = \sqrt{{\left(7 - 6\right)}^{2} + {\left(4 - 2\right)}^{2}} = \sqrt{1 + 4} = \sqrt{5} = 2.2361$

Hence $s = \frac{1}{2} \left(1.4142 + 3.6056 + 2.2361\right) = \frac{1}{2} \times 7.2559 = 3.628$

and Area=sqrt(3.628xx(3.628-1.4142)xx(3.628-3.6056)xx(3.628-2.2361)

= $\sqrt{3.628 \times 2.2138 \times 0.0224 \times 1.3919} = \sqrt{0.2504} = 0.5004$

Hence volume of pyramid is $\frac{1}{3} \times 0.5004 \times 12 = 2.0016$