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

Oct 18, 2017

$20$

#### Explanation:

The triangle with vertices $\left(4 , 2\right)$, $\left(3 , 6\right)$ and $\left(7 , 5\right)$ can be drawn inside a $4 \times 4$ square with vertices $\left(3 , 2\right)$, $\left(7 , 2\right)$, $\left(7 , 6\right)$ and $\left(3 , 6\right)$, dividing it into $4$ triangles...

The two smaller triangles are right angled triangles with legs of lengths $1$ and $4$, so their total area is $4$.

The larger isosceles right angled triangle has area $\frac{1}{2} \cdot 3 \cdot 3 = \frac{9}{2}$

So the total area of the given triangle is:

${4}^{2} - 4 - \frac{9}{2} = \frac{15}{2}$

The pyramid has volume:

$\frac{1}{3} \cdot \text{base" * "height} = \frac{1}{3} \cdot \frac{15}{2} \cdot 8 = 20$

Oct 18, 2017

$20$

#### Explanation:

The area of a triangle with vertices $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$, $\left({x}_{3} , {y}_{3}\right)$ is given by the formula:

$\text{Area} = \frac{1}{2} \left\mid {x}_{1} {y}_{2} + {x}_{2} {y}_{3} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} - {x}_{2} {y}_{1} - {x}_{3} {y}_{2} \right\mid$

So putting:

$\left\{\begin{matrix}\left({x}_{1} {y}_{1}\right) = \left(4 2\right) \\ \left({x}_{2} {y}_{2}\right) = \left(3 6\right) \\ \left({x}_{3} {y}_{3}\right) = \left(7 5\right)\end{matrix}\right.$

we find that the base of our pyramid has area:

$\frac{1}{2} \left\mid \left(4\right) \left(6\right) + \left(3\right) \left(5\right) + \left(7\right) \left(2\right) - \left(4\right) \left(5\right) - \left(3\right) \left(2\right) - \left(7\right) \left(6\right) \right\mid$

$= \frac{1}{2} \left\mid 24 + 15 + 14 - 20 - 6 - 42 \right\mid$

$= \frac{1}{2} \left\mid - 15 \right\mid$

$= \frac{15}{2}$

Then the volume of the pyramid is:

$\frac{1}{3} \times \text{base" xx "height} = \frac{1}{3} \cdot \frac{15}{2} \cdot 8 = 20$