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

May 9, 2017

24.47

#### Explanation:

• First find the length of each line using formula $\sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}}$ where 1 and 2 are x and y coordinates of the two points

Distance between (6,4) and (2,5) = $\sqrt{{\left(5 - 4\right)}^{2} + {\left(2 - 6\right)}^{2}}$
= $\sqrt{17}$
Distance between (2,5) and (3,2) = sqrt((2-5)^2 +(3-2)^2
= $\sqrt{10}$
Distance between (3,2) and (6,5) = $\sqrt{{\left(5 - 2\right)}^{2} + {\left(6 - 3\right)}^{2}}$
= $3 \sqrt{2}$

• Then find base area
Cos $\theta$ = $\frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 \cdot b \cdot c}$
= $\frac{{\left(\sqrt{17}\right)}^{2} + {\left(\sqrt{10}\right)}^{2} - {\left(3 \sqrt{2}\right)}^{2}}{2 \cdot \sqrt{10} \cdot \sqrt{17}}$
$\theta$ = $C o {s}^{-} 1$(0.345134245)
= ${69.8}^{\circ}$
Area = $\frac{1}{2} \cdot \sqrt{17} \cdot \sqrt{10} S \in 69.8$
= 6.11822

• Calculate volume using formula:
volume = Base area * height

= $6.11822 \cdot 4$
= $24.47$