# The base of a triangular pyramid is a triangle with corners at (8 ,5 ), (2 ,7 ), and (3 ,1 ). If the pyramid has a height of 8 , what is the pyramid's volume?

Mar 7, 2016

${V}_{P r} = B A \times H$
${A}_{\Delta} = \frac{1}{2} \left(6.08 \cdot 6.40\right) \sin 60.8 \approx 17$
${V}_{P y} = \frac{1}{3} {V}_{P r}$
${V}_{P y} = \frac{1}{3} \left(17 \cdot 8\right) = 45.3$ cubic units
you can also use determinant to get the area of the triangle:
${A}_{\Delta} = \frac{1}{2} \det | \vec{{v}_{1}} \times \vec{{v}_{2}} | = \frac{1}{2} \det | \left(- 1 , 6\right) , \left(5 , 4\right) | = 17$ #### Explanation:

This is a straightforward implementation of the Pyramid volume formula: ${V}_{P y} = \frac{1}{3} {V}_{P r}$ where
${V}_{P y} =$ Volume of Pyramid and ${V}_{P r} =$ Volume of Prism
with same base and same height.
Now the volume of any prism is straightforward accumulation of the base area over the dimension of the height of the prism, $H$.
${V}_{P r} = B A \times H$ where
$B A$ = Base Area of prism and $H$ = Height of prism.
Now find the vector that enclose the triangle and use the cross product between them to find the Area, ${A}_{\Delta} = \frac{1}{2} {A}_{\square}$.
The vectors are:
vec(v_1) = (2i + 7j) - (3i+ j) = -i +6j; |vec(v_1)|=6.08
vec(v_2) = (8i + 5j) - (3i+ j) = -5i +4j; |vec(v_2)|=6.40
vec(v_3) = (2i + 7j) - (8i+ 5j) = -6i +2j; |vec(v_3)|=6.32

Now the cross product of any of this three vectors will yield the area of a parallelogram, ${A}_{\square}$. Let's pick ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$
The angle between them is $\alpha = {60.8}^{0}$

${A}_{\Delta} = \frac{1}{2} {A}_{\square} = \frac{1}{2} | \vec{{v}_{1}} | \times | \vec{{v}_{2}} | \sin 60.8$
Theta now since ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ the area is now simply:
${A}_{\Delta} = \frac{1}{2} \left(6.08 \cdot 6.40\right) \sin 60.8 \approx 17$
Now we know height, $H = 8$
Thus the ${V}_{P y} = \frac{1}{3} \left(17 \cdot 8\right) = 45.3 c u b i c u n i t s$

You can use vectors ${\vec{v}}_{2}$ and ${\vec{v}}_{3}$ and get the same
area $\approx 17$
The angle between them is $\beta = {57.09}^{0}$
${A}_{\Delta} = \frac{1}{2} \left(6.32 \cdot 6.40\right) \sin 57.09 \approx 17$ as before

The other way is simply determine the cross product using the determinant. Again let's consider $\vec{{v}_{1}}$ and $\vec{{v}_{2}}$
Then ${A}_{\Delta} = \frac{1}{2} \det | \vec{{v}_{1}} \times \vec{{v}_{2}} | = \frac{1}{2} \det | \left(- 1 , 6\right) , \left(5 , 4\right) | = 17$