# The base of a triangular pyramid is a triangle with corners at (2 ,1 ), (5 ,2 ), and (8 ,7 ). If the pyramid has a height of 18 , what is the pyramid's volume?

##### 1 Answer
Nov 17, 2017

$36$ units cubed

#### Explanation:

Volume of a triangular pyramid is:

$V = \frac{1}{3} A h$

Where $A =$ area of base, and $h =$ height.

Dimensions of triangle:

Let the angles be at:

$A = \left(2 , 1\right)$

$B = \left(5 , 2\right)$

$C = \left(8 , 7\right)$

Using the distance formula:

Length $A B$

$A B = \sqrt{{\left(2 - 5\right)}^{+} {\left(1 - 2\right)}^{2}} = \sqrt{10}$

$B C = \sqrt{{\left(5 - 8\right)}^{2} + {\left(2 - 7\right)}^{2}} = \sqrt{34}$

$A C = \sqrt{{\left(8 - 2\right)}^{2} + {\left(7 - 1\right)}^{2}} = \sqrt{72} = 6 \sqrt{2}$

Finding $\sin \left(A\right)$ of angle $A$ using the cosine rule:

$\cos \left(A\right) = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}$

$\cos \left(A\right) = \frac{{\left(6 \sqrt{2}\right)}^{2} + {\left(\sqrt{10}\right)}^{2} - {\left(\sqrt{34}\right)}^{2}}{2 \left(6 \sqrt{2}\right) \left(\sqrt{10}\right)}$

$\to = \frac{72 + 10 - 34}{24 \sqrt{5}} = \frac{48}{24 \sqrt{5}} = \frac{2}{\sqrt{5}}$

$\sin \left(A\right) = \sin \left({\cos}^{-} 1 \left(\frac{2}{\sqrt{5}}\right)\right) = \frac{\sqrt{5}}{5}$

Area of triangle from diagram:

$\frac{1}{2} \left(\sqrt{10}\right) \sin \left(A\right) b$

$A r e a = \frac{1}{2} \left(\sqrt{10}\right) \left(\frac{\sqrt{5}}{5}\right) \left(6 \sqrt{2}\right) = \frac{60}{10} = 6$

Area of pyramid:

$\frac{1}{3} \left(6\right) \left(18\right) = 36 \textcolor{w h i t e}{88}$