# The base of a triangular pyramid is a triangle with corners at (6 ,7 ), (3 ,1 ), and (4 ,2 ). If the pyramid has a height of 8 , what is the pyramid's volume?

Mar 4, 2018

V=4

#### Explanation:

$V = \frac{1}{6} \cdot a \cdot {h}_{a} \cdot H$
$H$ is the hight into the 3-dimension.
${p}_{1} = \left(6 , 7\right)$
${p}_{2} = \left(3 , 1\right)$
${p}_{3} = \left(4 , 2\right)$
$H = 8$
To find the distanc $a$ between ${p}_{1}$ and ${p}_{2}$ we use pythagoras:
${a}^{2} + {b}^{2} = {c}^{2}$
Notice that the $c$ from pythagoras equals our $a$.
${\left({\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}\right)}^{\frac{1}{2}} = a$
${\left({\left(6 - 3\right)}^{2} + {\left(7 - 1\right)}^{2}\right)}^{\frac{1}{2}} = a$
${\left(9 + 36\right)}^{\frac{1}{2}} = a$
$\sqrt{45} = a$
${\left({\left({x}_{1} - {x}_{3}\right)}^{2} + {\left({y}_{1} - {y}_{3}\right)}^{2}\right)}^{\frac{1}{2}} = b$
$\sqrt{29} = b$
${\left({\left({x}_{2} - {x}_{3}\right)}^{2} + {\left({y}_{2} - {y}_{3}\right)}^{2}\right)}^{\frac{1}{2}} = c$
$\sqrt{2} = c$

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(\gamma\right)$
$2 = 45 + 29 - 2 \sqrt{45 \cdot 29} \cos \left(\gamma\right)$
$2 - 45 - 29 = - 2 \sqrt{1305} \cos \left(\gamma\right)$
${\cos}^{-} 1 \left(\frac{36}{\sqrt{1305}}\right) = \gamma$
4,76°~~gamma
$\sin \left(\gamma\right) = {h}_{a} / b | \cdot b$
$\sin \left(\gamma\right) \cdot b = {h}_{a}$
$0.45 \approx {h}_{a}$

$V = \frac{1}{6} \cdot \sqrt{45} \cdot 0.45 \cdot 8 \approx 4$