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

Jul 18, 2018

Volume of Pyramid $\approx 3.3333$ cubic units.

Explanation:

Using distance formula and Heron’s formula we can calculate the base Area of the pyramid. Multiplying it by one third the height will give the volume.

Distance formula $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Heron’s formula ${A}_{t} = \sqrt{s \cdot \left(s - a\right) \cdot \left(s - b\right) \cdot \left(s - c\right)}$

Where s is the semiperemter of the triangle and a, b, c are the three sides of the base triangle.

Let $A \left(3 , 1\right) , B \left(2 , 7\right) C \left(3 , 6\right)$

$a = \sqrt{{\left(2 - 3\right)}^{2} + {\left(7 - 6\right)}^{2}} = 1.4142$

$b = \sqrt{{\left(3 - 3\right)}^{2} + {\left(6 - 1\right)}^{2}} = 5$

$c = \sqrt{{\left(3 - 2\right)}^{2} + {\left(1 - 7\right)}^{2}} = 6.0828$

Semiperemeter $s = \frac{a + b + c}{2} = \frac{12.497}{2} = 6.2485$

${A}_{t} = \sqrt{6.2485 \cdot \left(6.2485 - 1.4142\right) \cdot \left(6.2485 - 5\right) \cdot \left(6.2485 - 6.0828\right)}$

${A}_{t} \approx 2.5$ sq. units

Volume of pyramid ${V}_{p} = \left(\frac{1}{3}\right) \cdot 2.5 \cdot 4 \approx 3.3333$ cubic units.