# The base of a triangular pyramid is a triangle with corners at (3 ,1 ), (4 ,9 ), and (5 ,7 ). If the pyramid has a height of 7 , what is the pyramid's volume?

Jan 12, 2018

$11.69 u n i t {s}^{3}$

#### Explanation:

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

$\therefore = \sqrt{{\left(5 - 3\right)}^{2} + {\left(7 - 1\right)}^{2}}$

$\therefore = \sqrt{{\left(2\right)}^{2} + {\left(6\right)}^{2}}$

$\therefore = \sqrt{\left(4\right) + \left(36\right)}$

$\therefore = \sqrt{\left(4\right) + \left(36\right)}$

$\therefore = \sqrt{40}$

side a$= 6.325$units

$\therefore = \sqrt{{\left(4 - 3\right)}^{2} + {\left(9 - 1\right)}^{2}}$

$\therefore = \sqrt{{\left(1\right)}^{2} + {\left(8\right)}^{2}}$

$\therefore = \sqrt{\left(1\right) + \left(64\right)}$

$\therefore = \sqrt{65}$

side b$= 8.062$units

$\therefore = \sqrt{{\left(5 - 4\right)}^{2} + {\left(7 - 9\right)}^{2}}$

$\therefore = \sqrt{{\left(1\right)}^{2} + {\left(2\right)}^{2}}$

$\therefore = \sqrt{\left(1\right) + \left(4\right)}$

$\therefore = \sqrt{5}$

side $c = 2.236$

Hero's formula:-

Area of $\triangle = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

where $s = \frac{a + b + c}{2}$

$\therefore s = \frac{6.325 + 8.062 + 2.236}{2}$

$\therefore s = \frac{16.623}{2}$

$\therefore s = 8.312$

:.=sqrt(8.312(8.312-6.325)(8.312-8.062)(8.312-2.236)))

$\therefore = \sqrt{\left(8.312\right) \left(1.987\right) \left(0.25\right) \left(6.076\right)}$

$\therefore = \sqrt{25.08771894}$

Area of $\triangle = 5.01 u n i t {s}^{2}$

Volume of triangular prism$= \frac{1}{3} A \times H$

$A =$triangular base and H= height of pyramid

$\therefore = \frac{1}{3} \times 5.01 \times 7$

$\therefore = 11.69 u n i t {s}^{3}$