The base of a triangular pyramid is a triangle with corners at $(8 ,5 )$ , $(7 ,2 )$ , and $(4 ,6 )$ . If the pyramid has a height of $7$ , what is the pyramid's volume?

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Answer 1 · 2018-07-30T03:08:28

$color(green)(V = 1/3 A_b h = (1/3) * 7.1171 * 7 = 16.6065, " cubic units"$

Given : $A (8,5), B (7,2), C (4,6)$ , h = 7#

Using distance formula we can calculate the lengths of sides a, b, c.

$d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)$

$a = sqrt((7-4)^2+(2-6)^2) = 5$

$b = sqrt((4-8)^2+(6-5)^2) = sqrt 17$

$c = sqrt((8-7)^2+(5-2)^2) = sqrt 10$

Semi perimeter of base triangle $s = (a+b+c)/2$

$s = (5 + sqrt 17 + sqrt 10)/2 = 12.2854/2 ~~ 6.2427$

Area of triangle $A_t = sqrt(s (s-a) (s-b) (s-c))$

$A_t = sqrt(6.2427 * (6.2427-5)*(6.2427- sqrt 17)*(6.2427-sqrt 10))$

#A_t ~~ 7.1171

Volume of pyramid $V = (1/3) * A_t * h$

$color(green)(V = (1/3) * 7.1171 * 7 = 16.6065, " cubic units"$

The base of a triangular pyramid is a triangle with corners at (8 ,5 )(8 ,5 ), (7 ,2 )(7 ,2 ), and (4 ,6 )(4 ,6 ). If the pyramid has a height of 7 7 , what is the pyramid's volume?