The base of a triangular pyramid is a triangle with corners at $(7 ,5 )$ , $(6 ,9 )$ , and $(3 ,4 )$ . If the pyramid has a height of $4$ , what is the pyramid's volume?

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Answer 1 · 2018-06-09T01:56:13

$color(crimson)("Volume of Pyramid " V_p = (1/3) * A_b * h = 11.36 " cubic units"$

$color(violet)("Volume of Pyramid " V_p = (1/3) * A_b * h$

$Area of base triangle " A_b = sqrt(s (s-a) (s-b) (s-c)), " using Heron's formula$

$A(7,5), B(6,9), C(3,4), h = 4$

$a = sqrt((6-3)^2 + (9-4)^2) = 5.83$

$b = sqrt((7-3)^2 + (5-4)^2) = 4.12$

$c = sqrt((6-7)^2 + (9-5)^2) = 4.12$

It's an isosceles triangle with sides b & c equal.

$"Semi-perimeter " s = (a + b + c) / 2$

$s = (5.83 + 4.12 + 4.12) / 2 ~~ 7.04$

$A_b = sqrt(7.04 * (7.04 - 5.83) * (7.04 - 4.12) * (7.04 - 4.12)) = 8.52$

$color(crimson)("Volume of Pyramid " V_p = (1/3) * A_b * h = (1/3) * 8.52 * 4 = 11.36 " cubic units"$

The base of a triangular pyramid is a triangle with corners at (7 ,5 )(7 ,5 ), (6 ,9 )(6 ,9 ), and (3 ,4 )(3 ,4 ). If the pyramid has a height of 4 4 , what is the pyramid's volume?