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"#