The base of a triangular pyramid is a triangle with corners at #(6 ,8 )#, #(2 ,4 )#, and #(7 ,3 )#. If the pyramid has a height of #2 #, what is the pyramid's volume?

1 Answer
Mar 19, 2016

#V_(Delta_(pyr)) = BA*H = 24*2 = 48#

Explanation:

Given: Base Triangle Vertices: #A(2,4,0), B(7,3,0), C(6,8,0)#
Using Distance formula find the length of the sides of the base triangle, #bar(AB), bar(BC), bar(CA)#:
#bar(AB) = sqrt((2-7)^2 + (4-3)^2 )=sqrt(26)#
#bar(BC) = sqrt((7-6)^2 + (3-8)^2 )=sqrt(26)#
#bar(CA) = sqrt((6-2)^2 + (8-4)^2 )=4sqrt(2)#

Now you can use the Pyramid volume formula to calculate the Volume. #V = 1/3 BA*H#
where #BA#, and #H# are Base Area and Height of the pyramid respectively.
We are going to use Heron Formula to calculate the area of the triangle:
#A_(Delta) = sqrt(p(p-bar(AB))(p-bar(BC))(p-bar(CA)) #
Where #p=P/2=(bar(AB)+bar(BC)+bar(CA))/2#
#p= (sqrt(26)+sqrt(26)+4sqrt(2))/2#
#A_(Delta) = sqrt(((sqrt(26)+sqrt(26)+4sqrt(2))/2)((sqrt(26)+sqrt(26)+4sqrt(2))/2-sqrt(26))((sqrt(26)+sqrt(26)+4sqrt(2))/2-sqrt(26))((sqrt(26)+sqrt(26)+4sqrt(2))/2-4sqrt(2)) #
#A_Delta = sqrt(((sqrt(26)+sqrt(26)+4sqrt(2)))/2(4sqrt(2))(4sqrt(2))((sqrt(26)+sqrt(26)-4sqrt(2))/2 )#
#A_Delta = sqrt( (2sqrt(26)+4sqrt(2) )(2sqrt(26)-4sqrt(2))(4*2) )#
#A_Delta = sqrt( (4*26-16*2 )(4*2) ) = 24#
and the Volume is:
#V_(Delta_(pyr)) = BA*H = 24*2 = 48#

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