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?

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Answer 1 · 2018-01-12T14:08:04

#11.69 units^3#

#sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#:.=sqrt((5-3)^2+(7-1)^2)#

#:.=sqrt((2)^2+(6)^2)#

#:.=sqrt((4)+(36))#

#:.=sqrt(40)#

side a#=6.325#units

#:.=sqrt((4-3)^2+(9-1)^2)#

#:.=sqrt((1)^2+(8)^2)#

#:.=sqrt((1)+(64))#

#:.=sqrt(65)#

side b#=8.062#units

#:.=sqrt((5-4)^2+(7-9)^2)#

#:.=sqrt((1)^2+(2)^2)#

#:.=sqrt((1)+(4))#

#:.=sqrt(5)#

side #c=2.236#

Hero's formula:-

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

where #s=(a+b+c)/2#

#:.s=(6.325+8.062+2.236)/2#

#:.s=16.623/2#

#:.s=8.312#

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

#:.=sqrt((8.312)(1.987)(0.25)(6.076))#

#:.=sqrt(25.08771894)#

Area of #triangle=5.01units^2#

Volume of triangular prism#=1/3AxxH#

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

#:.=1/3xx5.01xx7#

#:.=11.69 units^3#