The base of a triangular pyramid is a triangle with corners at #(8 ,5 )#, #(6 ,7 )#, and #(5 ,1 )#. If the pyramid has a height of #8 #, what is the pyramid's volume?

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Answer 1 · 2017-01-08T08:44:51

Pyramid's volume is #56/3#

Let they be

#A(8;5);B(6;7);C(5;1);h=8#

First, you would calculate the area of the triangle ABC, by getting the lenght of a side:

#AC=sqrt((x_C-x_A)^2+(y_C-y_A)^2)=sqrt((5-8)^2+(1-5)^2)=sqrt(9+16)=5#

and then the height relative to AC, by finding the distance between the point B and the line AC:

#m_(AC)=(y_C-y_A)/(x_C-x_A)=(1-5)/(5-8)=4/3#

Then the equation of the line AC is

#y-y_A=m_(AC)(x-x_A)#

that's

#y-5=4/3(x-8)#

#y-5=4/3x-32/3#

#3y-15=4x-32#

#4x-3y-17=0# in the form #ax+by+c=0#

Then let's calculate the distance:

#d=|ax_B+by_B+c|/sqrt(a^2+b^2)#

#d=|4*6-3*7-17|/sqrt(4^2+3^2)=|24-21-17|/sqrt25=14/5#

Now let's calculate the area of the triangle:

#Area=(side AC)*(d)*1/2=cancel5*cancel14^7/cancel5*1/cancel2=7#

The pyramid's volume is obtained by:

#V=1/3Area_(base)*h=1/3*7*8=56/3#