The base of a triangular pyramid is a triangle with corners at #(2 ,6 )#, #(5 ,3 )#, and #(8 ,2 )#. If the pyramid has a height of #18 #, what is the pyramid's volume?

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Answer 1 · 2017-11-26T12:52:23

Volume of the pyramid with triangular base is #18#

Volume of triangular pyramid = #(1/3) AH# where A is the area of the triangular base and H is the height of the pyramid.

Area of triangular base = #(1/2) b h# where b is the base and h is the height of the triangle.

Base #= sqrt((5-2)^2 + (3-6)^2) = sqrt(3^2 + 3^2 )=color(blue)( 3 sqrt2)#

Eqn of Base =is
#((y-y_1)/(y_2-y_1)) = ((x-x_1)/(x_2-x_1))#

#((y-6)/(3-6))=((x-2)/(5-2))#

#(y-6) =( -x + 2)#
#y+x = 8# Eqn. (1)
Slope of base #m = (y_2 - y_1) / (x_2 - x_1)#
Slope #m = (3-6)/(5-2) = -1#

Slope of altitude #m_1 = -(1/m) = -(1/(-1)) = 1#

Eqn of Altitude is
#(y-y_3) = m_1(x-x_3)#

#y- 2 = 1 (x-8)#
#y-x = -6#. Eqn (2)

Solving Eqns (1) & (2), we get coordinates of the base of the altitude.
Coordinates of base of altitude are (7,1)

Height of altitude# = sqrt((8-7)^2 + (2-1)^2 )= color(red)(sqrt2)#

Area of triangular base #= (1/2) color(blue)(3 sqrt2) color(red )(sqrt2)#

Area #=3#

Volume of pyramid = #(1/3) * 3 * 18 = 18#