Question

Find the volume of a frustum of a right pyramid whose lower base is a square with a side 5 in. And whose upper base is a square with a side 3in and whose altitude is 12in

Answer 1

by similar triangles 3/5 = x / 12
x = 36/5 = 7.2

volume of frustum = 1/3 * 5^2 * 12 - 1/3 * 3^2 * 7.2

= 100-21.6

= 78.4 in^3

Answer 2

the volume of the frustum =(1/3)h[a1+a2]+√(a1 a2) where a1=bottom area,a2=top area and h=height
a1=100sa.inch
a2=25 sq.inch and
h=12inch
so the volume =(1/3)(12)[100+25+√(100*25)]
=4*[125+50]
=4*175=700 cubic inches or [inch cubed]