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
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]
