Question

show that the volume of the frustum 13/27 pi r²h​

Answer 1

It is given that in the frustum ABB'A' OA = r, OO' =h and OB' = √(h2 + r2 /9 )
In right triangle OO'B we get 
O'B'2 = OB'2 - OO'2O'B'2 
=h2 + r2 /9 - h2 
O'B'2 = r2 /9
O'B' = r / 3 
∴Volume of the frustum = (1/3) πh[ r2 + r2 /9 + r×(r/3)] 
V = (1/3) πh[ r2 + r2 /9 + r2 / 3]
V = (1/3) πh[13 r2 ]
V = (13 / 27) πr2h.