The height of a right circular cone and the radius of its circular base are respectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so as to divide it into two parts. The volume of the frustum (i.e., the lover part) of the cone is 44 cubic cm. The radius of the upper circular surface of the frustum (taking π = 22/7 ) is

let DO' = r cm

OO' = h cm

triangle ADO' and ABO are similar

therefore $$ \frac{AO'}{AO} = \frac{DO'}{BO'} $$

$$ \frac{9 - h}{9} = \frac{r}{3} $$

9 - h = 3r

h = 9 - 3r

volume of frustum = $$ \frac{1}{3} \pi h ( r1^2 + r2^2 + r1r2 ) $$

$$ 44 = \frac{1}{3} \times \frac{22}{7} \times (9 - 3r) (9 + r^2 + 3r) $$

$$ 44 = \frac{1}{3} \times \frac{22}{7} \times 3 (3 - r)(r^2 + 3r + 3^2) $$

$$ 44 = \frac{22}{7} (3 - r)(r^2 + 3r + 3^2) $$

$$ 44 = \frac{22}{7} (3^3 - r^3) $$

$$ 14 = 27 - r^3 $$

$$ r^3 = 13 $$

$$ r =\sqrt[3]{13} $$