Two right circular cones of height 16.4 cm and 17.2 cm respectively have base radius of 8.4 cm. These two cones are melted and recast into a sphere. Find the diameter of the sphere.

sum of volume of 2 cones = volume of sphere

$$ \frac{1}{3} \pi r1^2 h1 + \frac{1}{3} \pi r2^2 h2 = \frac{4}{3} \pi r^3 $$

h1 = 16.4 cm , h2 = 17.2 cm

r1 = r2 = 8.4 cm

substituting

$$ \frac{1}{3} \pi \times 8.4^2 \times 16.4 + \frac{1}{3} \pi \times 8.4^2 \times 17.2 = \frac{4}{3} \pi \times r^3 $$

$$ \frac{1}{3} \pi 8.4^2 (16.4 + 17.2 ) = \frac{4}{3} \pi r^3 $$

$$ 8.4^2 \times 33.6 = 4 \times r^3 $$

$$ r^3 = \frac{ 8.4 \times 8.4 \times 33.6}{4} $$

solving r = 8.4

diameter = $$ 8.4 \times 2 = 16.8 $$