Question

A solid metal sphere of radius 9 is melted and transformed into 3 identical spheres. What is the ratio of the surface area of one of these spheres to the surface area of the original sphere?

Answer 1

Answer:

$243^{2/3}:324$ or approximately 0.481

Step-by-step explanation:

Volume of a sphere: 4/3πr³

Surface area of a sphere: 4πr²

The initial sphere of radius 9 has a volume of $\displaystyle\\9^3\cdot \frac{4}{3}\cdot \pi=972\pi$

When melted into three identical spheres, each must have a volume of $\displaystyle \frac{972\pi}{3}=324\pi$

The radius of the smaller spheres must be:

$\frac{4}{3}\pi r_s^3=324\pi \implies r_s=\sqrt[3]{243}\approx 6.24$

Surface area of each smaller sphere:

$\displaystyle \\4\pi r^2\vert_{r=6.24}\approx 155.76\pi$

Surface area of initial sphere:

$\displaystyle \\4\pi r^2\vert_{r=9}\approx 324\pi$

The desired ratio is $155.76/324\approx \boxed{0.481}$

Exact: $243^{2/3}:324$

Answer 2

Answer:

The ratio is 1 to 2  

Step-by-step explanation:

Volume of the original sphere is ≈ 3,053.63 units³

The surface area of the original sphere is ≈ 1,017.88 units²

Volume of one of the small sphere is ≈ 1,017.88 units³

The radius of one of the small sphere is ≈ 6.24 units

The surface area of one of the small sphere is ≈ 489.3 units²

The ratio of the surface area of one of the small sphere to the surface area of the original sphere is

$\frac{489.3}{1017.88}$ ≈ $\frac{1}{2}$

489.3 / 1017.88 ≈ 1 / 2  

The ratio is 1 to 2