A sphere has been cut in half by a plane. How do the surface areas of the two spherical segments compare to each other if together they exceed the surface area of the original sphere by $25 \%$?
We can choose the radius of the sphere to be a unit length, since only the ratios need to be determined. The surface area of a spherical cap can be calculated using the following formula:

$$
A=2 \pi r m+\pi \varrho^{2}
$$

where $r$ is the radius of the sphere, $\varrho$ is the radius of the cut circle, and $m$ is the height of the spherical cap. In our case, we can write the following equation:

$$
2 \pi m+\pi \varrho^{2}+2 \pi(2-m)+\pi \varrho^{2}=\frac{5}{4} \cdot 4 \pi
$$

Performing the indicated operations, we get $2 \varrho^{2}=1, \varrho^{2}=\frac{1}{2}$.

From the right-angled triangle $O A F$ in the diagram: $\varrho^{2}=1-(1-m)^{2}$. Performing the operations and substituting $\varrho^{2}=\frac{1}{2}$, we get the quadratic equation: $2 m^{2}-4 m+1=0$, from which $m=\frac{2 \pm \sqrt{2}}{2} ; m_{1}=\frac{2+\sqrt{2}}{2} \approx 1.7$ and $m_{2}=\frac{2-\sqrt{2}}{2} \approx 0.2928$.

Both values are valid, giving the heights of the two spherical caps.

The ratio of the larger spherical cap to the smaller one is:

$$
\frac{2 \pi m_{1}+\pi \varrho^{2}}{2 \pi m_{2}+\pi \varrho^{2}}=\frac{2 \pi \frac{2+\sqrt{2}}{2}+\pi \frac{1}{2}}{2 \pi \frac{2-\sqrt{2}}{2}+\pi \frac{1}{2}}=\frac{\frac{\pi(5+2 \sqrt{2})}{2}}{\frac{\pi(5-2 \sqrt{2})}{2}}=\frac{5+2 \sqrt{2}}{5-2 \sqrt{2}} \approx \frac{7.8284}{2.1715} \approx 3.6
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_70c6b82c5bfd24c3298dg-1.jpg?height=423&width=436&top_left_y=1106&top_left_x=801)