A sphere with radius $R$ is placed inside a hemispherical "cap" (spherical segment), whose radius is $\frac{R \sqrt{3}}{2}$. By what percentage is the surface area of the resulting shape larger than the surface area of the sphere?
Let $\varrho$ denote the radius of the hemisphere: $\varrho=\frac{R \sqrt{3}}{2}$. $(\varrho<R)$. The surface area of the sphere with radius $R$ is $A_{1}=4 \pi R^{2}$, and that of the hemisphere with radius $\varrho$ is $2 \pi \varrho^{2}=\frac{3}{2} \pi R^{2}$. If we place the "cap" on the sphere, it covers a spherical segment of height $m$ on the sphere. (The cross-section formed by the symmetry plane of the resulting shape is shown in the figure.)

By the Pythagorean theorem, $(R-m)^{2}+\varrho^{2}=R^{2}$, from which $m=\frac{R}{2}$. The surface area of the spherical segment is $2 \pi R m$. (This formula can be found in the Function Tables. Of course, we must exclude $\pi \varrho^{2}$, since the base circle of the spherical segment, being inside the sphere, does not contribute to the surface area.) The area of the covered part is $\pi R^{2}$, so the increase in surface area is $\frac{3}{2} \pi R^{2}-\pi R^{2}=\frac{1}{2} \pi R^{2}$. This is exactly one-eighth of $A_{1}$. Therefore, the surface area of the resulting shape is 12.5% larger than that of the sphere.

Andrea Pál (Jászárokszállás, Deák Ferenc Gimn., II. o.t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_f0cd4f74ce5e5277a6edg-1.jpg?height=580&width=499&top_left_y=683&top_left_x=802)