25) From a semicircle of cardboard with a radius of $10 \mathrm{~cm}$, a circle of maximum diameter is cut out. From the two remaining segments, two circles of maximum diameter are cut out. What is the percentage of wasted cardboard?
(A) $10 \%$
(B) $20 \%$
(C) $25 \%$
(D) $30 \%$
(E) $50 \%$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c3d7afe478310a3de07bg-2.jpg?height=216&width=418&top_left_y=1671&top_left_x=2383)
25) The answer is (C).

Let $R$ be the radius of the initial semicircle, obviously the first circle cut has radius $R / 2$. If $r$ is the radius of one of the further circles, then, by the Pythagorean theorem,

$$
\left(C^{\prime} N\right)^{2}=\left(C^{\prime} C\right)^{2}-(C N)^{2}=\left(C^{\prime} O\right)^{2}-\left(C^{\prime} M\right)^{2}
$$

which means

$$
\left(\frac{R}{2}+r\right)^{2}-\left(\frac{R}{2}-r\right)^{2}=(R-r)^{2}-r^{2}
$$

from which we get $r=\frac{R}{4}$. Therefore, the total area of the three circles cut out is:

$$
\pi \frac{R^{2}}{4}+2 \pi \frac{R^{2}}{16}
$$

and thus the ratio with the area of the initial semicircle, which is $\frac{\pi R^{2}}{2}$, is $\frac{3}{4}$. Therefore, $\frac{1}{4}$ of the cardboard is wasted.