A quarter of a circular cake (a quarter circle with a radius of $15 \mathrm{~cm}$) was still there when the guests left. Then we cut a $1.5 \mathrm{~cm}$ wide strip parallel to each of the extreme radii and ate it. What fraction of the cake piece did we consume?

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Let's place one of the cut pieces according to the diagram. Notice that if we subtract the area of the circular segment $T_{2}$ from the area of the semicircle $T_{1}$, the difference almost gives the area of the burnt pie slices; it is only slightly more, by the area of a square with a side length of $1.5 \mathrm{~cm}$.

To calculate the area of $A B$, we need the angle $A O B$. We can find half of this angle from the right triangle $O A C$: $\cos \alpha=\frac{1.5}{15}=0.1$, hence $\alpha \approx 84.26^{\circ}$ and $\angle O A B = 2 \alpha \approx 168.52^{\circ}$.

The area of the circular segment $A B$ is given by the difference between the area of the sector $A O B$ and the area of the triangle $A O B$:

$$
\begin{gathered}
T_{2}=\frac{r^{2} \pi \alpha}{360^{\circ}}-\frac{r^{2} \sin 168.52^{\circ}}{2} \approx 330.89-22.39=308.5 \mathrm{~cm}^{2} \\
\quad T_{1}-T_{2} \approx \frac{r^{2} \pi}{2}-308.5=\frac{15^{2} \pi}{2}-308.5 \approx 44.929 \mathrm{~cm}^{2}
\end{gathered}
$$

From this difference, we need to subtract the area of the square with a side length of $1.5 \mathrm{~cm}$. Then we get:

$$
44.929 \mathrm{~cm}^{2}-2.25 \mathrm{~cm}^{2}=42.679 \mathrm{~cm}^{2}.
$$

This is the area of the burnt pie. Finally, the ratio of the two areas is:

$$
\frac{42.678}{15^{2} \pi / 4} \approx 0.2405
$$

which means we consumed almost a quarter of the pie slice.

Varga Attila (Tamási, Béri Balogh Á. Gymn., 11th grade)

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