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Let's solve the given problem.

We are provided with the following information:

- The radius of the fountain is 10 meters.
- The gardener is planting flowers 15 meters away from the center of the fountain.

The distance from the center of the fountain to the flower beds represents the radius of the larger circle. This radius will be:

\[
\text{Radius of larger circle} = 10 \, \text{meters} + 15 \, \text{meters} = 25 \, \text{meters}
\]

Now, the area \( A \) of a circle is given by the formula:

\[
A = \pi r^2
\]

Substitute \( r = 25 \, \text{meters} \):

\[
A = \pi (25)^2 = \pi \times 625 = 625\pi \, \text{square meters}
\]

Thus, the area of the larger circle is approximately:

\[
A \approx 625 \times 3.1416 = 1963.5 \, \text{square meters}
\]

### Final Answer:
\[
\boxed{1963.5 \, \text{square meters}}
\]

The area of the larger circle, which includes the flower beds, is approximately **1963.5 square meters**.

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Would you like more details or further explanation? Here are 5 additional related questions to think about:

1. How would the solution change if the radius of the fountain were different?
2. What would be the circumference of the larger circle?
3. How can you calculate the area of the flower beds alone (excluding the fountain area)?
4. What is the significance of \(\pi\) in calculating areas of circles?
5. How could the gardener adjust the planting distance to cover more or less area?

**Tip:** Always double-check the radius used in area formulas; small errors can lead to significant changes in results.

A circular fountain in a park has a radius of 10 meters. A gardener is planting flowers around the fountain, 15 meters away from the center. The gardener plans to install a sprinkler system along the circumference of a circle that passes through the flower beds. What is the area of the larger circle that includes the flower beds?

Learn how to find the area of a larger circle that includes flower beds around a fountain. This problem involves calculating the radius and area of a circle using the formula A = πr². Suitable for students in Grades 6-8.

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