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The problem presented in the image appears to describe a swimming pool, represented in both top and side views, with a synthetic grass area on one side and a path (jalan) on three sides. The pool has dimensions of 20 meters in length and 10 meters in width. The depth of the pool decreases from 2 meters to 1 meter, as indicated by the side view.

### Key Points:
1. The pool is rectangular, with dimensions of 20 meters by 10 meters.
2. The depth of the pool decreases from 2 meters at one end to 1 meter at the other end, forming a sloped bottom.
3. There is a "jalan" or path around three sides of the pool, with the width of this path denoted as "x."
4. The total area of the path (jalan) is 88 square meters.
5. The remaining area on the fourth side is used for synthetic grass (rumput sintetis), and the problem asks for the area of the grass if the path's area is already known to be 88 square meters.

### Step-by-Step Solution:

#### Step 1: Path (Jalan) Area
The total area of the path is given as 88 square meters. The path surrounds three sides of the pool, and the width of the path on each side is denoted as \(x\).

The three sides of the pool that have paths are:
- Two lengths of 20 meters each (both long sides).
- One width of 10 meters (one short side).

So, the total perimeter covered by the path is \(20 + 20 + 10 = 50\) meters.

The area of the path is therefore \(50x\), where \(x\) is the width of the path. Given that this area is 88 square meters, we can set up the equation:
\[
50x = 88
\]
Solving for \(x\):
\[
x = \frac{88}{50} = 1.76 \text{ meters}.
\]

#### Step 2: Grass Area (Rumput Sintetis)
The area available for the synthetic grass is along the remaining side of the pool, which is 10 meters long (the other width), with a width of \(x = 1.76\) meters.

The area for the grass is therefore:
\[
\text{Grass Area} = 10 \times 1.76 = 17.6 \text{ square meters}.
\]

### Final Answer:
The area that will be planted with synthetic grass is 17.6 square meters.

Would you like more details or have any questions?

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Here are 5 related questions for further practice:
1. What would be the width of the path if the total area was 100 square meters instead of 88 square meters?
2. How would the area of the grass change if the pool were 25 meters long instead of 20 meters?
3. What is the total surface area of the pool's sloping bottom, given the depth variation?
4. How much synthetic grass would be needed if the width of the path was doubled?
5. If the depth of the pool was constant at 2 meters, how would that change the side view diagram?

**Tip:** When solving geometry problems, breaking down the problem into distinct areas or parts can help simplify the calculations.

A rectangular swimming pool measuring 20 m x 10 m, as shown in the top view, has a depth of 1 m, which decreases to 2 m, as shown in the side view. Three sides of the pool have a path with width x meters, and the remaining side is covered with synthetic grass, having the same width as the path. If the total area of the path is 88 m², calculate the area covered with synthetic grass.

Solve a geometry problem involving a rectangular pool and the calculation of the path and synthetic grass areas. The pool is 20 m x 10 m with varying depths, and the total area of the path around it is 88 m². Suitable for students in Grades 7-9.

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