3. A city wants to put a fence around a rectangular children's park. The park is $$ 32 \frac{1}{6} $$ yards by $$ 10 \frac{1}{3} $$ yards. How much fencing does the city need?

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To determine how much fencing the city needs to enclose the rectangular children's park, we need to calculate the **perimeter** of the rectangle. The perimeter $$ P $$ of a rectangle is given by the formula:

$$
P = 2 \times (\text{Length} + \text{Width})
$$

Given:
- **Length** = $$ 32 \frac{1}{6} $$ yards
- **Width** = $$ 10 \frac{1}{3} $$ yards

Let's break it down step by step.

### Step 1: Convert Mixed Numbers to Improper Fractions

1. **Length**:
   $$
   32 \frac{1}{6} = \frac{32 \times 6 + 1}{6} = \frac{192 + 1}{6} = \frac{193}{6} \text{ yards}
   $$

2. **Width**:
   $$
   10 \frac{1}{3} = \frac{10 \times 3 + 1}{3} = \frac{30 + 1}{3} = \frac{31}{3} \text{ yards}
   $$

### Step 2: Calculate the Sum of Length and Width

$$
\text{Length} + \text{Width} = \frac{193}{6} + \frac{31}{3}
$$

To add these fractions, they need a common denominator. The least common denominator between 6 and 3 is **6**.

$$
\frac{31}{3} = \frac{31 \times 2}{3 \times 2} = \frac{62}{6}
$$

Now, add the fractions:

$$
\frac{193}{6} + \frac{62}{6} = \frac{193 + 62}{6} = \frac{255}{6} = 42 \frac{3}{6} = 42 \frac{1}{2} \text{ yards}
$$

### Step 3: Calculate the Perimeter

$$
P = 2 \times 42 \frac{1}{2} = 2 \times \left(42 + \frac{1}{2}\right) = 84 + 1 = 85 \text{ yards}
$$

### **Conclusion**

The city needs **85 yards** of fencing to enclose the children's park.

**Answer:**  
The city needs 85 yards of fencing to enclose the park.
The city needs 85 yards of fencing to enclose the park.

3. A city wants to put a fence around a rectangular children's park. The park is \( 32 \frac{1}{6} \) yards by \( 10 \frac{1}{3} \) yards. How much fencing does the city need?

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