Grade 6 math worksheet with perimeter word problems involving real-life scenarios like rooms, gardens, and fields.

Let's solve each of the word problems step by step.

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i. Sonia's Room Perimeter

Problem:
Sonia’s room is rectangular. Length = 9 feet, Width = 9.5 feet.
Find the perimeter.

Formula for perimeter of a rectangle:
$$
P = 2 \times (L + W)
$$

Where:
- $ L = 9 $ ft
- $ W = 9.5 $ ft

$$
P = 2 \times (9 + 9.5) = 2 \times 18.5 = 37 \text{ feet}
$$

✔ Answer: The perimeter of Sonia’s room is 37 feet.

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ii. Albert’s Terrace Garden

Problem:
Garden is 8 feet wide and 3 feet long.
Find:
1. Area
2. Perimeter (for net covering all sides)

#### Area:
$$
\text{Area} = \text{Length} \times \text{Width} = 3 \times 8 = 24 \text{ square feet}
$$

#### Perimeter:
$$
P = 2 \times (L + W) = 2 \times (3 + 8) = 2 \times 11 = 22 \text{ feet}
$$

✔ Answers:
- Area = 24 sq ft
- Perimeter = 22 feet (so he needs a net of 22 feet in length)

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iii. Rectangle with Given Width and Perimeter

Problem:
Width = 3.6 inches
Perimeter = 36 inches
Find the length.

Use the perimeter formula:
$$
P = 2(L + W)
$$
$$
36 = 2(L + 3.6)
$$

Divide both sides by 2:
$$
18 = L + 3.6
$$

Subtract 3.6 from both sides:
$$
L = 18 - 3.6 = 14.4 \text{ inches}
$$

✔ Answer: The length is 14.4 inches.

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iv. Rectangular Field Fencing

Problem:
Field is 75 yards long and 25 yards wide.
Fence needed:
- Around the perimeter
- To divide the field into five sections

We need to find total fencing in feet.

#### Step 1: Find the perimeter
$$
P = 2(L + W) = 2(75 + 25) = 2 \times 100 = 200 \text{ yards}
$$

#### Step 2: Divide into five sections

To divide a rectangle into five equal sections, we assume it's divided along the length (i.e., parallel to the width). So we need 4 internal fences (since 5 sections require 4 dividers).

Each divider is as wide as the width of the field → 25 yards.

So total internal fencing = $ 4 \times 25 = 100 $ yards

#### Total fencing needed:
$$
\text{Perimeter} + \text{Internal dividers} = 200 + 100 = 300 \text{ yards}
$$

But the question asks for feet.

Convert yards to feet:
$ 1 \text{ yard} = 3 \text{ feet} $

$$
300 \text{ yards} \times 3 = 900 \text{ feet}
$$

✔ Answer: 900 feet of fencing are needed.

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| i. | 37 feet |
| ii. | Area = 24 sq ft, Perimeter = 22 feet |
| iii. | 14.4 inches |
| iv. | 900 feet |

Let me know if you'd like these explained visually or with diagrams!