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Grade 6 Math worksheet focusing on perimeter word problems involving rectangles and fields.

Grade 6 Math worksheet on perimeter word problems featuring four questions about calculating the perimeter of rectangles and fields.

Grade 6 Math worksheet on perimeter word problems featuring four questions about calculating the perimeter of rectangles and fields.

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Show Answer Key & Explanations Step-by-step solution for: Perimeter Word Problems Worksheets | Grade1to6.com
Let’s solve each problem one by one, step by step.

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Problem i:

Sonia’s room is a rectangle.
Length = 9 feet
Width = 9.5 feet

Perimeter of a rectangle = 2 × (length + width)
So, perimeter = 2 × (9 + 9.5) = 2 × 18.5 = 37 feet

Check: 9 + 9.5 = 18.5; times 2 = 37 → correct.

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Problem ii:

Albert’s terrace garden:
Width = 8 feet
Length = 3 feet

First, area = length × width = 3 × 8 = 24 square feet

Second, he needs to cover all sides with a net → that means we need the perimeter.
Perimeter = 2 × (length + width) = 2 × (3 + 8) = 2 × 11 = 22 feet

Check: Area: 3×8=24 ✔️ Perimeter: 2×(3+8)=22 ✔️

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Problem iii:

We are told:
Width = 3.6 inches
Perimeter = 36 inches
Find the length.

Formula: Perimeter = 2 × (length + width)
So, 36 = 2 × (length + 3.6)

Divide both sides by 2:
18 = length + 3.6

Subtract 3.6 from both sides:
length = 18 - 3.6 = 14.4 inches

Check: 2 × (14.4 + 3.6) = 2 × 18 = 36 → correct.

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Problem iv:

Rectangular field:
Length = 75 yards
Width = 25 yards

Fence needed for perimeter → first find perimeter in yards:
Perimeter = 2 × (75 + 25) = 2 × 100 = 200 yards

But wait — it also says fencing is needed to divide the field into five sections.

How? The problem doesn’t say how the divisions are made, but usually in such problems, if you’re dividing a rectangle into 5 sections along the length or width, you add internal fences parallel to the shorter side.

Assuming the most common setup: dividing the field into 5 equal parts along the length → so you’d need 4 extra fences running across the width (because 5 sections need 4 dividers).

Each divider would be as long as the width → 25 yards.

So total extra fencing = 4 × 25 = 100 yards

Total fencing = outer perimeter + inner dividers = 200 + 100 = 300 yards

BUT — the question asks for feet, not yards!

Remember: 1 yard = 3 feet
So 300 yards = 300 × 3 = 900 feet

Check:
Outer fence: 2*(75+25)=200 yards
Inner fences: 4 * 25 = 100 yards
Total: 300 yards → 300*3=900 feet → correct.

Note: If the divisions were done differently (e.g., along the width), the answer might change — but since it’s not specified, the standard assumption is dividing along the longer side (length) with fences parallel to the width. This gives 4 internal fences of 25 yards each.

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Final Answer:

i. 37 feet
ii. Area: 24 square feet; Perimeter: 22 feet
iii. 14.4 inches
iv. 900 feet
Parent Tip: Review the logic above to help your child master the concept of area and perimeter worksheet 6th grade.
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