Perimeter Word Problems Worksheets | Grade1to6.com - Free Printable

Let’s solve each problem one by one, step by step.

---

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.

---

Problem ii:

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

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

Now, he wants 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 ✔️

---

Problem iii:

Width = 3.6 inches
Perimeter = 36 inches
We need to 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 ✔️

---

Problem iv:

Rectangular field:
Length = 75 yards
Width = 25 yards

Fence needed for perimeter → first calculate perimeter:
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 equal sections along the length or width, you add internal fences.

Assume the most common case: dividing the field into 5 equal parts along the length, meaning you run 4 fences parallel to the width (since 5 sections need 4 dividers).

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

So, extra fencing = 4 × 25 = 100 yards

Total fencing = perimeter + internal fences = 200 + 100 = 300 yards

BUT — the question asks: “How many feet of fencing are needed?”

We have everything in yards. Convert to feet:
1 yard = 3 feet
So, 300 yards = 300 × 3 = 900 feet

Wait — let’s double-check the assumption.

Alternative: Maybe they mean dividing into 5 sections along the width? Then dividers would be parallel to the length → each 75 yards, and still 4 dividers → 4 × 75 = 300 yards → total fencing = 200 + 300 = 500 yards = 1500 feet.

Which is correct?

The problem says: “divide the field into five sections” — without specifying direction. But in standard math problems like this, unless specified, we assume the divisions are made parallel to the shorter side to minimize material — but actually, no, that’s not always true.

Actually, looking at typical textbook problems, when they say “divide into n sections”, and don’t specify, they often mean running fences parallel to the width (i.e., cutting the length into parts). So 4 fences across the width.

But let’s think again: If the field is 75 yards long and 25 yards wide, and you want 5 sections, you could either:

- Cut the length into 5 parts → each section 15 yards long → need 4 fences running the full width (25 yards each) → 4 × 25 = 100 yards extra.

OR

- Cut the width into 5 parts → each section 5 yards wide → need 4 fences running the full length (75 yards each) → 4 × 75 = 300 yards extra.

Which one is intended? The problem doesn’t specify. But note: the question says “fencing is also needed to divide the field into five sections.” It doesn’t say “equal” sections, but we assume equal.

In many similar problems, especially at Grade 6 level, they expect you to assume the divisions are made parallel to the width (so you’re splitting the length), because that uses less fencing — but actually, no, the problem doesn’t hint at minimizing.

Wait — perhaps I should look for clues. The answer might depend on interpretation.

But let me check online or recall: In most curriculum materials, when not specified, they assume the divisions are made parallel to the shorter dimension? No, actually, let's think logically.

If you have a rectangle 75x25, and you want 5 sections, the most straightforward way is to make 4 cuts parallel to the 25-yard side, so each section is 15x25. That makes sense.

Moreover, if you cut parallel to the 75-yard side, you’d get very narrow sections (5 yards wide), which might not be practical, but mathematically possible.

However, since the problem is from a Grade 6 worksheet, and likely expects a single answer, I think the intended interpretation is: divide the length into 5 equal parts, so add 4 fences parallel to the width.

Thus, extra fencing = 4 × 25 = 100 yards.

Total fencing = 200 (perimeter) + 100 (internal) = 300 yards.

Convert to feet: 300 × 3 = 900 feet.

But let me confirm units: the field dimensions are given in yards, and the question asks for feet — so conversion is necessary.

Another thought: maybe "divide into five sections" means something else? Like, perhaps radial? No, for a rectangle, it's almost always straight lines parallel to sides.

I think 900 feet is the expected answer.

To be thorough, let’s see if there’s another common interpretation.

Sometimes, “divide into five sections” might mean creating 5 separate areas, which could require more complex fencing, but that’s too advanced for Grade 6.

I’ll go with 4 internal fences parallel to the width.

So total fencing in yards: 200 + 100 = 300 yards → 900 feet.

✔ Final check: Perimeter 2*(75+25)=200 yards. Internal: 4*25=100 yards. Total 300 yards. 300*3=900 feet. Correct.

---

Final Answer:

i. 37 feet
ii. Area: 24 square feet; Perimeter: 22 feet
iii. 14.4 inches
iv. 900 feet