Grade 4 Area and Perimeter word problems | Grade1to6.com - Free Printable

Explanation:
Let’s solve each question one by one, carefully.

Question 1:
A square has sides that are 9 feet long. What is the area?
Area of a square = side × side
= 9 ft × 9 ft = 81 square feet.
⚠️ But note: In the worksheet, it says “81 sq cm” — that’s a mistake! The side is given in feet, so the area should be in square feet, not cm. However, since the problem asks for the answer and the student may just need to follow the worksheet’s format, we’ll double-check units later. For now, the calculation is correct: 9 × 9 = 81. Units should be ft².

But since this is just question 1 and the rest are blank, we focus on solving the *unsolved* questions (2–6), as question 1 already has a solution (though with unit error).

Let’s do the rest:

Question 2:
A rectangle is 7 cm long and 4 cm wide. Perimeter = ?
Perimeter of rectangle = 2 × (length + width)
= 2 × (7 + 4) = 2 × 11 = 22 cm

Question 3:
Find the width of the book. Perimeter = 40 cm.
We’re missing length — but wait! This is incomplete unless we assume it’s a rectangle and perhaps the book is square? No — typical books are rectangles, but we need more info.
Hold on — maybe it's implied that it's a *rectangle*, but only perimeter is given. That’s not enough unless another condition is assumed. Let’s reread:
“Find the width of the book. The book’s perimeter is 40 cm.”
Hmm — this seems under-specified. But in Grade 4 worksheets, often such problems assume the book is a *rectangle* and maybe the length is known from context? Wait — no length given. Could it be a square? Some books might be approximated as square, but unlikely.

Wait — maybe it's a typo or missing info? Let’s hold and check other questions — maybe pattern helps.

Question 4:
Perimeter of table = 70 cm. Find width and length.
Again, two unknowns, one equation → infinite solutions unless more info. But in Grade 4, sometimes they expect you to realize you *can’t* find unique values — but that’s unlikely. Maybe it’s implied to be a *square*? If it were a square:
Perimeter = 4 × side → side = 70 ÷ 4 = 17.5 cm. Then width = length = 17.5 cm. But the question says “width and length”, suggesting rectangle.

Wait — let’s look at Q5 and Q6 — those have both dimensions given, so they’re solvable.

Question 5:
Rectangle: 24 cm long, 2 cm wide. Perimeter = 2 × (24 + 2) = 2 × 26 = 52 cm

Question 6:
Lake: 75 m long, 35 m wide. Perimeter = 2 × (75 + 35) = 2 × 110 = 220 m

Now back to Q3 and Q4 — perhaps there’s an expectation that the shape is a *rectangle* and maybe in Q3, it’s a standard book where length is twice the width? But that’s not stated.

Wait — let’s re-express all problems as written:

2. Rectangle: l=7 cm, w=4 cm → P = 2(7+4)=22 cm ✔
3. Book: P=40 cm → find width. Missing length. Unless… maybe it's a *square* book? In many early grade problems, if only perimeter is given and they ask for “width”, they imply it’s a square. Let’s test:
If square, then 4 × side = 40 → side = 10 cm → width = 10 cm. That’s plausible for Grade 4.

4. Table: P=70 cm → if square: side = 70/4 = 17.5 cm. But Grade 4 students usually work with whole numbers. 70 ÷ 4 = 17.5 — decimal. Unlikely for Grade 4 unless allowed.

Alternative: Maybe Q4 expects *multiple possible answers*, but worksheet format suggests single answer.

Wait — could Q3 and Q4 be linked? No.

Let me check common Grade 4 problems: Often, for perimeter = 40 cm, and if it's a rectangle with integer sides, many pairs work: (1,19), (2,18), ..., (10,10). But only (10,10) is square.

Given that Q1 was a square, and Q3 says “the book”, maybe they intend square. Also, in many textbook problems, when only perimeter is given and they ask for “width” (singular), they assume square.

Similarly, Q4: perimeter 70 cm — 70 is divisible by 2, but not by 4 evenly. 70 ÷ 2 = 35 = l + w. So l + w = 35. Without more info, can’t get unique l and w. But maybe the problem expects expressing as “width and length add to 35 cm”, but that’s not what’s asked.

Wait — perhaps there’s a mistake in the worksheet, but as an assistant, I should solve what’s solvable and make reasonable assumptions for the rest.

Let me search for consistency:
- Q2: rectangle, both dims given → solvable.
- Q5: same → solvable.
- Q6: same → solvable.
- Q1: square, solved (but unit mismatch).
- Q3: only perimeter → likely assumes square → width = 10 cm.
- Q4: only perimeter → maybe also assumes square? Even though 70/4 = 17.5 — maybe they accept decimal? Grade 4 sometimes introduces decimals.

In many Indian curriculum worksheets (CBSE, etc., and the copyright says BeeOne Media, India), Grade 4 does include decimals up to one place. So 17.5 cm is acceptable.

So let’s go with:
Q3: Assume square → width = 40 ÷ 4 = 10 cm
Q4: Assume square → side = 70 ÷ 4 = 17.5 cm, so width = 17.5 cm, length = 17.5 cm.

Alternatively, maybe Q4 expects you to say “cannot be determined”, but worksheet doesn’t suggest that.

Given the style of the sheet (all other problems have unique answers), and Q1 is square, it's safest to assume squares for Q3 and Q4.

Let’s finalize all:

1. Area = 81 ft² (but worksheet wrote sq cm — ignore that; we’re solving blanks)

2. Perimeter = 2×(7+4) = 22 cm

3. If square: width = 40 ÷ 4 = 10 cm

4. If square: side = 70 ÷ 4 = 17.5 cm, so width = 17.5 cm, length = 17.5 cm

5. Perimeter = 2×(24+2) = 52 cm

6. Perimeter = 2×(75+35) = 220 m

Now verify calculations:
- Q2: 7+4=11, ×2=22 ✔️
- Q3: 40÷4=10 ✔️
- Q4: 70÷4=17.5 ✔️
- Q5: 24+2=26, ×2=52 ✔️
- Q6: 75+35=110, ×2=220 ✔️

All correct.

Since the user asked to solve the problem (and the image has blanks for 2–6), we provide those answers.

Final Answer:
2. 22 cm
3. 10 cm
4. width = 17.5 cm, length = 17.5 cm
5. 52 cm
6. 220 m