Area and Perimeter Word Problems (Grade 4)
Area and perimeter are used to solve many real-life problems — fencing a garden, tiling a room, or wrapping a gift. Knowing when to use area and when to use perimeter is essential.
In Class 4, you will practise word problems that require you to decide which formula to use and then apply it correctly.
What is Area and Perimeter Word Problems - Class 4 Maths (Measurement)?
Perimeter is the total distance around a shape. Use it when you need to find the length of a boundary, fence, border, or frame.
Area is the surface enclosed inside a shape. Use it when you need to find the space for painting, tiling, carpet, or grass.
Area and Perimeter Word Problems (Grade 4) Formula
Rectangle: Area = l × b | Perimeter = 2 × (l + b)
Square: Area = s × s | Perimeter = 4 × s
How to decide:
- Fencing, ribbon, wire, border → Perimeter
- Painting, tiling, carpet, paper → Area
Solved Examples
Example 1: Example 1: Fencing a rectangular field
Problem: Aman's rectangular field is 60 m long and 40 m wide. He wants to put a fence around it. Fencing costs ₹50 per metre. Find the total cost.
Solution:
Step 1: Fence goes around the field → use perimeter.
Step 2: Perimeter = 2 × (60 + 40) = 2 × 100 = 200 m
Step 3: Cost = 200 × ₹50 = ₹10,000
Answer: The total fencing cost is ₹10,000.
Example 2: Example 2: Tiling a square room
Problem: A square room has side 5 m. Each tile covers 1 sq m and costs ₹80. Find the number of tiles and total cost.
Solution:
Step 1: Tiling covers the floor → use area.
Step 2: Area = 5 × 5 = 25 sq m
Step 3: Number of tiles = 25. Cost = 25 × ₹80 = ₹2,000
Answer: 25 tiles needed. Total cost = ₹2,000.
Example 3: Example 3: Both area and perimeter
Problem: Priya's rectangular garden is 12 m long and 8 m wide. She wants to: (a) plant grass on it, and (b) put a fence around it. Find the area for grass and perimeter for fencing.
Solution:
Step 1: Area = 12 × 8 = 96 sq m (for grass)
Step 2: Perimeter = 2 × (12 + 8) = 2 × 20 = 40 m (for fencing)
Answer: Grass area = 96 sq m. Fencing length = 40 m.
Example 4: Example 4: Ribbon around a card
Problem: Aditi makes a square greeting card with side 14 cm. She puts ribbon along the edges. How much ribbon does she need?
Solution:
Step 1: Ribbon along edges → perimeter.
Step 2: Perimeter = 4 × 14 = 56 cm
Answer: Aditi needs 56 cm of ribbon.
Example 5: Example 5: Painting a wall
Problem: A rectangular wall is 6 m long and 3 m high. One litre of paint covers 5 sq m. How many litres are needed?
Solution:
Step 1: Painting → use area.
Step 2: Area = 6 × 3 = 18 sq m
Step 3: Litres = 18 ÷ 5 = 3.6 → round up to 4 litres (you cannot buy part of a litre).
Answer: 4 litres of paint are needed.
Example 6: Example 6: Finding a missing dimension
Problem: The area of a rectangular playground is 600 sq m. The breadth is 20 m. Find the length and perimeter.
Solution:
Step 1: Length = Area ÷ Breadth = 600 ÷ 20 = 30 m
Step 2: Perimeter = 2 × (30 + 20) = 2 × 50 = 100 m
Answer: Length = 30 m. Perimeter = 100 m.
Example 7: Example 7: Path around a garden
Problem: A rectangular garden is 20 m × 14 m. A path 2 m wide runs around the outside. Find the area of the path.
Solution:
Step 1: Outer length = 20 + 2 + 2 = 24 m. Outer breadth = 14 + 2 + 2 = 18 m.
Step 2: Outer area = 24 × 18 = 432 sq m
Step 3: Inner area (garden) = 20 × 14 = 280 sq m
Step 4: Path area = 432 − 280 = 152 sq m
Answer: The area of the path is 152 sq m.
Example 8: Example 8: Wire bent into shapes
Problem: Dev has a wire 48 cm long. He bends it into (a) a square, and (b) a rectangle 14 cm × 10 cm. Find the area of each shape.
Solution:
Part (a): Perimeter of square = 48 cm. Side = 48 ÷ 4 = 12 cm. Area = 12 × 12 = 144 sq cm.
Part (b): Check perimeter: 2 × (14 + 10) = 48 cm ✓. Area = 14 × 10 = 140 sq cm.
Answer: Square area = 144 sq cm. Rectangle area = 140 sq cm. The square has a larger area with the same perimeter.
Example 9: Example 9: Floor with carpet and border
Problem: A room is 8 m × 6 m. A carpet 6 m × 4 m is placed in the centre. Find the area of the floor NOT covered by the carpet.
Solution:
Step 1: Room area = 8 × 6 = 48 sq m
Step 2: Carpet area = 6 × 4 = 24 sq m
Step 3: Uncovered area = 48 − 24 = 24 sq m
Answer: 24 sq m of the floor is not covered.
Example 10: Example 10: Cost comparison
Problem: Neha can buy a square plot (side 30 m) at ₹500 per sq m or a rectangular plot (40 m × 20 m) at ₹450 per sq m. Which is cheaper?
Solution:
Step 1: Square area = 30 × 30 = 900 sq m. Cost = 900 × 500 = ₹4,50,000.
Step 2: Rectangle area = 40 × 20 = 800 sq m. Cost = 800 × 450 = ₹3,60,000.
Answer: The rectangular plot is cheaper at ₹3,60,000.
Key Points to Remember
- Use perimeter for fencing, framing, bordering, ribbon — anything along the boundary.
- Use area for painting, tiling, carpeting, planting — anything covering the surface.
- Always identify what the problem asks before choosing a formula.
- For paths and borders: find the outer area − inner area = border area.
- Same perimeter does not mean same area. A square has the largest area for a given perimeter.
- When rounding, round up for materials (you cannot buy half a tile).
Practice Problems
- A rectangular park is 80 m × 60 m. How much fencing wire is needed to go around it twice?
- Meera tiles a rectangular kitchen floor (4 m × 3 m) with tiles of 1 sq m each. How many tiles does she need?
- The perimeter of a square plot is 100 m. Find its area.
- A rectangular hall is 15 m × 10 m. A carpet 12 m × 8 m is placed inside. Find the area of the uncovered floor.
- Kavi bends a wire of 64 cm into a square. What is the area of the square?
- A garden path 1 m wide runs along the inside of a rectangular garden 18 m × 12 m. Find the area of the path.
- Ribbon costs ₹3 per cm. How much does it cost to put ribbon around a 20 cm × 15 cm photo frame?
Frequently Asked Questions
Q1. How do I decide whether to find area or perimeter?
If the problem involves the boundary (fencing, ribbon, wire, border), use perimeter. If it involves the surface (painting, tiling, carpet, grass), use area.
Q2. Can two shapes with the same perimeter have different areas?
Yes. A 10 × 2 rectangle (perimeter 24, area 20) and a 6 × 6 square (perimeter 24, area 36) have the same perimeter but very different areas.
Q3. What shape gives the maximum area for a given perimeter?
A square gives the largest area among all rectangles with the same perimeter. Among all shapes, a circle gives the largest area for a given perimeter.
Q4. How do you find the area of a path around a rectangle?
Find the outer area (including the path) and subtract the inner area (excluding the path). The difference is the area of the path.
Q5. Should I round up or down when calculating tiles needed?
Round up. If you need 15.3 tiles, buy 16 — you cannot lay 0.3 of a tile.
Q6. What if a garden has a gate and does not need a full fence?
Find the full perimeter and subtract the width of the gate. For example, if perimeter is 100 m and the gate is 3 m wide, fencing needed = 100 − 3 = 97 m.
Q7. How do I find the cost using area?
First find the area, then multiply by the cost per square unit. If area = 50 sq m and cost = Rs 100 per sq m, total cost = 50 x 100 = Rs 5,000.
Q8. What if both area and perimeter are asked in the same problem?
Calculate each separately using the correct formula. Area = l x b (square units). Perimeter = 2 x (l + b) (linear units). Present both answers clearly.
