Area Word Problems (Class 8)
Area word problems require you to apply the area formulas for various shapes — trapezium, rhombus, parallelogram, quadrilateral, circle, triangle — to solve real-life situations.
These problems involve fields, gardens, rooms, walls, and other practical scenarios where you need to calculate the area of a region and use it to determine cost, material needed, or other quantities.
The key is to identify the shape, recall the correct formula, and substitute the given values carefully.
What is Area Word Problems (Class 8)?
Area formulas needed:
- Rectangle: A = l × b
- Square: A = s²
- Triangle: A = ½ × b × h
- Parallelogram: A = b × h
- Trapezium: A = ½ × (a + b) × h (where a, b are parallel sides)
- Rhombus: A = ½ × d₁ × d₂ (where d₁, d₂ are diagonals)
- General quadrilateral: A = ½ × d × (h₁ + h₂) (d = diagonal, h₁ and h₂ are perpendicular distances)
- Circle: A = πr²
Area Word Problems (Class 8) Formula
Always: Identify shape → Select formula → Substitute values → Calculate
Types and Properties
Types of area word problems:
- Finding area: Given dimensions, calculate the area directly.
- Finding cost: Area × rate per unit area = total cost (painting, tiling, etc.).
- Finding dimensions: Given area and one dimension, find the other.
- Composite shapes: Break the figure into simpler shapes and add/subtract areas.
- Remaining area: Total area minus area of cut-out = remaining area (e.g., a circular pond in a square garden).
Solved Examples
Example 1: Example 1: Area of a trapezium-shaped field
Problem: A field is in the shape of a trapezium. The parallel sides are 25 m and 15 m, and the perpendicular distance between them is 12 m. Find the area.
Solution:
Given: a = 25 m, b = 15 m, h = 12 m.
- Area = ½ × (a + b) × h
- = ½ × (25 + 15) × 12
- = ½ × 40 × 12
- = 240 m²
Answer: Area = 240 m².
Example 2: Example 2: Cost of painting a wall
Problem: A rectangular wall is 8 m long and 4.5 m high. It has a door of size 2 m × 1 m. Find the cost of painting the wall at ₹50 per m².
Solution:
- Area of wall = 8 × 4.5 = 36 m²
- Area of door = 2 × 1 = 2 m²
- Area to paint = 36 − 2 = 34 m²
- Cost = 34 × 50 = ₹1,700
Answer: Cost of painting = ₹1,700.
Example 3: Example 3: Rhombus-shaped tile
Problem: A rhombus-shaped tile has diagonals of 12 cm and 16 cm. Find its area. How many tiles are needed to cover a floor of area 9600 cm²?
Solution:
- Area of one tile = ½ × d₁ × d₂ = ½ × 12 × 16 = 96 cm²
- Number of tiles = 9600/96 = 100
Answer: Area of tile = 96 cm². Tiles needed = 100.
Example 4: Example 4: General quadrilateral
Problem: A quadrilateral-shaped plot has a diagonal of 20 m. The perpendicular distances from the opposite vertices to this diagonal are 8 m and 6 m. Find the area.
Solution:
- Area = ½ × d × (h₁ + h₂)
- = ½ × 20 × (8 + 6)
- = ½ × 20 × 14
- = 140 m²
Answer: Area = 140 m².
Example 5: Example 5: Circular garden with path
Problem: A circular garden has radius 14 m. A path of width 3.5 m runs around the outside. Find the area of the path.
Solution:
- Outer radius = 14 + 3.5 = 17.5 m
- Area of outer circle = π × 17.5² = 22/7 × 306.25 = 962.5 m²
- Area of inner circle = π × 14² = 22/7 × 196 = 616 m²
- Area of path = 962.5 − 616 = 346.5 m²
Answer: Area of path = 346.5 m².
Example 6: Example 6: Parallelogram with missing height
Problem: The area of a parallelogram is 360 cm². If the base is 24 cm, find the corresponding height.
Solution:
- Area = base × height
- 360 = 24 × h
- h = 360/24 = 15 cm
Answer: Height = 15 cm.
Example 7: Example 7: Composite figure — rectangle with triangle
Problem: A rectangular park is 40 m × 30 m. A triangular flower bed with base 10 m and height 6 m is at one corner. Find the area of the park excluding the flower bed.
Solution:
- Area of park = 40 × 30 = 1200 m²
- Area of flower bed = ½ × 10 × 6 = 30 m²
- Remaining area = 1200 − 30 = 1170 m²
Answer: Area excluding flower bed = 1170 m².
Example 8: Example 8: Tiling cost for trapezium floor
Problem: A room has a trapezium-shaped floor with parallel sides 10 m and 7 m, and height 5 m. Tiles cost ₹120 per m². Find the total cost.
Solution:
- Area = ½ × (10 + 7) × 5 = ½ × 17 × 5 = 42.5 m²
- Cost = 42.5 × 120 = ₹5,100
Answer: Total cost = ₹5,100.
Example 9: Example 9: Area of a field divided by a diagonal
Problem: A quadrilateral field has a diagonal of 30 m. The two triangles formed have heights 8 m and 12 m from the diagonal. Find the total area.
Solution:
- Area = ½ × d × (h₁ + h₂) = ½ × 30 × (8 + 12) = ½ × 30 × 20 = 300 m²
Answer: Area = 300 m².
Example 10: Example 10: Finding the side of a square
Problem: A square field has the same area as a rectangle with length 64 m and breadth 25 m. Find the side of the square.
Solution:
- Area of rectangle = 64 × 25 = 1600 m²
- Area of square = s² = 1600
- s = √1600 = 40 m
Answer: Side of square = 40 m.
Real-World Applications
Where area calculations are used:
- Construction: Calculating material for flooring, painting, plastering.
- Agriculture: Measuring field areas for farming and irrigation.
- Interior design: Calculating carpet, wallpaper, or tile requirements.
- Land surveying: Measuring plot areas for property records.
- Gardening: Planning flower beds, lawns, and paths.
Key Points to Remember
- Read the problem carefully and identify the shape.
- Use the correct area formula for that shape.
- For composite figures, break into simpler shapes and add/subtract areas.
- For cost problems: Total cost = Area × Rate per unit area.
- When a region is cut out (door, window, pond), subtract its area from the total.
- Keep units consistent — convert if needed (1 m = 100 cm, 1 m² = 10000 cm²).
- For circular paths: Area = π(R² − r²) where R is outer radius and r is inner radius.
- Always state the answer with correct units (m², cm², etc.).
Practice Problems
- A trapezium-shaped plot has parallel sides 30 m and 20 m with height 8 m. Find the cost of levelling at ₹25 per m².
- A rhombus has diagonals 18 cm and 24 cm. Find its area.
- A rectangular hall 15 m × 10 m has two windows each 1.5 m × 1 m and a door 2 m × 1 m. Find the area to whitewash.
- A circular pond of radius 7 m is surrounded by a 2 m wide path. Find the area of the path.
- A quadrilateral has a diagonal of 40 m with perpendicular heights 10 m and 14 m. Find the area.
- How many square tiles of side 20 cm are needed to tile a floor of 8 m × 6 m?
- A triangular garden has base 16 m and height 9 m. Find the cost of planting grass at ₹15 per m².
Frequently Asked Questions
Q1. How do you find the area of a trapezium?
Area = ½ × (sum of parallel sides) × height. The height is the perpendicular distance between the parallel sides.
Q2. How do you find area of a general quadrilateral?
Draw one diagonal. It divides the quadrilateral into two triangles. Area = ½ × diagonal × (sum of perpendicular heights from opposite vertices).
Q3. How do you handle composite shapes?
Break the shape into rectangles, triangles, or other standard shapes. Calculate each area separately, then add them together.
Q4. What if a shape has a cut-out region?
Calculate the area of the whole shape and subtract the area of the cut-out. For example, wall with window = wall area − window area.
Q5. How do you convert between cm² and m²?
1 m = 100 cm, so 1 m² = 10,000 cm². To convert cm² to m², divide by 10,000.
Q6. What formula is used for a circular path?
Area of path = π(R² − r²) where R = outer radius and r = inner radius. This can also be written as π(R+r)(R−r).
Q7. How do you find the number of tiles needed?
Number of tiles = Total area of floor ÷ Area of one tile. Make sure both areas are in the same unit.
Q8. What is the area formula for a rhombus?
Area = ½ × d₁ × d₂, where d₁ and d₂ are the two diagonals. Alternatively, Area = base × height.
