We’ve covered individual shapes — now let’s put them together. Real exam problems rarely give us a clean rectangle or circle. Instead, they give us shaded regions, pathways, tiling problems, and composite figures. This is where mensuration gets practical and fun. It’s all about knowing which formula to pull out and when to add or subtract areas.
The Complete 2D Formula Reference
| Shape | Area | Perimeter |
|---|---|---|
| Square (side a) | a² | 4a |
| Rectangle (l × b) | l × b | 2(l + b) |
| Triangle (base b, height h) | ½ × b × h | a + b + c |
| Equilateral Triangle (side a) | (√3/4) × a² | 3a |
| Parallelogram | base × height | 2(a + b) |
| Rhombus (diag d₁, d₂) | ½ × d₁ × d₂ | 4 × side |
| Trapezium (parallel sides a, b) | ½ × (a + b) × h | sum of all sides |
| Circle (radius r) | πr² | 2πr |
| Semicircle | ½ × πr² | πr + 2r |
| Ring (radii R, r) | π(R² − r²) | 2π(R + r) |
Important note on semicircle perimeter: It’s πr + 2r (the curved part + the diameter), NOT just πr. This is a common trap.
Key Formulas
Composite Shapes — The Add/Subtract Strategy
The golden rule for composite shapes: break the figure into basic shapes, then add or subtract their areas.
- Shaded region inside a larger shape: Subtract the smaller area from the larger area
- Combined figure: Add the areas of the parts
- Overlapping figures: Add both areas and subtract the overlap
In simple language: if we can see rectangles, triangles, circles, or semicircles hiding inside the figure, we calculate each one and combine them.
Pathway Problems
These come up all the time. A path runs around (or inside) a rectangular garden/field/room.
Path Outside a Rectangle
If a rectangle (l × b) has a path of width w running outside it:
- Outer dimensions: (l + 2w) × (b + 2w)
- Path area = (l + 2w)(b + 2w) − l × b = 2w(l + b + 2w)
Path Inside a Rectangle
If the path runs inside the rectangle:
- Inner dimensions: (l − 2w) × (b − 2w)
- Path area = l × b − (l − 2w)(b − 2w) = 2w(l + b − 2w)
Path Around a Circle
If a circular path of width w surrounds a circle of radius r:
- Outer radius R = r + w
- Path area = πR² − πr² = π(R² − r²)
Trick: This can also be written as π(R + r)(R − r) = π(2r + w)(w).
Carpet and Tiling Problems
These are essentially “how many small shapes fit in a big shape” problems.
Number of tiles = Area of floor / Area of one tile
The catch: make sure both areas are in the same units! Convert everything to the same unit before dividing.
Cost of carpeting/painting = Area × Rate per unit area
Wire Bending Problems
When a wire is bent from one shape to another:
- The perimeter stays the same (the wire length doesn’t change)
- The area changes (different shapes with the same perimeter have different areas)
Among all shapes with the same perimeter, a circle has the maximum area. Among rectangles with the same perimeter, a square has the maximum area.
Worked Examples
Example 1: Pathway Problem
A rectangular garden is 40 m long and 30 m wide. A path 2 m wide runs outside around it. Find the area of the path.
Path area = 2w(l + b + 2w) = 2 × 2 × (40 + 30 + 2 × 2) = 4 × (70 + 4) = 4 × 74 = 296 m²
Let’s verify: Outer dimensions = 44 × 34 = 1496 m². Garden = 40 × 30 = 1200 m². Path = 1496 - 1200 = 296 m² ✓
Example 2: Composite Shape
A rectangular sheet of paper is 20 cm × 14 cm. A semicircle is cut out from each of the shorter sides (diameter = 14 cm). Find the area of the remaining paper.
Area of rectangle = 20 × 14 = 280 cm²
Two semicircles with diameter 14 (radius 7) = one full circle = π × 7² = 49π = 154 cm² (using π = 22/7)
Remaining area = 280 - 154 = 126 cm²
Example 3: Tiling Problem
A room is 6 m × 4 m. How many tiles of size 25 cm × 20 cm are needed to cover the floor?
First, convert to same units. Room = 600 cm × 400 cm.
Number of tiles = (600 × 400) / (25 × 20) = 240000 / 500 = 480 tiles
Example 4: Wire Bending
A wire is bent into a square of side 11 cm. If the same wire is bent into a circle, find the radius of the circle.
Wire length (perimeter of square) = 4 × 11 = 44 cm
This becomes the circumference: 2πr = 44 r = 44 / (2 × 22/7) = 44 × 7 / 44 = 7 cm
Bonus: Area of square = 121 cm². Area of circle = 22/7 × 49 = 154 cm². See? The circle has more area for the same perimeter!
Example 5: Painting Cost
A rectangular room is 12 m long, 8 m wide, and 4 m high. It has 2 doors (2 m × 1.5 m each) and 3 windows (1.5 m × 1 m each). Find the cost of painting the walls at Rs. 15 per m².
Total wall area = 2 × (l + b) × h = 2 × (12 + 8) × 4 = 160 m²
Door area = 2 × (2 × 1.5) = 6 m² Window area = 3 × (1.5 × 1) = 4.5 m² Total to subtract = 10.5 m²
Paintable area = 160 - 10.5 = 149.5 m² Cost = 149.5 × 15 = Rs. 2242.50
Common Exam Variations
- Shaded region between a circle and inscribed/circumscribed square
- Four circles placed inside a square (touching each other and the sides) — find the uncovered area
- Path around a rectangular pool/garden — inside vs outside path (read the question carefully!)
- Wire bent into different shapes — comparing areas
- Fencing problems — perimeter-based with one side being a wall (so only 3 sides need fencing)
Shortcut: Fencing with One Side as Wall
When one side of a rectangle is against a wall:
- Fencing needed = l + 2b (one length + two breadths)
- For maximum area with given fencing F: l = F/2 and b = F/4 (the optimal shape is when l = 2b)
Practice Problems
Q1: A circular garden of radius 21 m has a 3.5 m wide path running outside around it. Find the area of the path. (Use π = 22/7)
Q2: How many square tiles of side 20 cm are needed to pave a rectangular courtyard 3 m long and 2 m wide? What would be the cost at Rs. 5 per tile?
Q3: A wire shaped as a rectangle 16 cm × 12 cm is reshaped into a square. Find the side of the square and compare the areas.
Answers
A1: Outer radius = 21 + 3.5 = 24.5 m. Path area = π(24.5² - 21²) = (22/7)(600.25 - 441) = (22/7)(159.25) = 22 × 22.75 = 500.5 m².
A2: Floor area = 300 × 200 = 60000 cm². Tile area = 20 × 20 = 400 cm². Tiles needed = 60000/400 = 150 tiles. Cost = 150 × 5 = Rs. 750.
A3: Perimeter of rectangle = 2(16 + 12) = 56 cm. Side of square = 56/4 = 14 cm. Rectangle area = 192 cm², Square area = 196 cm². The square has 4 cm² more area — same perimeter but greater area.