Square (சதுரம்)
Welcome beginners! Square problems are fundamental in competitive exams and appear in 15-20% of geometry questions. Mastering squares helps you solve complex area, perimeter, and diagonal problems quickly, building a strong foundation for advanced geometry topics in TNPSC, SSC, Banking, and UPSC exams. This requires solid understanding of basic arithmetic operations.
🧠 Foundation: Understanding the Core Concept (அடிப்படை கருத்துகள்)
- A square is a special rectangle where all four sides are equal and all angles are 90° (சதுரம் என்பது நான்கு பக்கங்களும் சமமான சிறப்பு செவ்வகம்).
- Real-world examples: Chess board squares 🏁, floor tiles 🏠, smartphone screens 📱, photo frames 🖼️.
Core Formula Box:
Area = Side × Side = Side² பரப்பளவு = பக்கம் × பக்கம் = பக்கம்²
Level 0: Formula Playground
| Property | English Formula | Tamil Formula |
|---|---|---|
| Area (பரப்பளவு) | Side² | பக்கம்² |
| Perimeter (சுற்றளவு) | 4 × Side | 4 × பக்கம் |
| Diagonal (மூலைவிட்டம்) | Side × √2 | பக்கம் × √2 |
Practice Drills:
-
Problem: Side = 5 cm, find area. Solution: Area = 5² = 25 cm²
-
Problem: Side = 8 m, find perimeter. Solution: Perimeter = 4 × 8 = 32 m
-
Problem: Side = 6 cm, find diagonal. Solution: Diagonal = 6 × √2 = 6 × 1.414 = 8.484 cm
⚙️ Unit Conversion Mastery (அலகு மாற்றம்)
Common Conversion Mistakes:
- Mixing cm and m without converting
- Forgetting to square units for area calculations
- Using wrong √2 approximation
This requires proficiency in percentage calculations for area increase/decrease problems.
Key Conversion Factors:
- 1 m = 100 cm
- 1 km = 1000 m
- √2 ≈ 1.414
- Area units: cm² → m² (divide by 10,000)
Conversion Practice:
-
Problem: Area = 2500 cm², convert to m². Solution: 2500 ÷ 10,000 = 0.25 m²
-
Problem: Side = 0.5 km, find area in m². Solution: Side = 500 m, Area = 500² = 250,000 m²
-
Problem: Perimeter = 160 cm, convert to m. Solution: 160 ÷ 100 = 1.6 m
Level 1: Problem Reading Strategy (கேள்வியை அணுகும் முறை)
G-F-V Framework (Given-Find-Verify):
- Given: What information is provided?
- Find: What needs to be calculated?
- Verify: Does the answer make logical sense?
Step-by-Step Examples:
-
Problem: A square garden has a side of 12 m. Find its area.
- Given: Side = 12 m
- Find: Area
- Solution: Area = 12² = 144 m²
- Verify: ✓ Reasonable for a garden
-
Problem: The perimeter of a square is 48 cm. Find the side length.
- Given: Perimeter = 48 cm
- Find: Side length
- Solution: Side = 48 ÷ 4 = 12 cm
- Verify: ✓ 4 × 12 = 48
-
Problem: A square has an area of 81 cm². Find its diagonal.
- Given: Area = 81 cm²
- Find: Diagonal
- Solution: Side = √81 = 9 cm, Diagonal = 9 × √2 = 12.73 cm
- Verify: ✓ Diagonal > side length
Level 2: Pattern Recognition (கணக்குகளின் வகையை அறிதல்)
| Pattern | English Keywords | Core Concept | Tamil Term |
|---|---|---|---|
| Direct Formula | "side is", "given side" | Apply formula directly | நேரடி கணக்கு |
| Reverse Calculation | "area is", "perimeter is" | Find side first | எதிர் கணக்கு |
| Cost Problems | "rate per", "cost of" | Multiply by rate | விலை கணக்கு |
| Comparison | "difference", "ratio" | Compare two squares | ஒப்பீட்டு கணக்கு |
Pattern 1: Direct Formula (நேரடி கணக்கு)
Keywords: "side is given", "find area/perimeter/diagonal" Core Insight: Apply the basic formulas directly
Visual: Square with side 's'
s
┌───┐
s │ │ s Area = s²
│ │ Perimeter = 4s
└───┐ Diagonal = s√2
s
Example: A square has a side of 15 cm. Find its area and perimeter.
- Area = 15² = 225 cm²
- Perimeter = 4 × 15 = 60 cm
Pattern 2: Reverse Calculation (எதிர் கணக்கு)
Keywords: "area is", "perimeter is", "diagonal is" Core Insight: Find the side first, then calculate other properties
Example: The area of a square is 144 m². Find its perimeter.
- Side = √144 = 12 m
- Perimeter = 4 × 12 = 48 m
Pattern 3: Cost Problems (விலை கணக்கு)
Keywords: "cost per unit", "rate", "price" Core Insight: Calculate area/perimeter first, then multiply by rate
Example: A square plot costs ₹500 per m². If the side is 20 m, find total cost.
- Area = 20² = 400 m²
- Total cost = 400 × 500 = ₹2,00,000
Pattern 4: Comparison Problems (ஒப்பீட்டு கணக்கு)
Keywords: "difference", "ratio", "times larger" Core Insight: Calculate for both squares and compare
Example: Square A has side 8 cm, Square B has side 12 cm. Find the ratio of their areas.
- Area A = 8² = 64 cm²
- Area B = 12² = 144 cm²
- Ratio = 64:144 = 4:9
Level 3: Advanced Scenarios (சிக்கலான கணக்குகள்)
Complex Multi-step Problems:
- When area increases by a factor, side increases by square root of that factor
- When side increases by n times, area increases by n² times
Example 1: If the side of a square is increased by 20%, by what percent does the area increase?
- Original side = s, New side = 1.2s
- Original area = s², New area = (1.2s)² = 1.44s²
- Increase = 44%
Example 2: Two squares have areas in the ratio 4:9. If the smaller square has a perimeter of 16 cm, find the side of the larger square.
- Smaller square: Perimeter = 16 cm, Side = 4 cm, Area = 16 cm²
- Area ratio = 4:9, so larger area = 36 cm²
- Larger square side = √36 = 6 cm
⚠️ Common Beginner Mistakes (பொதுவான தவறுகள்)
-
Unit Mixing:
- ❌ Wrong: Side in cm, area calculated in m²
- ✅ Correct: Convert all units first
-
Area vs Perimeter Confusion:
- ❌ Wrong: Using side² for perimeter calculations
- ✅ Correct: Perimeter = 4 × side, Area = side²
-
Diagonal Approximation:
- ❌ Wrong: Using √2 = 1.5
- ✅ Correct: √2 ≈ 1.414
-
Percentage Increase Logic:
- ❌ Wrong: If side increases by 20%, area increases by 20%
- ✅ Correct: If side increases by 20%, area increases by 44%
🚀 Key Takeaways (முக்கிய குறிப்புகள்)
- Master the three core formulas: Area = side², Perimeter = 4 × side, Diagonal = side × √2
- Always identify the pattern first: Direct, Reverse, Cost, or Comparison
- When area is given, find side first using square root
- Unit conversion is crucial - be consistent throughout
- Success Formula: Practice + Pattern Recognition = Speed + Accuracy
📝 Problems to Practice (பயிற்சிக் கேள்விகள்)
Pattern 1: Direct Formula Problems
Problem 1: A square has a side length of 25 cm. Calculate its area, perimeter, and diagonal.
Pattern Identification: Direct Formula - Given side, find other properties
Solution:
- Area = 25² = 625 cm²
- Perimeter = 4 × 25 = 100 cm
- Diagonal = 25 × √2 = 25 × 1.414 = 35.35 cm
Problem 2: The side of a square field is 0.8 km. Find its area in hectares. (1 hectare = 10,000 m²)
Pattern Identification: Direct Formula with unit conversion
Solution:
- Side = 0.8 km = 800 m
- Area = 800² = 640,000 m²
- Area in hectares = 640,000 ÷ 10,000 = 64 hectares
Problem 3: A square tile has a side of 30 cm. How many such tiles are needed to cover a floor of area 54 m²?
Pattern Identification: Direct Formula with practical application
Solution:
- Tile side = 30 cm = 0.3 m
- Tile area = 0.3² = 0.09 m²
- Number of tiles = 54 ÷ 0.09 = 600 tiles
Pattern 2: Reverse Calculation Problems
Problem 4: The area of a square is 169 cm². Find its perimeter and diagonal.
Pattern Identification: Reverse Calculation - Given area, find other properties
Solution:
- Side = √169 = 13 cm
- Perimeter = 4 × 13 = 52 cm
- Diagonal = 13 × √2 = 18.38 cm
Problem 5: A square has a perimeter of 144 m. Calculate its area.
Pattern Identification: Reverse Calculation - Given perimeter, find area
Solution:
- Side = 144 ÷ 4 = 36 m
- Area = 36² = 1,296 m²
Problem 6: The diagonal of a square is 28.28 cm. Find its area and perimeter.
Pattern Identification: Reverse Calculation - Given diagonal, find other properties
Solution:
- Side = 28.28 ÷ √2 = 28.28 ÷ 1.414 = 20 cm
- Area = 20² = 400 cm²
- Perimeter = 4 × 20 = 80 cm
Pattern 3: Cost Problems
Problem 7: A square plot of land with side 40 m is to be fenced. If fencing costs ₹75 per meter, find the total cost.
Pattern Identification: Cost Problem - Calculate perimeter then multiply by rate
Solution:
- Perimeter = 4 × 40 = 160 m
- Total cost = 160 × 75 = ₹12,000
Problem 8: The cost of carpeting a square room at ₹150 per m² is ₹24,000. Find the side length of the room.
Pattern Identification: Cost Problem with reverse calculation
Solution:
- Area = 24,000 ÷ 150 = 160 m²
- Side = √160 = 12.65 m
Problem 9: A square garden with side 15 m needs to be turfed. If turf costs ₹80 per m² and an additional 10% is needed for wastage, find the total cost.
Pattern Identification: Cost Problem with additional percentage
Solution:
- Area = 15² = 225 m²
- Area with wastage = 225 × 1.10 = 247.5 m²
- Total cost = 247.5 × 80 = ₹19,800
Pattern 4: Comparison Problems
Problem 10: Square A has a side of 12 cm and Square B has a side of 18 cm. Find the ratio of their areas and perimeters.
Pattern Identification: Comparison Problem - Calculate for both squares
Solution:
- Area A = 12² = 144 cm², Area B = 18² = 324 cm²
- Area ratio = 144:324 = 4:9
- Perimeter A = 48 cm, Perimeter B = 72 cm
- Perimeter ratio = 48:72 = 2:3
Problem 11: If the side of a square is doubled, by what factor do the area and perimeter increase?
Pattern Identification: Comparison Problem with scaling
Solution:
- Let original side = s
- Original area = s², New area = (2s)² = 4s²
- Area increases by factor of 4
- Original perimeter = 4s, New perimeter = 4(2s) = 8s
- Perimeter increases by factor of 2
Problem 12: Two squares have areas 225 cm² and 625 cm². Find the difference between their perimeters.
Pattern Identification: Comparison Problem with reverse calculation
Solution:
- Square 1: Side = √225 = 15 cm, Perimeter = 60 cm
- Square 2: Side = √625 = 25 cm, Perimeter = 100 cm
- Difference = 100 - 60 = 40 cm
Related Articles
Mathematical Foundations
- Simplification - Master basic arithmetic operations essential for area, perimeter, and diagonal calculations
- Percentage - Apply percentage calculations in area increase/decrease and comparison problems
- Ratio and Proportion - Use proportional reasoning in square side and area relationship problems
Geometric Applications
- HCF and LCM of Fractions - Apply fraction operations in complex geometric calculations and unit conversions
- Time and Distance - Practice formula-based problem-solving techniques similar to geometric calculations
Interest and Financial Applications
- Simple Interest - Use systematic calculation approaches similar to area and perimeter formulas
- Compound Interest - Apply growth calculation concepts to area scaling and geometric progression problems
Advanced Problem-Solving
- Direction Sense - Practice systematic logical reasoning approaches applicable to geometric problem analysis