Area and Perimeter
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Solved Examples
Study MaterialSolved Examples – Area and Perimeter
Solved examples help students understand the practical application of mensuration and geometry concepts in competitive examinations. These examples are designed from basic to advanced level and cover important questions frequently asked in SSC, Banking, Railway, CDS, NDA, CAT, Defence, and various aptitude examinations.
Topics Covered in Solved Examples
- Area and Perimeter of Rectangle
- Square Problems
- Triangle Area Questions
- Circle Mensuration Problems
- Sector and Arc Calculations
- Percentage Change in Area
- Geometry Property Questions
- Advanced Mensuration Problems
Example 1: Perimeter of Rectangle
Question: Find the perimeter of a rectangle whose length is 12 cm and breadth is 8 cm.
Solution:
Perimeter of rectangle:
= 2(l + b)
= 2(12 + 8)
= 2 × 20
= 40 cm
Answer: 40 cm
Example 2: Area of Rectangle
Question: Find the area of a rectangle having length 15 m and breadth 6 m.
Solution:
Area of rectangle:
= l × b
= 15 × 6
= 90 m²
Answer: 90 m²
Example 3: Area of Square
Question: Find the area of a square whose side is 11 cm.
Solution:
Area of square:
= a²
= 11²
= 121 cm²
Answer: 121 cm²
Example 4: Perimeter of Square
Question: Find the perimeter of a square whose side is 9 cm.
Solution:
Perimeter of square:
= 4a
= 4 × 9
= 36 cm
Answer: 36 cm
Example 5: Area of Triangle
Question: Find the area of a triangle whose base is 14 cm and height is 10 cm.
Solution:
Area of triangle:
= 1/2 × Base × Height
= 1/2 × 14 × 10
= 70 cm²
Answer: 70 cm²
Example 6: Perimeter of Triangle
Question: Find the perimeter of a triangle having sides 7 cm, 8 cm, and 9 cm.
Solution:
Perimeter:
= 7 + 8 + 9
= 24 cm
Answer: 24 cm
Example 7: Circumference of Circle
Question: Find the circumference of a circle whose radius is 7 cm.
Solution:
Circumference:
= 2πr
= 2 × 22/7 × 7
= 44 cm
Answer: 44 cm
Example 8: Area of Circle
Question: Find the area of a circle whose radius is 14 cm.
Solution:
Area:
= πr²
= 22/7 × 14 × 14
= 616 cm²
Answer: 616 cm²
Example 9: Diagonal of Rectangle
Question: Find the diagonal of a rectangle whose length is 9 cm and breadth is 12 cm.
Solution:
Diagonal:
= √(l² + b²)
= √(9² + 12²)
= √(81 + 144)
= √225
= 15 cm
Answer: 15 cm
Example 10: Diagonal of Square
Question: Find the diagonal of a square whose side is 10 cm.
Solution:
Diagonal:
= a√2
= 10√2 cm
Answer: 10√2 cm
Example 11: Heron's Formula
Question: Find the area of a triangle whose sides are 13 cm, 14 cm, and 15 cm.
Solution:
Semi-perimeter:
s = (13 + 14 + 15)/2
= 21
Using Heron's Formula:
Area = √[s(s−a)(s−b)(s−c)]
= √[21 × 8 × 7 × 6]
= √7056
= 84 cm²
Answer: 84 cm²
Example 12: Area of Rhombus
Question: Find the area of a rhombus whose diagonals are 12 cm and 16 cm.
Solution:
Area:
= 1/2 × d₁ × d₂
= 1/2 × 12 × 16
= 96 cm²
Answer: 96 cm²
Example 13: Area of Trapezium
Question: Find the area of a trapezium whose parallel sides are 10 cm and 14 cm and height is 8 cm.
Solution:
Area:
= 1/2 × (a+b) × h
= 1/2 × (10 + 14) × 8
= 12 × 8
= 96 cm²
Answer: 96 cm²
Example 14: Area of Sector
Question: Find the area of a sector of angle 90° and radius 14 cm.
Solution:
Sector Area:
= (θ/360) × πr²
= (90/360) × 22/7 × 14 × 14
= 1/4 × 616
= 154 cm²
Answer: 154 cm²
Example 15: Arc Length
Question: Find the arc length of a sector of angle 60° and radius 21 cm.
Solution:
Arc Length:
= (θ/360) × 2πr
= (60/360) × 2 × 22/7 × 21
= 22 cm
Answer: 22 cm
Example 16: Percentage Increase in Area
Question: If the side of a square increases by 20%, find the percentage increase in area.
Solution:
Formula:
Percentage increase in area:
= 2x + x²/100
= 2(20) + 20²/100
= 40 + 4
= 44%
Answer: 44%
Example 17: Pythagoras Theorem
Question: Find the hypotenuse of a right triangle whose perpendicular sides are 6 cm and 8 cm.
Solution:
Using Pythagoras theorem:
Hypotenuse² = 6² + 8²
= 36 + 64
= 100
Hypotenuse = 10 cm
Answer: 10 cm
Example 18: Comparison of Areas
Question: If the side ratio of two squares is 2:5, find the ratio of their areas.
Solution:
Area ratio:
= 2² : 5²
= 4 : 25
Answer: 4 : 25
Important Exam Tips
- Memorize all standard formulas.
- Use π = 22/7 wherever suitable.
- Learn important Pythagorean triplets.
- Draw diagrams for complex problems.
- Practice sector and arc questions regularly.
- Verify units carefully.
- Practice percentage change concepts thoroughly.
Practicing solved examples regularly improves conceptual clarity, logical thinking, and calculation speed in solving mensuration and geometry aptitude questions in competitive examinations.