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Area, Shapes and Perimeter
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Area, Shapes, and Perimeter
Questions related to area, shapes, and perimeter are common in the Quantitative Aptitude section of most placement tests and competitive exams. These problems test a candidate’s understanding of basic geometry, formulas, and numerical calculations.
Candidates are expected to:
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Apply standard geometric formulas
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Calculate area and perimeter of different shapes
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Solve problems involving missing dimensions
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Work with composite figures
Area and Perimeter of Basic Shapes
Rectangle
Formulas:
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Area = length × width
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Perimeter = 2 × (length + width)
Example 1
Length = 8 m, Width = 5 m
Area = 8 × 5 = 40 m²
Perimeter = 2(8 + 5) = 2 × 13 = 26 m
Example 2
Length = 12 cm, Perimeter = 38 cm
Perimeter = 2(l + w)
38 = 2(12 + w)
38 = 24 + 2w
2w = 14
w = 7 cm
Area = 12 × 7 = 84 cm²
Example 3
Perimeter = 30 m, Width = 6 m
30 = 2(l + 6)
30 = 2l + 12
2l = 18
l = 9 m
Area = 9 × 6 = 54 m²
Square
Formulas:
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Area = side²
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Perimeter = 4 × side
Example 1
Side = 6 cm
Area = 6² = 36 cm²
Perimeter = 4 × 6 = 24 cm
Example 2
Perimeter = 40 m
40 = 4s
s = 10 m
Area = 10² = 100 m²
Circle
Formulas:
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Area = πr²
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Circumference = 2πr
Example 1
Radius = 5 cm (π = 3.14)
Area = 3.14 × 5²
≈ 78.5 cm²
Circumference = 2 × 3.14 × 5
≈ 31.4 cm
Example 2
Area = 314 m² (π = 3.14)
314 = 3.14r²
r² = 100
r = 10 m
Circumference = 2 × 3.14 × 10
≈ 62.8 m
Area and Perimeter of Triangles
Right Triangle
Formulas:
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Area = (base × height) / 2
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Perimeter = sum of all sides
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Hypotenuse:
c² = a² + b²
Example 1
Base = 6 cm, Height = 8 cm
Area = (6 × 8) / 2
= 24 cm²
Hypotenuse = √(6² + 8²)
= √100
= 10 cm
Perimeter = 6 + 8 + 10
= 24 cm
Example 2
Base = 5 m, Height = 5 m
Area = (5 × 5) / 2
= 12.5 m²
Hypotenuse = √(25 + 25)
= √50 ≈ 7.07 m
Perimeter ≈ 5 + 5 + 7.07
≈ 17.07 m
Equilateral Triangle
Formulas:
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Area = (√3 / 4) × side²
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Perimeter = 3 × side
Example 1
Side = 6 cm
Area = (√3 / 4) × 36
= 9√3 cm²
Perimeter = 3 × 6
= 18 cm
Example 2
Perimeter = 45 m
45 = 3s
s = 15 m
Area = (√3 / 4) × 225
= 56.25√3 m²
Example 3
Area = 36√3 cm²
(√3 / 4)s² = 36√3
s² = 144
s = 12 cm
Perimeter = 3 × 12
= 36 cm
Composite Shapes
Composite shapes are figures made up of two or more basic shapes.
Steps to Solve
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Break the figure into simple shapes (rectangle, square, triangle, etc.).
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Find the area of each part.
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Add or subtract areas as required.
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For perimeter, add the lengths of the outer sides only.
Example
A shape consists of:
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Rectangle: length = 6 cm, width = 4 cm
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Square: side = 3 cm
Area of rectangle = 6 × 4 = 24 cm²
Area of square = 3 × 3 = 9 cm²
Total area = 33 cm²
Tips for Solving Area and Perimeter Problems
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Memorise standard geometry formulas.
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Convert units when necessary.
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Draw rough diagrams for better understanding.
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Break complex shapes into simpler ones.
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Check whether the question asks for area, perimeter, or both.
