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Quantitative Aptitude

Area & Perimeter

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Concepts (2)

Mastering Area & Perimeter formulas for rectangles, squares, circles, and triangles is crucial for SSC CGL. Focus on quick recall and applying shortcuts for percentage changes and composite figures.

Core Formula

Understanding the basic formulas for common 2D shapes is fundamental for quick problem-solving.

  • Rectangle:
    • Area (A) = length (L) × breadth (B)
    • Perimeter (P) = 2 × (L + B)
  • Square:
    • Area (A) = side (s)²
    • Perimeter (P) = 4 × s
  • Circle:
    • Area (A) = πr² (where r = radius)
    • Circumference (C) = 2πr
  • Triangle:
    • Area (A) = 0.5 × base (b) × height (h)
  • Parallelogram:
    • Area (A) = base (b) × height (h)
  • Rhombus:
    • Area (A) = 0.5 × diagonal1 (d1) × diagonal2 (d2)
  • Trapezium:
    • Area (A) = 0.5 × (sum of parallel sides) × height (h)

Worked Example 1

Question: The dimensions of a rectangular room are 4 m x 3 m x 3 m. Find the cost of tiling the floor at the rate of ₹ 25 per square meter. Solution:

  1. Identify relevant dimensions: For tiling the floor, we only need the length and breadth of the floor. The height (3m) is a distractor. So, L = 4 m, B = 3 m.
  2. Calculate the area of the floor: Area = L × B = 4 m × 3 m = 12 m².
  3. Calculate the total cost: Cost = Area × Rate = 12 m² × ₹ 25/m² = ₹ 300. Answer: The cost of tiling the floor is ₹ 300.

Worked Example 2

Question: The length of a hall is 5 m more than its breadth. If the area of the hall is 84 m², what is the perimeter of the hall? Solution:

  1. Define variables: Let the breadth of the hall be x meters. Then, the length of the hall will be x + 5 meters.
  2. Formulate the area equation: Area = Length × Breadth 84 = (x + 5) × x 84 = x² + 5x x² + 5x - 84 = 0
  3. Solve the quadratic equation: Factorize: (x + 12)(x - 7) = 0 Possible values for x are -12 or 7. Since breadth cannot be negative, x = 7 m.
  4. Find length and breadth: Breadth (B) = 7 m Length (L) = 7 + 5 = 12 m
  5. Calculate the perimeter: Perimeter = 2 × (L + B) Perimeter = 2 × (12 + 7) = 2 × 19 = 38 m. Answer: The perimeter of the hall is 38 m.

Shortcuts & Tricks

  1. Percentage Change in Area (for squares/circles/equilateral triangles): If a side/radius increases/decreases by x%, the percentage change in area can be found using the successive percentage formula: x + x + (x*x)/100. For example, if radius increases by 20%, percentage increase in area = 20 + 20 + (20*20)/100 = 40 + 4 = 44%.
  2. Pythagorean Triplets: For right-angled triangles, memorize common triplets (3,4,5; 5,12,13; 8,15,17; 7,24,25) to quickly find missing sides, which can then be used for area/perimeter calculations.
  3. Ratio Method for Similar Figures: If two figures are similar and the ratio of their corresponding sides is a:b, then the ratio of their areas is a²:b².

Common Mistakes

  1. Confusing Area and Perimeter: Always double-check what the question is asking for.
  2. Unit Conversion Errors: Ensure all dimensions are in the same units before calculation (e.g., cm to m).
  3. Ignoring Distractors: Extra information (like height for floor area) can be misleading.
  4. Incorrect Value of Pi: Use 22/7 or 3.14 as specified, or 22/7 for calculations divisible by 7.
  5. Calculation Errors: Simple arithmetic mistakes, especially with squares and square roots.

Derivation (brief)

Understanding the logical basis of formulas aids recall:

  • Rectangle Area: Conceptually, the area of a rectangle is the number of unit squares that can fit inside it. If you have 'L' units along the length and 'B' units along the breadth, you can arrange L squares in B rows, totaling L × B unit squares.
  • Circle Area (intuitive): Imagine cutting a circle into a large number of very thin sectors. If you arrange these sectors alternately, they form a shape that closely resembles a parallelogram. The base of this 'parallelogram' would be half the circle's circumference (πr), and its height would be the circle's radius (r). Thus, Area ≈ base × height = πr × r = πr².

Advanced Examples

  1. Question: A wire is bent in the form of a square of area 121 cm². If the same wire is bent in the form of a circle, what is the area of the circle? Solution:

    1. Find side of square: Area of square = s² = 121 cm² => s = √121 = 11 cm.
    2. Find perimeter of square (length of wire): Perimeter = 4s = 4 × 11 = 44 cm.
    3. Find radius of circle: The wire length is the circumference of the circle. 2πr = 44 2 × (22/7) × r = 44 44/7 × r = 44 => r = 7 cm.
    4. Find area of circle: Area = πr² = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm². Answer: The area of the circle is 154 cm².

  2. Question: The parallel sides of a trapezium are 10 cm and 16 cm, and its height is 5 cm. Find its area. Solution:

    1. Apply Trapezium Area Formula: Area = 0.5 × (sum of parallel sides) × height
    2. Substitute values: Area = 0.5 × (10 + 16) × 5
    3. Calculate: Area = 0.5 × 26 × 5 = 13 × 5 = 65 cm². Answer: The area of the trapezium is 65 cm².

Variation Types

  1. Composite Figures: Problems involving finding the area/perimeter of shapes made by combining or subtracting basic geometric figures (e.g., a rectangle with a semi-circle cut out, or a path around a garden).
  2. Pathways/Borders: Calculating the area of a path around a rectangular field or a border inside a frame, often requiring subtraction of areas.
  3. Inscribed/Circumscribed Figures: Questions involving one shape fitted inside another (e.g., a circle inscribed in a square, or a square inscribed in a circle), requiring knowledge of their dimensional relationships.
  4. Cost-based Problems: Practical applications like tiling, fencing, or painting costs, where area or perimeter calculation is the first step.
  5. Percentage Change: Determining the percentage change in area or perimeter when the dimensions of a figure are increased or decreased by a certain percentage.

Time-Saving Methods

  1. Visualisation: Always quickly sketch the figure described in the problem. This helps in understanding the problem, identifying relevant dimensions, and breaking down complex shapes.
  2. Approximation (for π): In multiple-choice questions, if the options are widely spaced, you can sometimes use π ≈ 3 or π ≈ 3.14 for quick estimation to eliminate incorrect options, especially when exact calculation isn't immediately obvious.
  3. Memorize Common Values: For circles, memorize the area and circumference for common radii like r=7 (Area=154, Circumference=44) and r=14 (Area=616, Circumference=88), as these frequently appear in exams.
  4. Option Elimination: Use the properties of shapes or rough estimations to eliminate options that are clearly incorrect, narrowing down your choices.
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Master calculating areas of paths around rectangles/squares, circular rings, and complex combined figures using direct formulas and decomposition for speed and accuracy in SSC CGL.

Core Formulas

  1. Area of Path around a Rectangle/Square:

    • Path outside a rectangle (Length L, Breadth B, width w): Area = 2w(L + B + 2w)
    • Path inside a rectangle (Length L, Breadth B, width w): Area = 2w(L + B - 2w)
    • For a square of side 'a', substitute L=a, B=a. E.g., Path outside a square: 2w(a + a + 2w) = 2w(2a + 2w) = 4w(a + w)

  2. Area of a Circular Ring:

    • For an outer circle with radius R and an inner circle with radius r: Area = πR² - πr² = π(R² - r²) = π(R - r)(R + r)

  3. Area of Combined Figures:

    • Break down complex shapes into simpler, known geometric figures (rectangles, squares, circles, triangles, sectors, etc.).
    • Calculate the area of each component figure.
    • Add or subtract these areas as required to find the area of the shaded or desired region.

Worked Example 1

Q: A rectangular park is 60 m long and 40 m wide. A path 2 m wide is built outside the park. Find the area of the path. A: Given: L = 60 m, B = 40 m, w = 2 m. Using the formula for path outside a rectangle: Area = 2w(L + B + 2w) Area = 2 * 2 (60 + 40 + 2 * 2) Area = 4 (100 + 4) Area = 4 * 104 Area = 416 sq m

Worked Example 2

Q: The outer and inner radii of a circular track are 21 m and 14 m respectively. Find the area of the track. A: Given: Outer radius R = 21 m, Inner radius r = 14 m. Using the formula for the area of a circular ring: Area = π(R - r)(R + r) Area = (22/7) * (21 - 14) * (21 + 14) Area = (22/7) * 7 * 35 Area = 22 * 35 Area = 770 sq m

Shortcuts & Tricks

  • Path Area: Instead of calculating outer and inner areas separately, use the direct formulas 2w(L + B ± 2w). This saves steps and reduces calculation errors. For cross paths of width 'w' in a rectangular field L x B, the area is w(L + B - w). This is a common variation.
  • Ring Area: Always use π(R - r)(R + r). The difference of squares (R² - r²) is much faster to calculate as (R-r)(R+r), especially when R and r are large or have common factors.
  • Combined Figures: Look for symmetry. Often, you can calculate the area of one part and multiply. Recognize standard shapes and their area formulas instantly. For example, if a square has a circle inscribed, the area of the region between them is Side² - π(Side/2)².
  • Approximation: If options are far apart and π is involved, use π ≈ 3 or π ≈ 3.14 for quick estimation, especially when dealing with non-multiples of 7.

Common Mistakes

  1. Confusing +2w and -2w: Students often mix up the formulas for paths outside (add 2w) and inside (subtract 2w) the main figure. Read the question carefully.
  2. Radius vs. Diameter: In circular problems, ensure you're using the correct value (radius is half of diameter). A common error is using diameter in πr².
  3. Incorrect Decomposition: For combined figures, misidentifying the component shapes or incorrectly adding/subtracting areas (e.g., double-counting an overlapping region or missing a part).
  4. Calculation Errors: Especially with π (using 22/7 vs 3.14), squaring numbers, or basic arithmetic. Practice mental math and verify calculations.

IMAGE-Diagrams of paths around rectangles/squares and a circular ring.

IMAGE-Examples of combined figures like a rectangle with a semi-circle or a square with a sector cut out.

Derivation (brief)

  • Path Area (outside rectangle): Imagine a rectangle with length L and breadth B. A path of width 'w' outside it creates a larger rectangle. The new length becomes L + w + w = L + 2w, and the new breadth becomes B + w + w = B + 2w. The area of the path is the area of the outer rectangle minus the area of the inner rectangle: (L + 2w)(B + 2w) - LB. Expanding this gives LB + 2wL + 2wB + 4w² - LB = 2wL + 2wB + 4w² = 2w(L + B + 2w). The derivation for an inside path follows similar logic, where the inner rectangle dimensions become L - 2w and B - 2w.
  • Ring Area: This is simply the area of the larger circle (πR²) minus the area of the smaller circle (πr²), leading to π(R² - r²). The factorization (R - r)(R + r) is an algebraic identity that simplifies calculation.

Advanced Examples

Q: A square field of side 20 m has two cross paths, each 1.5 m wide, running through its center, parallel to its sides. Find the area of the paths. A: Given: Side of square a = 20 m, width of path w = 1.5 m. For two cross paths in a rectangular/square field, the formula is w(L + B - w). Here, L=a, B=a. Area = w(a + a - w) Area = 1.5 (20 + 20 - 1.5) Area = 1.5 (40 - 1.5) Area = 1.5 * 38.5 Area = 57.75 sq m

Q: A semi-circular garden is attached to one side of a rectangular plot of land. The rectangular plot is 14 m long and 10 m wide. The semi-circle's diameter is equal to the width of the rectangular plot. Find the total area of the land. A: Area of rectangle = Length × Width = 14 m × 10 m = 140 sq m. Diameter of semi-circle = Width of rectangle = 10 m. Radius of semi-circle (r) = Diameter / 2 = 10 / 2 = 5 m. Area of semi-circle = (1/2) * πr² = (1/2) * (22/7) * 5² Area of semi-circle = (1/2) * (22/7) * 25 = (11/7) * 25 = 275/7 ≈ 39.28 sq m. Total Area = Area of rectangle + Area of semi-circle Total Area = 140 + 275/7 = (980 + 275)/7 = 1255/7 ≈ 179.28 sq m.

Variation Types

  • Path of variable width: Though less common in SSC, be prepared to calculate areas by breaking the path into smaller rectangles.
  • Path only on certain sides: If a path is only on one or two sides, calculate the area of those specific rectangular strips.
  • Finding dimensions: Sometimes, the area of the path or ring is given, and you need to find one of the dimensions (L, B, w, R, or r). This involves solving a linear or quadratic equation.
  • Shaded regions with sectors/triangles: Often, a square or circle will have a sector or triangle cut out, or vice-versa. The key is to identify the basic shapes and apply their area formulas. For example, area of a sector = (θ/360) * πr².

Time-Saving Methods

  • Memorize common values: Know π ≈ 3.14 and 22/7. Also, squares up to 30, and common Pythagorean triplets (3,4,5; 5,12,13; 8,15,17; 7,24,25) for quick triangle calculations.
  • Unit consistency: Always check if all dimensions are in the same unit. Convert if necessary before calculation.
  • Option elimination: In multiple-choice questions, often you can eliminate options based on rough estimation or the presence of π or factors of 7/11.
  • Practice mental arithmetic: For simple multiplications and additions, avoid writing down every step to save time.
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