Class 7 Notes

Class 7 Mathematics Notes Chapter 11 (Perimeter and Area) – Mathematics Book

Philoid

Philoid

— 5 min read

Mathematics
Alright class, let's dive into Chapter 11: Perimeter and Area. This is a fundamental chapter, and understanding these concepts well is crucial not just for your class exams but also for various government competitive exams where quantitative aptitude is tested. Pay close attention to the formulas and how they are applied.

Chapter 11: Perimeter and Area - Detailed Notes

1. Introduction

  • Perimeter: The total distance around a closed figure. Imagine walking along the boundary of a shape; the total distance you cover is its perimeter.
    • Units: Measured in units of length (e.g., cm, m, km).
  • Area: The measure of the surface enclosed by a closed figure. It tells us how much space the shape occupies.
    • Units: Measured in square units (e.g., cm², m², km², hectare).
    • Important Conversion: 1 Hectare (ha) = 10,000 m²

2. Squares

  • A square is a quadrilateral with four equal sides and four right angles (90°).
  • Let 's' be the length of the side of the square.
    • Perimeter of a Square (P): P = s + s + s + s = 4s
    • Area of a Square (A): A = side × side = s²
    • Side from Perimeter: s = P / 4
    • Side from Area: s = √A

3. Rectangles

  • A rectangle is a quadrilateral with opposite sides equal and parallel, and four right angles (90°).
  • Let 'l' be the length and 'b' be the breadth (or width) of the rectangle.
    • Perimeter of a Rectangle (P): P = l + b + l + b = 2l + 2b = 2(l + b)
    • Area of a Rectangle (A): A = length × breadth = l × b
    • Finding Length/Breadth:
      • l = A / b
      • b = A / l
      • l = (P/2) - b
      • b = (P/2) - l

4. Parallelograms

  • A parallelogram is a quadrilateral with opposite sides equal and parallel. (Note: Rectangles and Squares are special types of parallelograms).
  • Key: The area depends on the base and the corresponding perpendicular height.
  • Let 'b' be the base and 'h' be the corresponding height (the perpendicular distance from the opposite vertex to the base).
    • Area of a Parallelogram (A): A = base × height = b × h
    • Perimeter of a Parallelogram: P = 2 (sum of adjacent sides). If sides are 'a' and 'b', P = 2(a+b). Note: The height 'h' is NOT usually one of the sides unless it's a rectangle/square.
    • Important: Any side can be chosen as the base, but the height must be perpendicular to that chosen base.

5. Triangles

  • A triangle is a closed figure formed by three line segments.
  • Key: Like parallelograms, the area depends on the base and the corresponding perpendicular height (also called altitude).
  • Let 'b' be the base and 'h' be the corresponding height (the perpendicular distance from the opposite vertex to the base).
    • Area of a Triangle (A): A = (1/2) × base × height = (1/2)bh
    • Connection to Parallelograms: A triangle can be seen as half of a parallelogram with the same base and height. If you draw a diagonal in a parallelogram, it divides it into two congruent triangles.
    • Perimeter of a Triangle: P = Sum of all three sides.
    • Right-Angled Triangle: If it's a right-angled triangle, the two sides containing the right angle can be taken as base and height. Area = (1/2) × (product of sides containing the right angle).

6. Circles

  • A circle is a set of points equidistant from a fixed central point.
    • Radius (r): The distance from the center to any point on the circle.
    • Diameter (d): The distance across the circle passing through the center. d = 2r or r = d/2.
    • Circumference (C): The perimeter or the distance around the circle.
    • Pi (π): A mathematical constant, approximately equal to 3.14 or 22/7. The ratio of a circle's circumference to its diameter (C/d = π).
  • Formulas:
    • Circumference of a Circle (C):
      • C = πd
      • C = 2πr
    • Area of a Circle (A): A = πr²

7. Applications (Very Important for Competitive Exams)

  • Area of Paths:
    • Path outside a rectangle: Area of path = Area of (Outer Rectangle) - Area of (Inner Rectangle).
      • If path width is 'w', outer length L = l + 2w, outer breadth B = b + 2w.
      • Area = LB - lb = (l+2w)(b+2w) - lb
    • Path inside a rectangle: Area of path = Area of (Outer Rectangle) - Area of (Inner Rectangle).
      • If path width is 'w', inner length L = l - 2w, inner breadth B = b - 2w.
      • Area = lb - LB = lb - (l-2w)(b-2w)
    • Cross Paths: If two paths cross at the center of a rectangle (length 'l', breadth 'b'), one parallel to length (width w1) and one parallel to breadth (width w2).
      • Area of paths = (Area of length-wise path) + (Area of breadth-wise path) - (Area of common square/rectangle at the center)
      • Area = (l × w2) + (b × w1) - (w1 × w2)
  • Cost Calculations:
    • Cost of fencing = Perimeter × Rate per unit length
    • Cost of leveling/tiling/ploughing = Area × Rate per unit area
  • Unit Conversions: Be careful with units. Ensure all dimensions are in the same unit before calculating. Remember:
    • 1 m = 100 cm
    • 1 m² = 1 m × 1 m = 100 cm × 100 cm = 10,000 cm²
    • 1 km = 1000 m
    • 1 km² = 1 km × 1 km = 1000 m × 1000 m = 1,000,000 m²
    • 1 hectare = 10,000 m²

Multiple Choice Questions (MCQs)

  1. The perimeter of a square park is 120 m. What is its area?
    (A) 900 m²
    (B) 14400 m²
    (C) 30 m²
    (D) 600 m²

  2. The length and breadth of a rectangular field are 50 m and 30 m respectively. What is the cost of fencing it at ₹20 per meter?
    (A) ₹1600
    (B) ₹30000
    (C) ₹3200
    (D) ₹1500

  3. The area of a parallelogram is 48 cm². If its base is 8 cm, what is its corresponding height?
    (A) 6 cm
    (B) 12 cm
    (C) 384 cm
    (D) 4 cm

  4. The base of a triangle is 10 cm and its height is 7 cm. The area of the triangle is:
    (A) 70 cm²
    (B) 35 cm²
    (C) 17 cm²
    (D) 17.5 cm²

  5. What is the circumference of a circle with a diameter of 14 cm? (Use π = 22/7)
    (A) 44 cm
    (B) 88 cm
    (C) 154 cm
    (D) 22 cm

  6. The area of a circular garden is 154 m². Find its radius. (Use π = 22/7)
    (A) 14 m
    (B) 7 m
    (C) 22 m
    (D) 49 m

  7. A wire is bent in the shape of a rectangle of length 10 cm and breadth 6 cm. If the same wire is re-bent into a square, what will be the side of the square?
    (A) 8 cm
    (B) 16 cm
    (C) 4 cm
    (D) 60 cm

  8. The area of a right-angled triangle is 30 cm². If one of the sides containing the right angle is 5 cm, what is the length of the other side containing the right angle?
    (A) 6 cm
    (B) 12 cm
    (C) 10 cm
    (D) 15 cm

  9. A rectangular park is 60 m long and 40 m wide. A path 2 m wide is constructed outside the park. Find the area of the path.
    (A) 416 m²
    (B) 2400 m²
    (C) 200 m²
    (D) 400 m²

  10. How many square meters are there in 2.5 hectares?
    (A) 250 m²
    (B) 2500 m²
    (C) 25000 m²
    (D) 250000 m²

Answer Key for MCQs:

  1. (A) 900 m² (Side = 120/4 = 30 m. Area = 30² = 900 m²)
  2. (C) ₹3200 (Perimeter = 2(50+30) = 160 m. Cost = 160 × 20 = ₹3200)
  3. (A) 6 cm (Height = Area / Base = 48 / 8 = 6 cm)
  4. (B) 35 cm² (Area = (1/2) × 10 × 7 = 35 cm²)
  5. (A) 44 cm (Circumference = πd = (22/7) × 14 = 44 cm)
  6. (B) 7 m (Area = πr² => 154 = (22/7) × r² => r² = (154 × 7) / 22 = 7 × 7 = 49 => r = 7 m)
  7. (A) 8 cm (Perimeter of rectangle = 2(10+6) = 32 cm. This is the length of the wire. Perimeter of square = 32 cm. Side of square = 32/4 = 8 cm)
  8. (B) 12 cm (Area = (1/2) × base × height => 30 = (1/2) × 5 × height => height = (30 × 2) / 5 = 12 cm)
  9. (A) 416 m² (Outer length = 60+2+2 = 64 m. Outer breadth = 40+2+2 = 44 m. Outer Area = 64 × 44 = 2816 m². Inner Area = 60 × 40 = 2400 m². Path Area = 2816 - 2400 = 416 m²)
  10. (C) 25000 m² (1 hectare = 10000 m². So, 2.5 hectares = 2.5 × 10000 = 25000 m²)

Revise these formulas and concepts regularly. Practice solving problems involving different shapes and applications, especially those involving paths and cost calculations. Good luck!

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