Chapter notes: Area of a Trapezium and a Polygon | Mathematics Class 8 ICSE PDF Download

Introduction

The chapter "Area of Trapezium and a Polygon" introduces students to the concepts of measuring the perimeter and area of various shapes like triangles, rectangles, squares, trapeziums, parallelograms, rhombuses, and circles. It explains how to calculate the boundary length (perimeter) and the surface area enclosed by these shapes using specific formulas. The chapter also covers unit conversions for length and area measurements and provides methods to find dimensions or areas of shapes based on given conditions. Through step-by-step explanations, students learn to apply these formulas to solve real-world problems involving geometric figures.

REVIEW

• Perimeter: The total length of the boundary of a shape.
• Area: The amount of surface enclosed within the boundary of a shape.
• Units for Perimeter: Measured in meters (m), centimetres (cm), decimeters (dm), millimetres (mm), etc.
• Unit Conversions for Length:
  • 1 m = 100 cm, 1 cm = 0.01 m
  • 1 m = 10 dm, 1 dm = 0.1 m
  • 1 dm = 10 cm, 1 cm = 0.1 dm
  • 1 m = 1000 mm, 1 mm = 0.001 m
• Units for Area: Measured in square meters (m²), square centimetres (cm²), square millimetres (mm²), etc.
• Unit Conversions for Area:
  • 1 m² = 10,000 cm², 1 cm² = 0.0001 m²
  • 1 m² = 100 dm², 1 dm² = 0.01 m²
  • 1 dm² = 100 cm², 1 cm² = 0.01 dm²

PERIMETER AND AREA OF TRIANGLES

• Perimeter: Sum of all three sides of a triangle (a + b + c).

• Area Using Sides (Heron’s Formula):
  • Step 1: Calculate semi-perimeter, s = (a + b + c) / 2.
  • Step 2: Area = √[s(s - a)(s - b)(s - c)].
• Area Using Base and Height:
  • Area = (1/2) × base × height.

Example : The base of an isosceles triangle is 12 cm, and its perimeter is 32 cm. Find its area.
Solution:

• Let each equal side be x cm. Perimeter = x + x + 12 = 32 → 2x = 20 → x = 10 cm.
• Sides are 10 cm, 10 cm, 12 cm. Semi-perimeter, s = (10 + 10 + 12) / 2 = 16 cm.
• Area = √[16(16 - 10)(16 - 10)(16 - 12)] = √[16 × 6 × 6 × 4] = √2304 = 48 cm².

Alternative Method: Draw perpendicular AD from vertex A to base BC, bisecting BC.

• BD = CD = 12 / 2 = 6 cm. In right triangle ABD, AD² = AB² - BD² = 10² - 6² = 64 → AD = 8 cm.
• Area = (1/2) × base × height = (1/2) × 12 × 8 = 48 cm².

PERIMETER AND AREA OF RECTANGLES

• Perimeter: 2 × (length + breadth) = 2(l + b).
• Area: length × breadth = l × b.
• Diagonal: √(length² + breadth²) = √(l² + b²) [Using Pythagoras' Theorem].

PERIMETER AND AREA OF SQUARES

• Perimeter: 4 × side = 4a.
• Area: side × side = a².
• Diagonal: √(side² + side²) = √(2a²) = a√2.

TRAPEZIUM

• Area: (1/2) × (sum of parallel sides) × height = (1/2) × (a + b) × h.
• Height: The perpendicular distance between the parallel sides.

PARALLELOGRAM

• Area: Base × corresponding height.
• Diagonals: Each diagonal divides the parallelogram into two equal triangles.

RHOMBUS

• Definition: A parallelogram with all sides equal.
• Perimeter: 4 × side = 4a.
• Area: (1/2) × product of diagonals = (1/2) × d1 × d2.
• Diagonals: Bisect each other at 90° and divide the rhombus into four equal triangles.
• Area Using Base and Height: Base × height (as it is a parallelogram).

Example : The diagonals of a rhombus are 16 cm and 12 cm. Find its area, side length, and perimeter.
Solution:

• Area = (1/2) × 16 × 12 = 96 cm².
• Half diagonals: OA = 16 / 2 = 8 cm, OB = 12 / 2 = 6 cm.
• Side AB = √(OA² + OB²) = √(8² + 6²) = √(64 + 36) = 10 cm.
• Perimeter = 4 × 10 = 40 cm.

CIRCLE

• Definition: A closed curve where all points are equidistant from a fixed center point.
• Radius: Distance from the center to any point on the circumference.
• Diameter: A line through the center joining two points on the circumference, d = 2r.
• Circumference: The perimeter of the circle = 2πr, where π = 22/7.
• Area: πr².