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Mensuration - Area of Trapezium and General Quadrilaterals

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A trapezium is a quadrilateral with one pair of parallel sides. Visually, imagine a four-sided figure where the top side aa and the bottom side bb are parallel lines, and the vertical gap between them represents the height hh.

The area of a trapezium is calculated as half the product of the sum of the parallel sides and the perpendicular distance between them. This formula essentially takes the average of the two parallel bases and multiplies it by the height.

A general quadrilateral can be divided into two triangles by drawing one of its diagonals. Visually, imagine a diagonal dd splitting the shape; we then draw two perpendicular lines (heights h1h_1 and h2h_2) from the opposite vertices to this diagonal.

The area of a general quadrilateral is the sum of the areas of the two triangles formed by the diagonal. This is expressed as 12×d×(h1+h2)\frac{1}{2} \times d \times (h_1 + h_2), where dd is the diagonal and h1,h2h_1, h_2 are the perpendicular offsets.

A rhombus is a special quadrilateral where all sides are equal and diagonals bisect each other at right angles. Visually, it resembles a diamond shape where the two internal diagonals d1d_1 and d2d_2 cross at a 9090^{\circ} angle.

The area of a rhombus can be found using its diagonals. Since the diagonals are perpendicular, the area is simply half the product of the lengths of the two diagonals.

The height or altitude of any quadrilateral is the perpendicular distance from a vertex to the opposite side or between two parallel sides. In geometry diagrams, this is denoted by a line meeting a base at a 9090^{\circ} angle, often marked with a small square symbol.

📐Formulae

Area of Trapezium=12×(a+b)×hArea\ of\ Trapezium = \frac{1}{2} \times (a + b) \times h

Area of General Quadrilateral=12×d×(h1+h2)Area\ of\ General\ Quadrilateral = \frac{1}{2} \times d \times (h_1 + h_2)

Area of Rhombus=12×d1×d2Area\ of\ Rhombus = \frac{1}{2} \times d_1 \times d_2

Area of Rhombus=base×altitudeArea\ of\ Rhombus = base \times altitude

💡Examples

Problem 1:

Find the area of a trapezium whose parallel sides are 14 cm14\ cm and 20 cm20\ cm, and the perpendicular distance between them is 8 cm8\ cm.

Solution:

  1. Identify given values: a=14 cma = 14\ cm, b=20 cmb = 20\ cm, h=8 cmh = 8\ cm
  2. Apply formula: Area=12×(a+b)×hArea = \frac{1}{2} \times (a + b) \times h
  3. Substitute: Area=12×(14+20)×8Area = \frac{1}{2} \times (14 + 20) \times 8
  4. Calculate: Area=12×34×8=17×8=136 cm2Area = \frac{1}{2} \times 34 \times 8 = 17 \times 8 = 136\ cm^2

Explanation:

We use the standard trapezium area formula by adding the parallel sides, multiplying by the height, and dividing by 2.

Problem 2:

The diagonal of a quadrilateral is 30 cm30\ cm and the perpendiculars dropped on it from the opposite vertices are 10.5 cm10.5\ cm and 12.5 cm12.5\ cm. Find the area.

Solution:

  1. Identify given values: d=30 cmd = 30\ cm, h1=10.5 cmh_1 = 10.5\ cm, h2=12.5 cmh_2 = 12.5\ cm
  2. Apply formula: Area=12×d×(h1+h2)Area = \frac{1}{2} \times d \times (h_1 + h_2)
  3. Substitute: Area=12×30×(10.5+12.5)Area = \frac{1}{2} \times 30 \times (10.5 + 12.5)
  4. Calculate sum: 10.5+12.5=2310.5 + 12.5 = 23
  5. Final calculation: Area=15×23=345 cm2Area = 15 \times 23 = 345\ cm^2

Explanation:

To find the area of a general quadrilateral, we treat it as two triangles sharing the same diagonal base and sum their individual areas.