ICSE Class 8 Maths Chapter 29 Area Propositions

Chapter 29: Area Propositions

29.1 Introduction

The area of a plane closed figure is the measure of the region (surface) enclosed by its boundary.

In each figure, given below, the shaded portion represents the area enclosed:

(i) A triangle with horizontal line shading

(ii) A trapezoid with horizontal line shading

(iii) A semicircle with horizontal line shading

In geometry, equal figures mean figures having equal areas.

Congruent figures are always equal in area.

But, figures equal in area are not necessarily congruent.

Teacher's Note

When you wrap a gift, understanding that different box shapes can hold the same amount of space helps optimize packaging.

29.2 Figures on the Same Base

The adjoining figure shows two quadrilaterals ABCD and DCEF with their one side DC common. We say that both of these quadrilaterals are on the same base DC.

Similarly, triangles ABC and DBC, as shown above, are one same base BC. Also, quadrilateral ABCD and triangle EDC, as shown above, are on same base DC.

Teacher's Note

Recognizing shapes that share a base is like understanding how different containers can sit on the same shelf space.

29.3 Figures on the Same Base and Between the Same Parallels

Figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.

Examples:

1. In the adjoining figure, trapezium ABCD and parallelogram BCEF are on the same base BC and between the same parallels BC parallel to AE.

2. Triangles ABC and DBC are on the same base BC and between the same parallels AD parallel to BC.

3. Parallelogram ABCD and triangle ABE are on the same base AB and between the same parallels AB parallel to DE.

Teacher's Note

Parallel lines create equal spacing, much like railroad tracks maintaining constant distance for safe train travel.

29.4 Some Important Propositions (without proof)

Proposition 1

Parallelograms on the same base and between the same parallels are equal in area.

The adjoining figure shows two parallelograms ABCD and ABEF on the same base AB and between the same parallels AB parallel to DE.

Area of parallelogram ABCD = Area of parallelogram ABEF.

Proposition 2

The area of a parallelogram is equal to the area of rectangle on the same base and between the same parallels.

The adjoining figure shows a parallelogram ABCD and a rectangle ABEF on the same base AB and between the same parallels AB parallel to FC.

Area of parallelogram ABCD = Area of rectangle ABEF.

Proposition 3

The area of a triangle is half the area of a parallelogram on the same base and between the same parallels.

(i) Triangle ABE = 1/2 of parallelogram ABCD

(ii) Triangle PQT = 1/2 of parallelogram PQRS

(iii) Triangle ABE = 1/2 of parallelogram ABCD.

Proposition 4

Triangles on the same base and between the same parallels are equal in area

(i) Triangle ABC = Triangle ABD

(ii) Triangle PQR = Triangle PQS

(iii) Triangle OLM = Triangle OLN

Teacher's Note

These propositions help architects design buildings where different room shapes can occupy the same floor area efficiently.

Test Yourself

1. Write, true or false:

(a) Parallelograms on the same base are equal in area

(b) Triangles between the same parallels are equal in area

(c) If two parallelograms are equal in area, they are on the same base and between the same parallels

(d) Area of a rectangle is equal to area of the parallelogram if both are on the same base and between the same parallels

2. In the adjoining figure, if area of triangle ABC = 40 cm2, the area of parallelogram ABCD = ___________

3. In the given figure

(a) Triangle ABC = ___________

(b) Triangle ABC - Triangle AOB = ___________

Example 1

In the given figure, ABCD is a rectangle, ABEF is a parallelogram and AB is parallel to DE. If the area of parallelogram ABEF is 60 cm2; find the area of:

(i) rectangle ABCD.

(ii) triangle BEF.

Solution

(i) Since, the rectangle ABCD and the parallelogram ABEF are on the same base AB and between the same parallels AB parallel to DE.

Area of the rectangle ABCD = Area of the parallelogram ABEF = 60 cm2

(ii) Since, the parallelogram ABEF and the triangle BEF are on the same base EF and between the same parallels EF parallel to AB.

Area of triangle BEF = 1/2 x Area of the parallelogram ABEF. = 1/2 x 60 cm2 = 30 cm2

Example 2

In the given figure, ABCD is a rectangle, ABEF is a parallelogram and AB is parallel to CF. If AB = 12 cm and BC = 8 cm; find:

(i) area of rectangle ABCD.

(ii) area of parallelogram ABEF.

(iii) area of triangle BEF.

Solution

(i) Area of the rectangle ABCD = Base x height = 12 cm x 8 cm = 96 cm2

(ii) Since, the parallelogram ABEF and the rectangle ABCD are on the same base AB and between the same parallels AB parallel to CF.

Area of the parallelogram ABEF = Area of the rectangle ABCD = 96 cm2

(iii) Since, the triangle BEF and the parallelogram ABEF are on the same base EF and between the same parallels EF parallel to AB.

Area of the triangle BEF = 1/2 x Area of the parallelogram ABEF = 1/2 x 96 cm2 = 48 cm2

Example 3

Given: A parallelogram ABCD. P is a point on side DC and Q is a point on side BC.

To prove: Triangle APB and Triangle AQD are equal in area.

Solution

Triangle APB and parallelogram ABCD are on the same base AB and between same parallels AB parallel to DC.

Area of Triangle APB = 1/2 of parallelogram ABCD ... I

Similarly, Triangle AQD and parallelogram ABCD are on the same base AD and between same parallels AD and BC.

Area of Triangle AQD = 1/2 of parallelogram ABCD ... II

Area of Triangle APB = Area of Triangle AQD

Hence Proved.

Example 4

In the given figure, ABCD is a trapezium in which AB is parallel to DC. If its diagonals AC and BD intersect at point O, prove that the triangles AOD and BOC are equal in area.

Solution

Since, the triangle ADB and the triangle ACB are on the same base AB and between the same parallels AB and DC, therefore the triangles ADB and ACB are equal in area.

i.e. Area of the Triangle ADB = Area of the Triangle ACB.

Subtracting the common portion (Triangle AOB) from both the sides, we get:

Area of Triangle ADB - Area of Triangle AOB = Area of Triangle ACB - Area of Triangle AOB.

Area of Triangle AOD = Area of Triangle BOC.

Hence Proved

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