ICSE Class 9 Maths Chapter 16 Area Theorems Proof and Use

Area Theorems

Proof And Use

16.1 Introduction

Area of a plane figure is the region bounded by it.

Students have already used formulae for finding the areas of different geometrical figures. For example:

area of a triangle = \(\frac{1}{2}\) base × height

area of a rectangle = length × breadth

area of a parallelogram = base × height and so on.

In the current chapter, we shall be comparing the areas of different geometrical figures, such as parallelograms, rectangles and triangles subject to certain conditions.

1. Equal figures mean, the figures equal in area.

2. Congruent figures are always equal in area, but the converse is not always true.

16.2 Figures Between The Same Parallels

If a parallelogram PQRS, a rectangle EFGH and a triangle LMN are so drawn that their bases lie on the same straight line (say, CD) and their other vertices lie on another straight line (say, AB) parallel to CD, then the parallelogram PQRS, the rectangle EFGH and the triangle LMN are said to be between the same parallels.

It is obvious that the parallelogram, the rectangle and the triangle between the same parallels have equal altitudes (height).

Note: 1. In the figure, given above, if QR = FG = MN, we say that the figures PQRS, EFGH and LMN are on equal bases and between the same parallels.

2. In the figure, given alongside, if PQ is parallel to AB, then the figures PAB, FABE, DABC, QAB, etc. are said to be on the same base and between the same parallels.

Theorem 19

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

Given: Parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and DE.

To Prove: Area of (parallelogram ABCD) = Area of (parallelogram ABEF).

Proof

Hence proved.

Corollary

Since, rectangle is a parallelogram also, the above theorem can also be stated as: "The area of a parallelogram is equal to the area of a rectangle on the same base and between the same parallels."

Theorem 20

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

Given: Triangle ABC and parallelogram ABDE on the same base AB and between the same parallels AB and ED.

To Prove: Area (triangle ABC) = \(\frac{1}{2}\) Area (parallelogram ABDE)

Construction

Complete the parallelogram ABFC.

Proof

Hence proved.

Teacher's Note

These theorems explain why triangular roof designs use less material than rectangular ones when built on the same base and height - a principle used in efficient architecture and construction to minimize costs.

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