ICSE Class 8 Maths Geometry Chapter 06 Areas of Geometrical Figures

Areas Of Geometrical Figures

The area of a closed bounded figure is the measure of the surface enclosed by its boundary. In this chapter, we will study the relations between the areas of some geometrical figures.

Two geometrical figures are called equal if they have equal areas. Such figures need not be congruent, however. For example, the rectangle and the square in the figure have equal areas but they are not congruent.

Theorem 1

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

In the figure, the parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and FC.

Therefore, area of the parallelogram ABCD = area of parallelogram ABEF.

Theorem 2

A parallelogram is equal in area to a rectangle on the same base and between the same parallels.

In the adjoining figure, the parallelogram ABCD and the rectangle ABPQ are on the same base AB and between the same parallel AB and QC.

Therefore, area of parallelogram ABCD = area of rectangle ABPQ = \(AB \times AQ = b \times h\)

Thus, the area of a parallelogram is the product of the base (b) and the perpendicular height (h).

Note - Theorem 2 follows directly from Theorem 1 as every rectangle is a parallelogram too.

Teacher's Note

Understanding parallelogram area helps in calculating land area for construction and real estate purposes, which is essential for property valuation.

Theorem 3

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

In each of the following figures, triangle ABC and parallelogram BCDE are on the same base BC and between the same parallels BC and XY.

In each case, area of \(\triangle ABC = \frac{1}{2} \times\) area of parallelogram BCDE

\(= \frac{1}{2} BC \times AM\) [since area of parallelogram = base \(\times\) height]

Area of a triangle = \(\frac{1}{2} \times\) base \(\times\) height (or altitude)

Example

In the figure, the area of the parallelogram ABCD is 264 cm\(^2\). Find the area of triangle EAB. Also, find DC if the height LM = 12 cm.

Solution

Area of \(\triangle EAB = \frac{1}{2} \times\) area of parallelogram ABCD

\(= \frac{1}{2} \times 264 \text{ cm}^2 = 132 \text{ cm}^2\)

Now, area of \(\triangle EAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times LM = \frac{1}{2} \times AB \times 12 \text{ cm}\)

\(\Rightarrow 132 \text{ cm}^2 = \frac{1}{2} \times AB \times 12 \text{ cm} \Rightarrow AB = \frac{132}{6} \text{ cm} = 22 \text{ cm}\)

Thus, \(DC = AB = 22 \text{ cm}\) [Opposite sides of a parallelogram are equal.]

Teacher's Note

Triangle area calculations are used in surveying land for agriculture and determining the material needed for triangular roof structures in construction.

Theorem 4

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

In the figure, triangle ABC and triangle DBC are on the same base BC and between the same parallels AD and BC.

Therefore, area of \(\triangle ABC =\) area of \(\triangle DBC\).

Solved Examples

Example 1

In the adjoining figure, find the areas of the parallelogram ABCD and the triangle PAD.

Solution

The area of the rectangle EFCD = \(DC \times CF\) = \(26 \times 18 \text{ cm}^2\).

Since the parallelogram ABCD and the rectangle EFCD are on the same base DC and between the same parallels DC and EF,

Therefore, the area of the parallelogram ABCD = the area of the rectangle EFCD = \(26 \times 18 \text{ cm}^2\).

Now, triangle PAD and parallelogram ABCD are on the same base AD and between the same parallels AD and BC.

Therefore, area of \(\triangle PAD = \frac{1}{2} \times\) area of parallelogram ABCD = \(\frac{1}{2} \times 26 \times 18 \text{ cm}^2 = 234 \text{ cm}^2\).

Example 2

In the adjoining figure, ABCD is a trapezium. Prove that the area of triangle PAD = the area of triangle PBC.

Solution

Triangle ABD and triangle ABC are on the same base AB and between the same parallels AB and DC.

Therefore, the area of \(\triangle ABD =\) the area of \(\triangle ABC\).

Subtracting the area of triangle ABP from both sides,

The area of \(\triangle ABD -\) the area of \(\triangle ABP =\) the area of \(\triangle ABC -\) the area of \(\triangle ABP\).

Therefore, the area of \(\triangle PAD =\) the area of \(\triangle PBC\).

Example 3

In the adjoining figure, DE \(\|\) BC. Prove that (i) the area of triangle BED = the area of triangle CED and (ii) the area of triangle ABE = the area of triangle ACD

Solution

(i) Triangle BED and triangle CED are on the same base DE and between the same parallels DE and BC.

Therefore, the area of \(\triangle BED =\) the area of \(\triangle CED\).

(ii) Adding the area of triangle ADE to both sides,

The area of \(\triangle BED +\) the area of \(\triangle ADE =\) the area of \(\triangle CED +\) the area of \(\triangle ADE\).

Therefore, the area of \(\triangle ABE =\) the area of \(\triangle ACD\).

Teacher's Note

These theorems about equal areas help in understanding land division and inheritance laws where properties must be divided into equal portions despite having different shapes.

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