ICSE Class 8 Maths Chapter 25 Theorems on Area

Chapter 25 - Theorems On Area

Area

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

In this section, we shall compare the area of two known figures which have certain relationship between them.

Equal Figures

Two plane figures are called equal if and only if they have equal areas.

Difference Between Congruent And Equal Figures

Since congruent figures are exact duplicate of each other, so they have equal areas. Thus, congruent figures are equal figures. However, two equal figures may not be congruent.

For example, consider the adjoining right angled triangles ABC and PQR.

Area of \(\triangle ABC = \left(\frac{1}{2} \times 6 \times 4\right) \text{ cm}^2\)

= 12 cm\(^2\)

Area of \(\triangle PQR = \left(\frac{1}{2} \times 8 \times 3\right) \text{ cm}^2\)

= 12 cm\(^2\)

\(\therefore\) Area of \(\triangle ABC\) = Area of \(\triangle PQR\)

\(\Rightarrow\) \(\triangle ABC\) and \(\triangle PQR\) are equal figures. Clearly, \(\triangle ABC\) and \(\triangle PQR\) are not congruent.

Teacher's Note

Understanding the difference between congruent and equal figures helps us recognize that two shapes can have the same area even if they look completely different - like how a rectangle and a triangle can cover the same amount of floor space.

Theorem 1

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

In the adjoining diagram, parallelograms ABCD and ABEF have the same base AB and are between the same parallel lines AB and DE.

\(\therefore\) Area of parallelogram ABCD = Area of parallelogram ABEF

Note. Parallelograms ABCD and ABEF are equal figures but these are not congruent figures.

Teacher's Note

This theorem shows us that two parallelograms sitting on the same foundation with the same height will always cover the same amount of space, which is useful in construction and land measurement.

Theorem 2

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

In the adjoining diagram, parallelogram ABCD and rectangle ABEF have the same base AB and are between the same parallel lines AB and FC.

\(\therefore\) Area of parallelogram ABCD = Area of rectangle ABEF.

Corollary

Area of a parallelogram = base \(\times\) height

By the above theorem,

area of parallelogram ABCD = area of rectangle ABEF

But area of rectangle = length \(\times\) breadth = AB \(\times\) BE

\(\therefore\) Area of parallelogram ABCD = AB \(\times\) BE = base \(\times\) height.

Teacher's Note

The formula for parallelogram area directly applies to real-world scenarios like calculating the area of slanted rooftops or irregular garden plots.

Theorem 3

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

In the adjoining diagram, \(\triangle ABC\) and parallelogram BCDE have the same base AB and are between the same parallel lines BC and ED.

\(\therefore\) Area of \(\triangle ABC = \frac{1}{2}\) area of parallelogram BCDE

Corollary

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

In the adjoining diagram, note that

base of \(\triangle ABC\) = BC = base of parallelogram ABDE and height of \(\triangle ABC\) = AN = height of parallelogram gm BCDE

By the above theorem,

area of \(\triangle ABC = \frac{1}{2}\) area of parallelogram BCDE

= \(\frac{1}{2} \times BC \times AN\)

= \(\frac{1}{2}\) base \(\times\) height.

Teacher's Note

The triangle area formula is essential for calculating spaces like triangular plots of land or the surface area of tent-shaped structures in everyday life.

Theorem 4

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

In the adjoining diagram, \(\triangle ABC\) and \(\triangle DBC\) have the same base BC and are between the same parallel lines BC and AD.

\(\therefore\) Area of \(\triangle ABC\) = Area of \(\triangle DBC\)

(Proofs of the above theorems are deferred to the next class.)

Example 1

In this diagram, ABCD is a rectangle with side AB = 8 cm and AD = 5 cm.

Find

(i) area of rectangle ABCD

(ii) area of parallelogram ABEF

(iii) area of \(\triangle EFG\)

Solution

(i) Area of rectangle ABCD = base \(\times\) height = (8 \(\times\) 5) cm\(^2\) = 40 cm\(^2\).

(ii) Area of parallelogram ABEF = area of rectangle ABCD (on the same base AB and between same parallels AB and DE) = 40 cm\(^2\).

(iii) Area of \(\triangle EFG = \frac{1}{2}\) area of parallelogram ABEF (on the same base EF and between same parallels EF and AG)

= \(\left(\frac{1}{2} \times 40\right)\) cm\(^2\) = 20 cm\(^2\).

Example 2

ABCD is a trapezium with AB || DC, and diagonals AC and BD intersect at O. Prove that area of \(\triangle ODA\) = area of \(\triangle OCB\).

Solution

Area of \(\triangle DAB\) = Area or \(\triangle CAB\)

(triangles on the same base AB and between the same parallel lines AB and DC)

\(\Rightarrow\) area of \(\triangle ODA\) + area of \(\triangle OAB\) = area of \(\triangle OBC\) + area of \(\triangle OAB\)

(from figure, area axiom)

\(\Rightarrow\) area of \(\triangle ODA\) = area of \(\triangle OBC\).

(subtracting same area of \(\triangle OAB\) from both sides)

Example 3

Prove that a median divides a triangle into two triangles of equal area.

Solution

Let ABC be any triangle and AD be its median i.e. D is mid-point of BC, so BD = DC. From A, draw AN perpendicular to BC.

Since area of a triangle = \(\frac{1}{2}\) base \(\times\) height,

area of \(\triangle ABD = \frac{1}{2} \times BD \times AN\)

and area of \(\triangle ADC = \frac{1}{2} \times DC \times AN\)

= \(\frac{1}{2} \times BD \times AN\)

(\(\therefore\) BD = DC)

\(\Rightarrow\) area of \(\triangle ADC\) = area of \(\triangle ABD\)

Hence, a median of a triangle divides it into two triangles of equal area.

Teacher's Note

The median property of triangles is useful in dividing properties fairly or balancing loads equally - such as dividing a triangular plot of land into two equal parts.

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