Welcome, Guest | 0 |

## ekShiksha : Privacy Setting

# 9. AREAS OF PARALLELOGRAMS AND TRIANGLES

## 9.1 INTRODUCTION

In Chapter 5, you have seen that the study of Geometry, originated with the measurement of earth (lands) in the process of recasting boundaries of the fields and dividing them into appropriate parts. For example, a farmer Budhia had a triangular field and she wanted to divide it equally among her two daughters and one son. Without actually calculating the area of the field, she just divided one side of the triangular field into three equal parts and joined the two points of division to the opposite vertex. In this way, the field was divided into three parts and she gave one part to each of her children. Do you think that all the three parts so obtained by her were, in fact, equal in area? To get answers to this type of questions and other related problems, there is a need to have a relook at areas of plane figures, which you have already studied in earlier classes.

You may recall that the part of the plane enclosed by a simple closed figure is called a planar region corresponding to that figure. The magnitude or measure of this planar region is called its area. This magnitude or measure is always expressed with the help of a number (in some unit) such as 5 cm^{2}, 8 m^{2}, 3 hectares etc. So, we can say that area of a figure is a number (in some unit) associated with the part of the plane enclosed by the figure.

We are also familiar with the concept of congruent figures from earlier classes and from Chapter 7. Two figures are called congruent, if they have the same shape and the same size. In other words, if two figures A and B are congruent (see Fig. 9.1) , then using a tracing paper,

**Fig. 9.1**

you can superpose one figure over the other such that it will cover the other completely. So if two figures A and B are congruent,they must have equal areas. However, the converse of this statement is not true. In other words,two figures having equal areas need not be congruent. For example, in Fig. 9.2, rectangles ABCD and EFGH have equal areas (9 x 4 cm2 and 6 x 6 cm2) but clearly they are not congruent. (Why?)

**Fig. 9.2**

Now let us look at Fig. 9.3 given below:

**Fig. 9.3**

You may observe that planar region formed by figure T is made up of two planar regions formed by figures P and Q. You can easily see that

You may denote the area of figure A as ar(A), area of figure B as ar(B), area of figure T as ar(T), and so on. Now you can say that *area of a figure is a number (in some unit) associated with the part of the plane enclosed by the figure with the following two properties:*

(1) If A and B are two congruent figures, then ar(A) = ar(B); and (2) if a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar(T) = ar(P) + ar(Q).

You are also aware of some formulae for finding the areas of different figures such as rectangle, square, parallelogram, triangle etc., from your earlier classes. In this chapter, attempt shall be made to consolidate the knowledge about these formulae by studying some relationship between the areas of these geometric figures under the condition when they lie on the same base and between the same parallels. This study will also be useful in the understanding of some results on 'similarity of triangles'.

## 9.2 FIGURES ON THE SAME BASE AND BETWEEN THE SAME PARALLELS

Look at the following figures:

**Fig. 9.4**

In Fig. 9.4(i), trapezium ABCD and parallelogram EFCD have a common side DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC. Similarly, in Fig. 9.4 (ii), parallelograms PQRS and MNRS are on the same base SR; in Fig. 9.4(iii), triangles ABC and DBC are on the same base BC and in Fig. 9.4(iv), parallelogram ABCD and triangle PDC are on the same base DC.

Now look at the following figures:

**Fig. 9.5**

In Fig. 9.5(i), clearly trapezium ABCD and parallelogram EFCD are on the same base DC. In addition to the above, the vertices A and B (of trapezium ABCD) opposite to base DC and the vertices E and F (of parallelogram EFCD) opposite to base DC lie on a line AF parallel to DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC and between the same parallels AF and DC. Similarly, parallelograms PQRS and MNRS are on the same base SR and between the same parallels PN and SR [see Fig.9.5 (ii)] as vertices P and Q of PQRS and vertices M and N of MNRS lie on a line PN parallel to base SR.In the same way, triangles ABC and DBC lie on the same base BC and between the same parallels AD and BC [see Fig. 9.5 (iii)] and parallelogram ABCD and triangle PCD lie on the same base DC and between the same parallels AP and DC [see Fig. 9.5(iv)].

So, *two 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.*

Keeping in view the above statement, you cannot say that Δ PQR and Δ DQR of Fig. 9.6(i) lie between the same parallels l and QR. Similarly, you cannot say that

**Fig. 9.6**

parallelograms EFGH and MNGH of Fig. 9.6(ii) lie between the same parallels EF and HG and that parallelograms ABCD and EFCD of Fig. 9.6(iii) lie between the though they have a common base DC and lie between the parallels AD and BC). So, it should clearly be noted that out of the two parallels, one must be the line containing the common base.Note that ΔABC and ΔDBE of Fig. 9.7(i) are not on the common base. Similarly, ?ABC and parallelogram PQRS of Fig. 9.7(ii) are also not on the same base.

**Fig. 9.7**

## 9.3 PARALLELOGRAMS ON THE SAME BASE AND BETWEEN THE SAME PARALLELS

Now let us try to find a relation, if any, between the areas of two parallelograms on the same base and between the same parallels. For this, let us perform the following activities:

Let us now try to prove this relation between the two such parallelograms.

**Fig. 9.11**

Let us now take some examples to illustrate the use of the above theorem.

In Fig. 9.13, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that:

(i) ar (ABCD) = ar (EFCD)

(ii) ar (ABCD) = DC x AL.

(i) As a rectangle is also a parallelogram, therefore, ar (ABCD) = ar (EFCD) (Theorem 9.1)

(ii) From above result, ar (ABCD) = DC x FC (Area of the rectangle = length x breadth) (1)

As AL ⊥ DC, therefore, AFCL is also a rectangle.

So, AL = FC (2)

Therefore, ar (ABCD) = DC x AL [From (1) and (2)]

**Fig. 9.13**

Can you see from the Result (ii) above* that area of a parallelogram is the product of its any side and the coresponding altitude. * Do you remember that you have studied this formula for area of a parallelogram in Class VII. On the basis of this formula, Theorem 9.1 can be rewritten as *parallelograms on the same base or equal bases and between the same parallels are equal in area.* Can you write the converse of the above statement? It is as follows:* Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.* Is the converse true? Prove the converse using the formula for area of the parallelogram.

If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.

Let Δ ABP and parallelogram ABCD be on the same base AB and between the same parallels AB and PC (see Fig. 9.14). You wish to prove that ar (PAB) = ½ar (ABCD).

Draw BQ || AP to obtain another parallelogram ABQP. Now parallelograms ABQP and ABCD are on the same base AB and between the same parallels AB and PC.

**Fig. 9.14**

Therefore, ar (ABQP) = ar (ABCD) (By Theorem 9.1) (1)

But Δ PAB ≅ Δ BQP (Diagonal PB divides parallelogram ABQP into two congruenttriangles.)

So, ar (PAB) = ar (BQP) (2)

Therefore, ar (PAB) = ½ar (ABQP) [From (2)] (3)

This gives ar (PAB) = ½ar (ABCD) [From (1) and (3)]

## 9.4 TRIANGLES ON THE SAME BASE AND BETWEEN THE SAME PARALLELS

Let us look at Fig. 9.18. In it, you have two triangles ABC and PBC on the same base BC and between the same parallels BC and AP. What can you say about the areas of such triangles? To answer this question, you may perform the activity of drawing several pairs of triangles on the same base and between the same parallels on the graph sheet and find their areas by the method of counting the squares. Each time, you will find that the areas of the two triangles are (approximately) equal. This activity can be performed using a geoboard also. You will again find that the two areas are (approximately) equal.

**Fig. 9.18**

To obtain a logical answer to the above question, you may proceed as follows:

In Fig. 9.18, draw CD || BA and CR || BP such that D and R lie on line AP(see Fig.9.19)

**Fig. 9.19**

From this, you obtain two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR.

Therefore, ar (ABCD) = ar (PBCR) (Why?)

Now Δ ABC ≅ Δ CDA and Δ PBC ≅ Δ CRP (Why?)

So, ar (ABC) = ar (ABCD) and ar (PBC) = ar (PBCR) (Why?)

Therefore, ar (ABC) = ar (PBC)

In this way, you have arrived at the following theorem:

For having equal corresponding altitudes, the triangles must lie between the same parallels. From this, you arrive at the following converse of Theorem 9.2 .

**Fig. 9.20**

Show that a median of a triangle divides it into two triangles of equal areas.

Let ABC be a triangle and let AD be one of its medians (see Fig. 9.21). You wish to show that ar (ABD) = ar (ACD).

Since the formula for area involves altitude, let us draw AN ⊥ BC.

Now ar(ABD) = ½xbase x altitude (of Δ ABD)

=½ x BD x AN

= ½xCDxAN (As BD = CD)

= ½x base x altitude (of Δ ACD)

= ar(ACD)

**Fig. 9.21**

In Fig. 9.22, ABCD is a quadrilateral and BE || AC and also BE meets DC produced at E. Show that area of Δ ADE is equal to the area of the quadrilateral ABCD.

Observe the figure carefully. Δ BAC and Δ EAC lie on the same base AC and between the same parallels AC and BE.

Therefore, ar(BAC) = ar(EAC) (By Theorem 9.2)

So, ar(BAC) + ar(ADC) = ar(EAC) + ar(ADC) (Adding same areas on both sides)

or ar(ABCD) = ar(ADE)

**Fig. 9.22**

In this chapter, you have studied the following points:

- Area of a figure is a number (in some unit) associated with the part of the plane enclosed by that figure.
- Two congruent figures have equal areas but the converse need not be true.
- If a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then ar (T) = ar (P) + ar (Q), where ar (X) denotes the area of figure X.
- Two 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.
- Parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
- Area of a parallelogram is the product of its base and the corresponding altitude.
- Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.
- If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
- Triangles on the same base (or equal bases) and between the same parallels are equal in area.
- Area of a triangle is half the product of its base and the corresponding altitude.
- Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.
- A median of a triangle divides it into two triangles of equal areas.