NCERT Class 9 Mathematics

158 MATHEMATICS

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

Example 1 : 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 × AL

Solution : (i) As a rectangle is also a parallelogram,

therefore, ar (ABCD) = ar (EFCD) (Theorem 9.1)

(ii) From above result,

ar (ABCD) = DC × FC (Area of the rectangle = length × breadth) (1)

As AL DC, therefore, AFCL is also a rectangle

So, AL = FC (2)

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

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.

Example 2 : 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.

Solution : Let ✁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) =

1

ar (ABCD)

2

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. 9.13

Fig. 9.14