AREA OF PARALLELOGRAMS AND TRIANGLES

Theorem 9.1 : Parallelograms on the same base and between the same parallels

are equal in area.

Proof : Two parallelograms ABCD and EFCD, on

the same base DC and between the same parallels

AF and DC are given (see Fig.9.12).

We need to prove that ar (ABCD) = ar (EFCD).

In Δ ADE and Δ BCF,

∠ DAE = ∠ CBF (Corresponding angles from AD

BC and transversal AF) (1)

∠ AED = ∠ BFC (Corresponding angles from ED

FC and transversal AF) (2)

Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3) Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

Now, ar (ABCD) = ar (ADE) + ar (EDCB)

= ar (BCF) + ar (EDCB) [From(5)]

= ar (EFCD)

So, parallelograms ABcD and EFCD are equal in area.

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 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) =
                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.
Therefore, ar (ABQP) = ar (ABCD) (By Theorem 9.1) (1)
But Δ PAB ≅ Δ BQP (Diagonal PB divides parallelogram ABQP into two congruent
triangles.)
So, ar (PAB) = ar (BQP) (2)
Therefore, ar (PAB) =
1
2
ar (ABQP) [From (2)] (3)
This gives ar (PAB) =
1 ar (ABCD)
2
[From (1) and (3)]

Theorem 9.2 : Two triangles on the same base (or equal bases) and between the
same parallels are equal in area.Now, suppose ABCD is a parallelogram whose one of the diagonals is AC
(see Fig. 9.20). Let AN⊥ DC. Note that
Δ ADC ≅ Δ CBA (Why?)
So, ar (ADC) = ar (CBA) (Why?)
Therefore, ar (ADC) =
1 ar (ABCD)
2
=
1 (DC AN)
2
× (Why?)
So, area of Δ ADC =
1
2 × base DC × corresponding altitude AN
In other words, area of a triangle is half the product of its base (or any side) and
the corresponding altitude (or height). Do you remember that you have learnt this
formula for area of a triangle in Class VII ? From this formula, you can see that two
triangles with same base (or equal bases) and equal areas will have equal
corresponding altitudes.
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 .
Theorem 9.3 : Two triangles having the same base (or equal bases) and equal
areas lie between the same parallels.
Let us now take some examples to illustrate the use of the above results.

FOR REFERENCE OF DIAGRAMS PLEASE REFER THE NCERT BOOK OF CLASS 9TH.....WE HAVE STATED ABOUT THE FIGURE.