Theorem 9.1 Class 9 Maths Explanation with Proof

Today, we will discuss theorem 9.1 class 9 Maths which is related to Chapter 9 Areas of Parallelograms and Triangles  Class 9 Mathematics. After understanding theorem 9.1, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

 

 Theorem 9.1 Class 9

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

Given 

Two parallelograms ABCD and EFCD have the same base CD and lie between two parallels AF and DC

To prove 

Areas ABCD= area EFCD

Proof 

In ∆ADE and ∆BCF,

∠DAE= ∠CBF (corresponding angles from AD∥BC and transversal AF)………………………(i)

∠AED= ∠BFC (corresponding angles from ED∥FC and transversal AF)………………………(ii)

Therefore, ∠ADE= ∠BFC (Angles sum property of a triangle)…………………………..(iii)

Also AD=BC(opposite sides of the parallelograms ABCD)…..(iv)

So, ∆ADE≅∆BCF (By ASA rule using (i),(iii), and (iv))

Therefore arADE=arBCF(congruent figures have equal areas)………..(v)

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

                            = ar(BCF)+ar(EDCB)

                            =ar(EFCD) 

So , parallelograms ABCD and EFCD are equal in area.

Hence proved

Related Topics

1. Theorem 9.2

2. Theorem 9.3