Theorem 9.2 Class 9 Maths Explanation with Proof

Today, we will discuss theorem 9.2 class 9 Maths which is related to Chapter 9 Areas of Parallelograms and Triangles  Class 9 Mathematics. After understanding theorem 9.2, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

 

 Theorem 9.2 Class 9

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Given 

Two parallelograms ABCD and ABEF on the same base BC and ar(||gm ABCD)= ar(||gmABEF)

To Prove 

AB||DE

Construction

Draw DL⊥AB and FM⊥AB.

Proof 

We know that

Area of ||gm=(base ×height)sq units

Area of ||gm ABCD=(AB×DL) sq units………….(i)

Area of ||gm ABEF= (AB×FM) sq units……………(ii)

From (i) and (ii) we have

AB×DL=AB×FM    [∵ar||gm ABCD=ar||gmABEF]

⇒DL=FM 

Also , DL||FM           (∵Dl and FM are both perpendicular to AB)

Now,  in quadrilateral DLMF, we have

DL=FM and DL||FM( A pair of opposite sides is equal and parallel)

∴  DLMF is a parallelogram

⇒LM∥DF (∵opposite sides of a parallelogram are parallel)

∴AB||DE 

Hence proved

Related Topics

1. Theorem 9.1

2. Theorem 9.3