## Revision Notes of Chapter 9 Area of Parallelograms and Triangles Class 9th Math

**Topics in the Chapter**

- Area of Plane Figures
- Fundamentals
- Theorems

**Areas of Plane Figures**

- Two congruent figures have equal areas.
- A diagonal of a parallelogram divides it into two triangles of equal area.
- Parallelograms on the same base (or equal base) and between the same parallels are equal in area.
- Triangles on the same base (or equal bases) and between the same parallels are equal in area.
- Area of a parallelogram = base × height.
- Area of a triangle = 1/2 ×base × height
- A median of a triangle divides it into two triangles of equal area.
- Diagonals of a parallelogram divides it into four triangles of equal area.

**Fundamentals**

- 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.

Examples:

- Area of triangle is half the product of its base (or any side) and the corresponding altitude (or height).
- Two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.
- A median of a triangle divides it into two triangles of equal areas.

**Theorems**

**Theorem 1:**

__Statement:__Parallelograms on the same base and between the same parallels are equal in area.

__Given:__Two parallelograms ABCD and EBCF on the same base BC and between the same parallels BC and AF.

__To prove:__ar (||gm ABCD) = ar (||gm EBCF)

__Proof:__

In Î”ABE and Î”DCF,

EAB = ∠FDC (pair of corresponding angles are equal as BE||CF, AF is the transversal) --- (i)

FDC =

∠BEA = ∠CFD (pair of corresponding angles are equal as BE||CF, AF is the transversal) --- (ii)

Now,

∠EAB + ∠ABE + ∠BEA = ∠FDC + ∠DCF + ∠CFD = 180° (sum of measure of the interior angles of a triangle is 180°)

∴ ∠ABE = ∠DCF --- (iii) [using (i) and (ii)]

Also, AB = DC (Opposite side of a ||gm are equal) --- (iv)

Using (i), (iii) and (iv),

Î”ABE ⩭ Î”DCF (ASA rule)

∴ ar (Î”ABE) = ar (Î”DCF)

∴ ar (||gm ABCD) = ar (Î”ABE) + ar (trapezium EBCD)

= ar (Î”DCF) + ar (trapezium EBCD)

= ar (||gm EBCD)

Hence, ar (Il gm ABCD) = ar (||gm EBCF)

**Theorem 2:**

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

__Given:__Two triangles Î”ABC and Î”DBC are on the same base BC and between the same parallels EF and BC.

__To prove:__ar(Î”ABC) = ar(Î”DBC)

__Construction:__Through B, draw BE || AC, intersecting the line AD produced in E and through C, draw CF || BD, intersecting the line AD produced in F.

__Proof :__

EACB and DFCB are parallelograms (since two pairs of opposite sides are parallel).

Also ||gm EACB and ||gm DFCB are on the same base BC and between the same parallels EF and BC.

ar (||gm EACB) = ar (||gm DFCB) --- (i)

Now,

AB is the diagonal of ||gm EACB

∴ ar(Î”EAB) = ar(Î”ABC)

∴ ar (Î”ABC) = (1/2) ar(||gm EACB) --- (ii)

Similarly,

ar (Î”DBC) = (1/2) ar(||gm DFCB) --- (iii)

From equations (i), (ii) and (iii), we get,

ar(Î”ABC) = ar(Î”DBC)

**Theorem 3:**

__Statement:__Two triangles on the same base (or equal bases) and equal areas lie between the same parallels.