# CBSE Class 9 Maths Revision Notes Chapter 9

## Class 9 Mathematics Revision Notes for Areas of Parallelograms and Triangles of Chapter 9

Extramarks offers a detailed explanation of Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles, in which different concepts like the definition of a triangle, a parallelogram, and different theorems related to finding areas of these shapes are explored. Subject-matter experts in an easy-to-understand manner create these notes, to allow simpler understanding and higher retention of mathematical concepts in students’ minds. These notes are easily accessible, helping students build a successful strategy for their exam preparation.

## Class 9 Mathematics Revision Notes for Areas of Parallelograms and Triangles of Chapter 9

### Area of a Closed Shape

The planar area of any enclosed figure is the portion of the plane enclosed by that figure. The measure of this planar area in the form of numbers in suitable units is termed ‘area’. Ex: 5 m2.

### Properties of the Area of a Figure

• If two figures are similar in size and shape, they are said to be congruent. And if two figures are congruent, then their area must also be the same.
• The reverse is not true, i.e. if two figures have the same area, then they need not be congruent.

### Area of Parallelogram

Area of a parallelogram=base ✕ height

The height is perpendicular to the base.

### Figures on the Same Base and Between the Same Parallels

If two geometric figures share a common side as their base and their opposite vertices are located on the base’s parallel, they are said to be on the same base and between the same parallels.

### Parallelograms on the Same Base and Between the Same Parallels

Two parallelograms with the same base, and between the same parallels, are equal in area.

### Triangles on the Same Base and Between the Same Parallels

Two triangles that share the same base and their opposite vertex lie on the same parallel line (parallel to the base) have equal areas.

Here, ABC and PBC are two triangles having a common base BC and are situated between the same parallel lines XY and BC.

Then ar(ABC)=ar(PBC)

### Area of Triangle

Area of a triangle = 12*base×height

### Median of a Triangle

• The line segment drawn from a vertex of a triangle to the midpoint of the opposite side is known as the median.
• Every triangle has three medians.
• The centroid is the point where the medians intersect.
• Any median divides the triangle into two parts of equal areas.

### 1. Prove that triangles of equal areas, having one side of one of either triangles equal to one side of the other, have their corresponding altitudes equal.

Given: Two triangles ABC and DEF, such that:

1. arr(ABC)=arr(DEF)
2. BC=EF

AM and DN are altitudes of ABC and, respectively.

To prove: AM=DN

Proof:

In AM is the altitude, BC is the base.

ABC=12*BC*AM

DEF Here, DN is the altitude and BC is the base

In DEF, DN is the altitude, EF is the base.

DEF = 12 *EF * DN

12*BC*AM=12*EF * DN

Also,… given

Thus AM = DN

Hence, the theorem is proved.

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