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

Please refer to Areas of Parallelograms and Triangles Class 9 Mathematics Notes and important questions below. The Class 9 Mathematics Chapter wise notes have been prepared based on the latest syllabus issued for the current academic year by CBSE. Students should revise these notes and go through important Class 9 Mathematics examination questions given below to obtain better marks in exams

## Areas of Parallelograms and Triangles Class 9 Mathematics Notes and Questions

The below Class 9 Areas of Parallelograms and Triangles notes have been designed by expert Mathematics teachers. These will help you a lot to understand all the important topics given in your NCERT Class 9 Mathematics textbook.

Refer to Chapter 9 Areas of Parallelograms and Triangles Notes below which have been designed as per the latest syllabus issued by CBSE and will be very useful for upcoming examinations to help clear your concepts and get better marks in examinations.

1. Figures on the same Base and Between the same Parallels
2. Parallelograms on the same Base and between the same Parallels
3. Triangles on the same Base and between the same Parallels

Area of a figure is a number (in square unit) associated with the part of the plane enclosed by that figure.
· Two congruent figures have equal areas but the converse is not true.
· Area of a parallelogram = (base X height )
Area of a triangle = 1/2 ´base ´ height
Area of a trapezium = 1/2 (sum of parallel sides) dis tan ce between them
Area of rhombus = 1/2 product of diagonals
Parallelogram on the same base and between the same parallels are equal in area.
A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
Triangles on the same base and between the same parallels are equal in area.
If a triangle and parallelogram are on the same base and between the same parallels, then.
(area of the parallelogram) = 1/2 Area of triangle
A diagonal of parallelogram divides it into two triangles of equal areas.
In parallelogram ABCD, we have Area of ΔABD = area of ΔACD

· The diagonals of a parallelogram divide it into four triangles of equal areas therefore ar(ΔAOB)=ar(ΔCOD)=ar(ΔAOD)=ar(ΔBOC)

If a parallelogram and a triangle are on the same base and between the same parallel, then area of the triangle is equal to one half area of the parallelogram.
A median AD of a ΔABC divides it into two triangles of equal areas. Therefore ar(ΔABD)=ar(ΔCD)
If the medians of a intersect at G, then ar(ΔAGB)=ar(AGC)=ar(ΔBGC)=1/3(ΔABC)

Triangles with equal bases and equal areas have equal corresponding altitude.