Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

Mathematics (Maths) Class 9

Class 9 : Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

The document Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. Which of the following figures lie on the same base and between the same parallels.
 In such a case, write the common base and the two parallels.


Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

Solution: The figures (i), (iii), and (v) lie on the same base and between the same parallels.

 Common baseTwo parallels
Fig. (i)
Fig. (iii)
Fig. (v)
DC
QR
AD
DC and AB
QR and PS
AD and BQ

 

PARALLELOGRAMS ON THE SAME BASE AND BETWEEN THE SAME PARALLELS

Let us recall that “A diagonal of a parallelogram divides it into two triangles of equal area.” In case of a parallelogram, any side can be called its base, whereas the length of the line segment which is perpendicular to the base from the opposite side is called the altitude or height of the parallelogram corresponding to the given base.
In the figure,

(i) DP is the altitude of parallelogram ABCD, corresponding to the base AB.

Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

(ii) DQ is the altitude of parallelogram ABCD, corresponding to the base BC.

(iii) BR is the altitude of parallelogram ABCD, corresponding to the base AD.

(iv) CS is the altitude of parallelogram ABCD, corresponding to the base AB.
 

Theorem: Parallelograms on the same base (or on equal bases) and between the same parallels are equal in area.
Proof: Let us have two parallelogram ABCD and ABEF on the same base AB and between the same parallel lines AB and FC.

Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

In ΔADF and ΔBCE, AD = BC      [Opposite sides of a parallelogram]
AF = BE                          [Opposite sides of a parallelogram]
∠DAF = ∠CBE                 [Since AD || BC and AF || BE]
∴ ΔADF ≌ ΔBCE [SAS criteria]

Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

But congruent triangles are equal in area. ∴
ar (ΔADF) = ar (ΔBCE) Adding ar (ABED) to both sides, we have
ar (ΔADF) + ar (ABED) + ar (BCE) + ar (ABED)
⇒ ar (parallelogram ABEF) = ar (parallelogram ABCD).

REMEMBER

I . Area of a parallelogram is the product of its any side and the corresponding altitude.
II. Parallelograms on the same base (or equal bases) and between the same parallels are equal in area.
III. Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of Class 9

Dynamic Test

Content Category

Related Searches

Important questions

,

ppt

,

Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

,

Previous Year Questions with Solutions

,

Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

,

Exam

,

Extra Questions

,

mock tests for examination

,

shortcuts and tricks

,

Summary

,

practice quizzes

,

pdf

,

Semester Notes

,

MCQs

,

Free

,

video lectures

,

study material

,

Objective type Questions

,

Viva Questions

,

Sample Paper

,

past year papers

,

Ex 9.1 NCERT Solutions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev

;