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Edited By Safeer PP | Updated on Jul 29, 2022 06:17 PM IST

NCERT Solutions for Class 9 Maths chapter 9 exercise 9.2 has six questions based on the notion of a parallelogram with the same base and parallels. The questions consist of proving the area of two figures equal, finding any side, etc. NCERT book Exercise 9.2 Class 9 Maths has various standard questions that help to have in-depth knowledge of the topic. NCERT syllabus Class 9 Maths chapter 9 exercise 9.2. also has one problem-based question that gives a practical understanding of the topic,

NCERT solutions for Class 9 Maths exercise 9.2 focuses on the parallelograms on the same base, and between the same parallels, its solution gives a new perspective to solve the questions. The solution of this exercise provides the students with confidence in this topic.

A parallelogram is a figure in which two opposite sides are parallel to each other whose diagonals are not equal. In Class 9 Maths chapter 9 the following exercises are also present.

- Areas Of Parallelograms And Triangles Exercise 9.1
- Areas Of Parallelograms And Triangles Exercise 9.3
- Areas Of Parallelograms And Triangles Exercise 9.4

** Q1 ** In Fig. , ABCD is a parallelogram, and . If , and , find AD.

**Answer: **

We have,

AE DC and CF AD

AB = 16 cm, AE = 8 cm and CF = 10 cm

Since ABCD is a parallelogram,

therefore, AB = DC = 16 cm

We know that, area of parallelogram (ABCD) = base . height

= CD AE = (16 8 )

SInce, CF AD

therefore area of parallelogram = AD CF = 128

= AD = 128/10

= ** AD = 12.8 cm **

Thus the required length of AD is 12.8 cm.

** Q2 ** If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that

** Answer: **

Join GE and HE,

Since F, F, G, H are the mid-points of the parallelogram ABCD. Therefore, GE || BC ||AD and HF || DC || AB.

It is known that if a triangle and the parallelogram are on the same base and between the same parallel lines. then the area of the triangle is equal to the half of the area of the parallelogram.

Now, EFG and ||gm BEGC are on the same base and between the same parallels EG and BC.

Therefore, ar ( EFG) = ar (||gm BEGC)...............(i)

Similarly, ar ( EHG) = 1/2 . ar(||gm AEGD)..................(ii)

By adding eq (i) and eq (ii), we get

ar ( EFG) + ar ( EHG) = 1/2 (ar (||gm BEGC) + ar(||gm AEGD))

ar (EFGH) = 1/2 ar(ABCD)

Hence proved

** Answer: **

We have,

ABCD is a parallelogram, therefore AB || CD and BC || AD.

Now, APB and ||gm ABCD are on the same base AB and between two parallels AB and DC.

Therefore, ar ( APB) = 1/2 . ar(||gm ABCD)...........(i)

Also, BQC and ||gm ABCD are on the same base BC and between two parallels BC and AD.

Therefore, ar( BQC) = 1/2 . ar(||gmABCD)...........(ii)

From eq(i) and eq (ii), we get,

ar ( APB) = ar( BQC)

Hence proved.

** Q4 (i) ** In Fig. , P is a point in the interior of a parallelogram ABCD. Show that

[ ** Hint ** : Through P, draw a line parallel to AB.]

**Answer: **

We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB

Now, APB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.

Therefore, ar ( APB) = 1/2 . ar(ABEF)...............(i)

Similarly, ar ( PCD ) = 1/2 . ar (EFDC) ..............(ii)

Now, by adding both equations, we get

Hence proved.

** Q4 (ii) ** In Fig. , P is a point in the interior of a parallelogram ABCD. Show that

[ ** Hint**: Through P, draw a line parallel to AB.]

**Answer: **

We have a ||gm ABCD and AB || CD, AD || BC. Through P, draw a line parallel to AB

Now, APD and ||gm ADGHare on the same base AD and between the same parallels GH and AD.

Therefore, ar ( APD) = 1/2 . ar(||gm ADGH).............(i)

Similarily, ar ( PBC) = 1/2 . ar(||gm BCGH)............(ii)

By adding the equation i and eq (ii), we get

Hence proved.

** Q5 ** In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i)

(ii)

**Answer: **

(i) Parallelogram PQRS and ABRS are on the same base RS and between the same parallels RS and PB.

Therefore, ............(i)

Hence proved

(ii) AXS and ||gm ABRS are on the same base AS and between same parallels AS and RB.

Therefore, ar ( AXS) = 1/2 . ar(||gm ABRS)............(ii)

Now, from equation (i) and equation (ii), we get

Hence proved.

** Answer: **

We have a field in the form of parallelogram PQRS and a point A is on the side RS. Join AP and AQ. The field is divided into three parts i.e, APS, QAR and PAQ.

Since APQ and parallelogram, PQRS is on the same base PQ and between same parallels RS and PQ.

Therefore, ............(i)

We can write above equation as,

ar (||gm PQRS) - [ar ( APS) + ar( QAR)] = 1/2 .ar(PQRS)

from equation (i),

Hence, she can sow wheat in APQ and pulses in [ APS + QAR] or wheat in [ APS + QAR] and pulses in APQ.

Class 9 Maths chapter 9 exercise 9.2 is about the parallelogram in the same base and same parallels. It has six questions, among which one question is application-based, all in all, the exercise is packed with standard questions aiming to give students in-depth knowledge. Exercise 9.2 Class 9 Maths -Parallelogram is a geometric figure in which the opposite sides are parallel; using this very concept, the questions on this exercise could be tackled efficiently. The NCERT solutions for Class 9 Maths exercise 9.2 mainly focuses on the various properties exhibited by the parallelogram. Exercise 9.2 Class 9 Maths primary focus is to enlighten and give good practice to the topic; that is why it has a nice set of standard questions.

**Also Read| **Areas Of Parallelograms And Triangles Class 9 Notes

• The NCERT solutions for Class 9 Maths exercise 9.2 assists students in solving and revising all of the problems in this exercise.

• If you go over the NCERT solutions for Class 9 Mathematics chapter 9 exercise 9.2, you will be able to gain more marks, and if you practice it completely, it will help you score well in maths in examinations.

• Parallelograms on the same base and between the same parallels are the basis of exercise 9.2 Class 9 Maths.

**Also, See**

1. What are the characteristics of the Parallelogram?

It has the following characteristics:

Its two opposite sides are parallel and equal

Opposite angles are equal.

The Sum of all angles is 360.

2. What are the properties of the diagonals of the Parallelogram?

The diagonal of the parallelogram has the following characteristics:

They bisect each other.

They divide the parallelogram into two congruent triangles.

3. Are rectangle, square, and rhombus not Parallelogram?

Every figure whose opposite sides are parallel is a parallelogram; therefore, rectangle, square, and rhombus are parallelograms with a particular category.

4. Is the number of vertices and the sides of the parallelogram fixed?

Yes, it has four sides and four vertices.

5. How can the height of the parallelogram be found?

The height of the parallelogram is the perpendicular distance between the baseline and topline.

6. Is Parallelogram a quadrilateral?

The quadrilateral is any closed figure; hence parallelogram is a quadrilateral.

7. Can trapezium be a parallelogram?

No, since its two opposite sides are not parallel to each other.

8. What is the diagonal of the parallelogram?

Diagonal is any line that joins the vertically opposite angles of the parallelogram.

9. What concept is illustrated in NCERT solutions for Class 9 Maths chapter 9 exercise 9.2?

In this exercise, the properties of the parallelogram are furnished. Here the particular topic is considered, that is, the parallelograms in the same base and between the same parallels.

10. What kinds of questions do the NCERT answers for Class 9 Mathematics chapter 9 exercise 9.2 cover?

This exercise consists of 6 questions focused on the properties of the parallelograms in the same base and between the same parallels. It consists of standard questions consisting of one application-type question to enhance the knowledge of the topic.

Sep 06, 2024

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