case study for quadrilaterals class 9

Class 9 Maths Case Study Questions of Chapter 8 Quadrilaterals PDF Download

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Class 9 Maths Case Study Questions Chapter 8  are very important to solve for your exam. Class 9 Maths Chapter 8 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Maths Chapter 8 Quadrilaterals

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These case study questions challenge students to apply their knowledge of quadrilaterals in practical scenarios, enhancing their problem-solving abilities. This article provides the Class 9 Maths Case Study Questions of Chapter 8: Quadrilaterals, enabling students to practice and excel in their examinations.

Quadrilaterals Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 8 Quadrilaterals

Case Study/Passage-Based Questions

Case Study 1: Laveena’s class teacher gave students some colorful papers in the shape of quadrilaterals. She asked students to make a parallelogram from it using paper folding. Laveena made the following parallelogram.

How can a parallelogram be formed by using paper folding? (a) Joining the sides of a quadrilateral (b) Joining the mid-points of sides of a quadrilateral (c) Joining the various quadrilaterals (d) None of these

Answer: (b) Joining the mid-points of sides of quadrilateral

Which of the following is true? (a) PQ = BD (b) PQ = 1/2 BD (c) 3PQ = BD (d) PQ = 2BD

Answer: (b) PQ = 1/2 BD

Which of the following is correct combination? (a) 2RS = BD (b) RS = 1/3 BD (c) RS = BD (d) RS = 2BD

Answer: (a) 2RS = BD

Which of the following is correct? (a) SR = 2PQ (b) PQ = SR (c) SR = 3PQ (d) SR = 4PQ

Answer: (b) PQ = SR

Case Study/Passage Based Questions

Case Study 2: Anjali and Meena were trying to prove mid-point theorem. They draw a triangle ABC, where D and E are found to be the midpoints of AB and AC respectively. DE was joined and extended to F such that DE = EF and FC is also joined.

▲ADE and ▲CFE are congruent by which criterion? (a) SSS (b) SAS (c) RHS (d) ASA

Answer: (b) SAS

∠EFC is equal to which angle? (a) ∠DAE (b) ∠EDA (c) ∠AED (d) ∠DBC

Answer: (b)∠EDA

∠ECF is equal to which angle? (a) ∠EAD (b) ∠ADE (c) ∠AED (d) ∠B

Answer: (a) ∠EAD

CF is equal to (a) EC (b) BE (c) BC (d) AD

Answer: (d) AD

CF is parallel to (a) AE (b) CE (c) BD (d) AC

Answer: (c) BD

Case Study 3. A group of students is exploring different types of quadrilaterals. They encountered the following scenario:

Four friends, Aryan, Bhavana, Chetan, and Divya, participated in a geometry project. They constructed a figure with four sides and made the following observations:

• The opposite sides of the figure are parallel.
• The opposite angles of the figure are congruent.
• The figure has two pairs of congruent adjacent sides.
• The sum of the measures of the interior angles of the figure is 360 degrees.

Based on this information, the students were asked to analyze the properties of the quadrilateral they constructed. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The type of quadrilateral formed by their figure is: (a) Parallelogram (b) Rhombus (c) Rectangle (d) Square

Answer: (a) Parallelogram

Q2. The measure of each angle in the figure is: (a) 90 degrees (b) 120 degrees (c) 135 degrees (d) 180 degrees

Answer: (d) 180 degrees

Q3. The figure is an example of a quadrilateral that satisfies the: (a) Opposite sides are equal condition (b) Opposite angles are congruent condition (c) Diagonals bisect each other condition (d) None of the above

Answer: (b) Opposite angles are congruent condition

Q4. The sum of the measures of the exterior angles of the figure is: (a) 90 degrees (b) 180 degrees (c) 270 degrees (d) 360 degrees

Answer: (d) 360 degrees

Q5. The figure has rotational symmetry of: (a) Order 1 (b) Order 2 (c) Order 3 (d) Order 4

Answer: (a) Order 1

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Mathematics Chapter 8 Quadrilaterals with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Maths Quadrilaterals Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE Case Study Questions Class 9 Maths Chapter 8 Quadrilaterals PDF Download

CBSE Case Study Questions Class 9 Maths Chapter 8  are very important to solve for your exam. Class 9 Maths Chapter 8 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Case Study Questions Maths Chapter 8 Quadrilaterals

CBSE Case Study Questions Class 9 Maths Chapter 8

Case Study/Passage-Based Questions

Case Study 1. Laveena’s class teacher gave students some colorful papers in the shape of quadrilaterals. She asked students to make a parallelogram from it using paper folding. Laveena made the following parallelogram.

How can a parallelogram be formed by using paper folding? (a) Joining the sides of quadrilateral (b) Joining the mid-points of sides of quadrilateral (c) Joining the various quadrilaterals (d) None of these

Answer: (b) Joining the mid-points of sides of quadrilateral

Which of the following is true? (a) PQ = BD (b) PQ = 1/2 BD (c) 3PQ = BD (d) PQ = 2BD

Answer: (b) PQ = 1/2 BD

Which of the following is correct combination? (a) 2RS = BD (b) RS = 1/3 BD (c) RS = BD (d) RS = 2BD

Answer: (a) 2RS = BD

Which of the following is correct? (a) SR = 2PQ (b) PQ = SR (c) SR = 3PQ (d) SR = 4PQ

Answer: (b) PQ = SR

Case Study/Passage Based Questions

Case Study 2. Anjali and Meena were trying to prove mid-point theorem. They draw a triangle ABC, where D and E are found to be the midpoints of AB and AC respectively. DE was joined and extended to F such that DE = EF and FC is also joined.

▲ADE and ▲CFE are congruent by which criterion? (a) SSS (b) SAS (c) RHS (d) ASA

Answer: (b) SAS

∠EFC is equal to which angle? (a) ∠DAE (b) ∠EDA (c) ∠AED (d) ∠DBC

Answer: (b)∠EDA

∠ECF is equal to which angle? (a) ∠EAD (b) ∠ADE (c) ∠AED (d) ∠B

Answer: (a) ∠EAD

CF is equal to (a) EC (b) BE (c) BC (d) AD

Answer: (d) AD

CF is parallel to (a) AE (b) CE (c) BD (d) AC

Answer: (c) BD

Case Study 3. A group of students is exploring different types of quadrilaterals. They encountered the following scenario:

Four friends, Aryan, Bhavana, Chetan, and Divya, participated in a geometry project. They constructed a figure with four sides and made the following observations:

• The opposite sides of the figure are parallel.
• The opposite angles of the figure are congruent.
• The figure has two pairs of congruent adjacent sides.
• The sum of the measures of the interior angles of the figure is 360 degrees.

Based on this information, the students were asked to analyze the properties of the quadrilateral they constructed. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The type of quadrilateral formed by their figure is: (a) Parallelogram (b) Rhombus (c) Rectangle (d) Square

Answer: (a) Parallelogram

Q2. The measure of each angle in the figure is: (a) 90 degrees (b) 120 degrees (c) 135 degrees (d) 180 degrees

Answer: (d) 180 degrees

Q3. The figure is an example of a quadrilateral that satisfies the: (a) Opposite sides are equal condition (b) Opposite angles are congruent condition (c) Diagonals bisect each other condition (d) None of the above

Answer: (b) Opposite angles are congruent condition

Q4. The sum of the measures of the exterior angles of the figure is: (a) 90 degrees (b) 180 degrees (c) 270 degrees (d) 360 degrees

Answer: (d) 360 degrees

Q5. The figure has rotational symmetry of: (a) Order 1 (b) Order 2 (c) Order 3 (d) Order 4

Answer: (a) Order 1

Hope the information shed above regarding Case Study and Passage Based Questions for Case Study Questions Class 9 Maths Chapter 8 Quadrilaterals with Answers Pdf free download has been useful to an extent. If you have any other queries about Case Study Questions Class 9 Maths Chapter 8 Quadrilaterals and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible.

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CBSE Case Study Questions for Class 9 Maths Quadrilaterals Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 9 Maths Quadrilaterals  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 9 Maths Quadrilaterals PDF

Checkout our case study questions for other chapters.

• Chapter 6 Lines and Angles Case Study Questions
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• Chapter 9 Areas of Parallelograms and Triangles Case Study Questions
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Case Study Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

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Here we are providing case study questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles. Students are suggested to solve the questions by themselves first and then check the answers. This will help students to check their grasp on this particular chapter Triangles.

Case Study Questions:

Questions 1:

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Quadrilaterals Case Study Questions (CSQ’s)

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• NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals
• NCERT Solutions

NCERT Solutions for Maths Class 9 Chapter 8 Quadrilaterals - Free PDF

Class 9 maths lays the foundation for practical math applications. Understanding problems is crucial, and Vedantu’s Class 9 Maths Chapter 8 Solutions greatly assist students. Regular practice enhances analytical skills, leading to improved exam scores. Proficient use of NCERT Solutions for Class 9 Maths Chapter 8 can be a significant advantage. Free PDF downloads make studying on Vedantu easy and enjoyable. For additional help, NCERT Solution for Class 9 Science is also available for download, aiding in scoring higher marks in exams.

Important Topics under NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT Class 9 Maths Chapter 8 syllabus is on Quadrilaterals. It is a very important chapter covered in class 9 and is divided into 2 exercise sections and 5 main topics. Here is a list of key topics covered in NCERT Class 9 Maths Chapter 8 Quadrilaterals. We suggest students focus on these topics to fully utilize the benefits of Vedantu's solutions for this chapter.

Introduction to Quadrilaterals

The Angle Sum Property of Quadrilaterals

Types of Quadrilaterals

Properties of Parallelogram

The Mid-point Theorem

We suggest students take advantage of our solutions to achieve good marks in exams. Going through these five topics will help in practicing the provided solutions effortlessly.

Exercises under NCERT Solutions for Maths Class 9 Chapter 8 Quadrilaterals

NCERT Solutions for Maths Class 9 Chapter 8, "Quadrilaterals" comprises two exercises with a total of 19 questions. Here's a detailed explanation of the types of questions included in each exercise:

Exercise 8.1:

This exercise consists of 12 questions that cover a wide range of concepts related to quadrilaterals. The questions require students to identify and recognize the properties of different types of quadrilaterals. Here are the different types of questions you can expect to find in this exercise:

Identification of Quadrilaterals: In this type of question, students are given a diagram of a quadrilateral and are asked to identify the type of quadrilateral, such as a square, rectangle, parallelogram, or trapezium.

Checking Properties: In these questions, students need to confirm a given characteristic of a quadrilateral. For instance, they might be asked to confirm that the opposite sides of a parallelogram are equal.

Application of Properties: In this type of question, students need to apply the properties of quadrilaterals to solve problems. For instance, they may be asked to find the perimeter or area of a given quadrilateral.

Exercise 8.2:

This section has seven questions that emphasize applying the properties of quadrilaterals. The questions are more intricate, demanding a thorough grasp of the concepts from the chapter. Here are the types of questions you can anticipate in this exercise:

Proving Properties: In these questions, students are asked to prove a given property of a quadrilateral using the properties they have learned in the chapter.

Applying Properties: Similar to Exercise 8.1, these questions need students to use quadrilateral properties to solve problems. However, the questions in this exercise are more intricate, demanding a deeper understanding of the concepts.

Construction of Quadrilaterals: In some questions, students are asked to construct a quadrilateral based on certain given conditions, such as the length of the sides or the angles of the quadrilateral.

Access NCERT Answers for Class 9 Mathematics Chapter 8 – Quadrilaterals

1. The angles of the quadrilateral are in the ratio \[3:5:9:13\].Find all the angles of the quadrilateral.

Given: The quadrilateral angles are in the ratio \[3:5:9:13\] .

To find: Angles of a quadrilateral

Consider the angle's common ratio is \[x\] .

Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.

The sum of a quadrilateral's inner angles is \[360^{\circ} \] ,

$\therefore 3 x+5 x+9 x+13 x=360^{\circ} $

$30 x=360^{\circ} $

$x=12^{\circ}$

As a result, the angles can be calculated as

$3 x=3 \times 12=36^{\circ} $

$5 x=5 \times 12=60^{\circ} $

$9 x=9 \times 12=108^{\circ} $

$13 x=13 \times 12=156^{\circ}$

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Given: Diagonals of the parallelogram are the same.

To prove: It is a rectangle.

Consider ABCD be the given parallelogram.

Now we need to show that ABCD is a rectangle, by proving that one of its interior angles is .

In \[\Delta ABC\] and \[\Delta DCB\] ,

AB = DC (side opposite to the parallelogram are equal)

BC = BC (in common)

AC = DB (Given)

\[\therefore \Delta ABC \cong \Delta DCB\] (By SSS Congruence rule)

\[ \Rightarrow \angle ABC{\text{ }} = \angle DCB\]

The sum of the measurements of angles on the same side of a transversal is known to be \[{180^o}\] .

Hence, ABCD is a rectangle because it is a parallelogram with a \[{90^o}\] inner angle.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Given: The diagonals of a quadrilateral bisect at right angles.

To prove: It is a rhombus.

Consider ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle

i.e. \[OA{\text{ }} = {\text{ }}OC,{\text{ }}OB{\text{ }} = {\text{ }}OD\] , and .

In order to prove ABCD a rhombus, we need to prove ABCD is the parallelogram and all sides of the ABCD are the same.

In \[\Delta AOD\] and \[\Delta COD\] ,

OA = OC (Diagonals of the parallelogram bisect each other)

\[\angle AOD{\text{ }} = \angle COD\] (Given)

OD = OD (Common)

\[\therefore \Delta AOD \cong \Delta COD\] (By SAS congruence rule)

\[\therefore AD{\text{ }} = {\text{ }}CD\] … (1)

Similarly, it can be proved that

\[AD{\text{ }} = {\text{ }}AB\] and \[CD{\text{ }} = {\text{ }}BC\] … (2)

From Equations (1) and (2),

\[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}AD\]

To conclude, ABCD is a parallelogram because the opposite sides of the quadrilateral ABCD are equal. ABCD is a rhombus because all of the sides of a parallelogram ABCD are equal.  

4. Show that the diagonals of a square are equal and bisect each other at right angles.

Given: A square is given.

To find: The diagonals of a square are the same and bisect each other at ${90^o}$

Consider ABCD be a square.

Consider the diagonals AC and BD intersect each other at a point O.

We must first show that the diagonals of a square are equal and bisect each other at right angles,

\[{\text{AC  =  BD, OA  =  OC, OB  =  OD}}\] , and .

\[AB{\text{ }} = {\text{ }}DC\] (Sides of the square are equal)

\[\angle ABC{\text{ }} = \angle DCB\] (All the interior angles are of the value \[{90^o}\] )

\[BC{\text{ }} = {\text{ }}CB\] (Common side)

\[\therefore \Delta ABC \cong \Delta DCB\] (By SAS congruency)

\[\therefore AC{\text{ }} = {\text{ }}DB\] (By CPCT)

Hence, the diagonals of a square are equal in length.

In \[\Delta AOB\] and \[\Delta COD\] ,

\[\angle AOB{\text{ }} = \angle COD\] (Vertically opposite angles)

\[\angle ABO{\text{ }} = \angle CDO\] (Alternate interior angles)

AB = CD (Sides of a square are always equal)

\[\therefore \Delta AOB \cong \Delta COD\] (By AAS congruence rule)

\[\therefore AO{\text{ }} = {\text{ }}CO\] and \[OB{\text{ }} = {\text{ }}OD\] (By CPCT)

As a result, the diagonals of a square are bisected.

In \[\Delta AOB\] and \[\Delta COB\] ,

Because we already established that diagonals intersect each other,

\[AO{\text{ }} = {\text{ }}CO\]

\[AB{\text{ }} = {\text{ }}CB\] (Sides of a square are equal)

\[BO{\text{ }} = {\text{ }}BO\] (Common)

\[\therefore \Delta AOB \cong \Delta COB\] (By SSS congruency)

\[\therefore \angle AOB{\text{ }} = \angle COB\] (By CPCT)

However,  (Linear pair)

As a result, the diagonals of a square are at right angles to each other.

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Given: The diagonals of a quadrilateral are equal and bisect at ${90^o}$

To find: It is a square

Consider the quadrilateral ABCD, in which the diagonals AC and BD cross at point O.

The diagonals of ABCD are equal and bisect each other at right angles, which is a given.

Therefore, \[AC{\text{ }} = {\text{ }}BD,{\text{ }}OA{\text{ }} = {\text{ }}OC,{\text{ }}OB{\text{ }} = {\text{ }}OD\] , and .

We need to prove that ABCD is a parallelogram, \[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}AD\] , and one of its interior angles being .

\[AO{\text{ }} = {\text{ }}CO\] (Diagonals bisect each other)

\[OB{\text{ }} = {\text{ }}OD\] (Diagonals bisect each other)

\[\therefore \Delta AOB \cong \Delta COD\] (SAS congruence rule)

$AB = CD$ (By CPCT) ... (1)

And, \[\angle OAB{\text{ }} = \angle OCD\] . (By CPCT)

However, for lines AB and CD, they are alternate interior angles, and alternate interior angles are equal only when the two lines are parallel.

\[AB{\text{ }}||{\text{ }}CD\] ... (2)

From (1) and (2), we obtain

ABCD is a parallelogram.

In \[\Delta AOD{\text{ , }}\Delta COD\] ,

\[\angle AOD{\text{ }} = \angle COD\] (Given that each is )

\[OD{\text{ }} = {\text{ }}OD\] (Common)

\[\therefore \Delta AOD \cong \Delta COD\] (SAS congruence rule)

\[\therefore AD{\text{ }} = {\text{ }}DC\] ... (3)

However, \[AD{\text{ }} = {\text{ }}BC\] and \[AB{\text{ }} = {\text{ }}CD\] (Opposite sides of the parallelogram ABCD)

\[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}DA\]

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In \[\Delta ADC\] and \[\Delta BCD\] ,

\[AD{\text{ }} = {\text{ }}BC\] (Already proved)

\[AC{\text{ }} = {\text{ }}BD\] (Given)

\[DC{\text{ }} = {\text{ }}CD\] (Common)

\[\therefore \Delta ADC \cong \Delta BCD\] (SSS Congruence rule)

\[\therefore \angle ADC{\text{ }} = \angle BCD\] (By CPCT)

However, \[\angle ADC{\text{ }} + \angle BCD{\text{ }} = {\text{ }}180^\circ \] (Co-interior angles)

\[\begin{array}{*{20}{l}} {\angle ADC{\text{ }} + \angle ADC{\text{ }} = {\text{ }}180^\circ } \\ {2\angle ADC{\text{ }} = {\text{ }}180^\circ } \\ {\therefore \angle ADC{\text{ }} = {\text{ }}90^\circ } \end{array}\]

A right angle is one of the quadrilateral ABCD's inner angles.

As a result, we can see that ABCD is a parallelogram, \[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}AD\] and one of its interior angles is .

Hence , ABCD is a square.  

6. Diagonal AC of a parallelogram ABCD is bisecting \[\angle A\](see the given figure). Show that

(i) It is bisecting \[\angle C\]also,

(ii) ABCD is a rhombus

Answer:  

Given: Diagonal AC of a parallelogram ABCD is bisecting \[\angle A\]

To find: (i) It is bisecting \[\angle C\] also,

 (i) ABCD is a parallelogram.

\[\angle DAC{\text{ }} = \angle BCA\] (Alternate interior angles) ... (1)

And, \[\angle BAC{\text{ }} = \angle DCA\] (Alternate interior angles) ... (2)

However, it is given that AC is bisecting  \[\angle A\] .

\[\angle DAC{\text{ }} = \angle BAC\] ... (3)

From Equations (1), (2), and (3), we obtain

\[\angle DAC{\text{ }} = \angle BCA{\text{ }} = \angle BAC{\text{ }} = \angle DCA\] ... (4)

\[\angle DCA{\text{ }} = \angle BCA\]

Hence, AC is bisecting \[\angle C\] .

(ii) From Equation (4), we obtain

\[\angle DAC{\text{ }} = \angle DCA\]

\[DA{\text{ }} = {\text{ }}DC\] (Side opposite to equal angles are equal)

However, \[DA{\text{ }} = {\text{ }}BC\] and \[AB{\text{ }} = {\text{ }}CD\] (Opposite sides of a parallelogram)

As a result, ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC is bisecting \[\angle A\] as well as \[\angle C\] and diagonal BD is bisecting \[\angle B\] as well as \[\angle D\].

Given: ABCD is a rhombus.

To find: Diagonal AC is bisecting \[\angle A\] as well as \[\angle C\] and diagonal BD is bisecting \[\angle B\] as well as \[\angle D\].

Let us now join AC.

In \[\Delta ABC\] ,

\[BC{\text{ }} = {\text{ }}AB\] (Sides of the rhombus are equal to each other)

\[\angle 1{\text{ }} = \angle 2\] (Angles opposite to equal sides of a triangle are equal)

However, \[\angle 1{\text{ }} = \angle 3\] (Parallel lines AB and CD have different interior angles.)

\[\angle 2{\text{ }} = \angle 3\]

Also, \[\angle 2{\text{ }} = \angle 4\] (Alternate interior angles for \[||\] lines BC and DA)

\[\angle 1{\text{ }} = \angle 4\]

Hence, AC bisects \[\angle A\] .

Also, it can be proved that BD is bisecting \[\angle B\] and \[\angle D\] as well.

8. ABCD is a rectangle in which diagonal AC bisects \[\angle A\] as well as \[\angle C\]. Show that:

(i) ABCD is a square

(ii) Diagonal BD bisects \[\angle B\] as well as \[\angle D\].

Given: ABCD is a rectangle where the diagonal AC bisects \[\angle A\] as well as \[\angle C\] .

To find: (i) ABCD is a square

              (ii) Diagonal BD bisects \[\angle B\] as well as \[\angle D\] .

It is given that ABCD is a rectangle. \[\angle A{\text{ }} = \angle C\]\[\]

$ \Rightarrow \dfrac{1}{2}\angle A = \dfrac{1}{2}\angle C$ (AC bisects \[\angle A\] and \[\angle C\] )

$ \Rightarrow \angle DAC = \dfrac{1}{2}\angle DCA$

CD = DA (Sides that are opposite to the equal angles are also equal)

Also, \[DA{\text{ }} = {\text{ }}BC\] and \[AB{\text{ }} = {\text{ }}CD\] (Opposite sides of the rectangle are same)

ABCD is a rectangle with equal sides on all sides.

Hence, ABCD is a square.

(ii) Let us now join BD.

In \[\Delta BCD\] ,

\[BC{\text{ }} = {\text{ }}CD\] (Sides of a square are equal to each other)

\[\angle CDB{\text{ }} = \angle CBD\] (Angles opposite to equal sides are equal)

However, \[\angle CDB{\text{ }} = \angle ABD\] (Alternate interior angles for \[AB{\text{ }}||{\text{ }}CD\] )

\[\angle CBD{\text{ }} = \angle ABD\]

BD bisects \[\angle B.\]

Also, \[\angle CBD{\text{ }} = \angle ADB\] (Alternate interior angles for \[BC{\text{ }}||{\text{ }}AD\] )

\[\angle CDB{\text{ }} = \angle ABD\]

BD bisects \[\angle D\] and \[\angle B\] .

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that:

(i) \[\Delta APD \cong \Delta CQB\]

(ii) \[AP{\text{ }} = {\text{ }}CQ\]

(iii) \[\Delta AQB \cong \Delta CPD\]

(iv) \[AQ{\text{ }} = {\text{ }}CP\]

(v) APCQ is a parallelogram

Given: A parallelogram is given.

To prove: (i) \[\Delta APD \cong \Delta CQB\]

                  (ii) \[AP{\text{ }} = {\text{ }}CQ\]

      (iii) \[\Delta AQB \cong \Delta CPD\]

      (iv) \[AQ{\text{ }} = {\text{ }}CP\]

      (v) APCQ is a parallelogram

(i) In \[\Delta APD\] and \[\Delta CQB\] ,

\[\angle ADP{\text{ }} = \angle CBQ\] (Alternate interior angles for \[BC{\text{ }}||{\text{ }}AD\] )

\[AD{\text{ }} = {\text{ }}CB\] (Opposite sides of the parallelogram ABCD)

\[DP{\text{ }} = {\text{ }}BQ\] (Given)

\[\therefore \Delta APD \cong \Delta CQB\] (Using SAS congruence rule)

(ii) As we had observed that \[\Delta APD \cong \Delta CQB\] ,

\[\therefore AP{\text{ }} = {\text{ }}CQ\] (CPCT)

(iii) In \[\Delta AQB\] and \[\Delta CPD\] ,

\[\angle ABQ{\text{ }} = \angle CDP\] (Alternate interior angles for \[AB{\text{ }}||{\text{ }}CD\] )

\[AB{\text{ }} = {\text{ }}CD\] (Opposite sides of parallelogram ABCD)

\[BQ{\text{ }} = {\text{ }}DP\] (Given)

\[\therefore \Delta AQB \cong \Delta CPD\] (Using SAS congruence rule)

(iv) Since we had observed that \[\Delta AQB \cong \Delta CPD\] ,

\[\therefore AQ{\text{ }} = {\text{ }}CP\] (CPCT)

(v) From the result obtained in (ii) and (iv),

\[AQ{\text{ }} = {\text{ }}CP\] and

\[AP{\text{ }} = {\text{ }}CQ\]

APCQ is a parallelogram because the opposite sides of the quadrilateral are equal.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that

(i) \[\Delta APB \cong \Delta CQD\]

(i) In \[\Delta APB\] and \[\Delta CQD\] ,

\[\angle APB{\text{ }} = \angle CQD\] (Each 90°)

\[AB{\text{ }} = {\text{ }}CD\] (The opposite sides of a parallelogram ABCD)

\[\angle ABP{\text{ }} = \angle CDQ\] (Alternate interior angles for \[AB{\text{ }}||{\text{ }}CD\] )

\[\therefore \Delta APB \cong \Delta CQD\] (By AAS congruency)

(ii) By using

\[\therefore \Delta APB \cong \Delta CQD\] , we obtain

\[AP{\text{ }} = {\text{ }}CQ\] (By CPCT)

11. In \[\Delta ABC\] and\[\Delta DEF\], \[AB{\text{ }} = {\text{ }}DE,{\text{ }}AB{\text{ }}||{\text{ }}DE,{\text{ }}BC{\text{ }} = {\text{ }}EF\] and \[BC{\text{ }}||{\text{ }}EF\]. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that

(i) Quadrilateral ABED is a parallelogram

(ii) Quadrilateral BEFC is a parallelogram

(iii) \[AD{\text{ }}||{\text{ }}CF\] and AD = CF

(iv)Quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) \[\Delta ABC \cong \Delta DEF\].

Given: \[\Delta ABC\] and \[\Delta DEF\] are two triangles.

To prove: Quadrilateral ABED is a parallelogram

(vi) \[\Delta ABC \cong \Delta DEF\] .

(i) It is given that \[AB{\text{ }} = {\text{ }}DE\] and \[AB{\text{ }}||{\text{ }}DE\] .

A parallelogram is formed when two opposite sides of a quadrilateral are equal and parallel to one other.

As a result, ABED is a parallelogram in quadrilateral form.

(ii) Again, BC = EF and \[BC{\text{ }}||{\text{ }}EF\]

Therefore, quadrilateral  BCEF  is a parallelogram.

(iii) As we had observed that ABED and BEFC are parallelograms, therefore

AD = BE and \[AD{\text{ }}||{\text{ }}BE\]

(Opposite sides of a parallelogram are equal and parallel)

And, BE = CF and \[BE{\text{ }}||{\text{ }}CF\]

\[\therefore AD{\text{ }} = {\text{ }}CF\] and \[AD{\text{ }}||{\text{ }}CF\]

(iv) We know it's a parallelogram since one set of opposing sides (AD and CF) of quadrilateral ACFD are equal and parallel to one other.

(v) Due to the fact that ACFD is a parallelogram, the pair of opposing sides will be equal and parallel to one another.

\[\therefore AC{\text{ }}||{\text{ }}DF\] and \[AC{\text{ }} = {\text{ }}DF\]

(vi) In \[\Delta ABC\] and \[\Delta DEF\] ,

\[AB{\text{ }} = {\text{ }}DE\] (Given)

\[BC{\text{ }} = {\text{ }}EF{\text{ }}\] (Given)

\[AC{\text{ }} = {\text{ }}DF\] (ACFD is a parallelogram)

\[\therefore \Delta ABC \cong \Delta DEF\] (By SSS congruence rule)

12. ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) \[\Delta ABC \cong \Delta BAD\]

(iv) diagonal AC = diagonal BD

(Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.)

Given: ABCD is a trapezium.

To find: (i) ∠A = ∠B

Let us extend AB by drawing a line through C, which is parallel to AD, intersecting AE at point

E. It is clear that AECD is a parallelogram.

(i) \[AD{\text{ }} = {\text{ }}CE\] (Opposite sides of parallelogram AECD)

However, \[AD{\text{ }} = {\text{ }}BC\] (Given)

Therefore, \[BC{\text{ }} = {\text{ }}CE\]

\[\angle CEB{\text{ }} = \angle CBE\] (Angle opposite to the equal sides are also equal)

Consideing parallel lines AD and CE. AE is the transversal line for them.

 (Angles on a same side of transversal)

 (Using the relation ∠CEB = ∠CBE) ... (1)

However,  (Linear pair angles) ... (2)

From Equations (1) and (2), we obtain

\[\angle A{\text{ }} = \angle B\]

 (ii) \[AB{\text{ }}||{\text{ }}CD\]

 (Angles on a same side of the transversal)

Also, \[\angle C{\text{ }} + \angle B{\text{ }} = {\text{ }}180^\circ \] (Angles on a same side of a transversal)

\[\therefore \angle A{\text{ }} + \angle D{\text{ }} = \angle C{\text{ }} + \angle B\]

However, \[\angle A{\text{ }} = \angle B\] (Using the result obtained in (i))

\[\therefore \angle C{\text{ }} = \angle D\]

 (iii) In \[\Delta ABC\] and \[\Delta BAD\] ,

\[AB{\text{ }} = {\text{ }}BA\] (Common side)

\[BC{\text{ }} = {\text{ }}AD\] (Given)

\[\angle B{\text{ }} = \angle A\] (Proved before)

\[\therefore \Delta ABC \cong \Delta BAD\] (SAS congruence rule)

(iv) We had seen that,

\[\Delta ABC \cong \Delta BAD\]

\[\therefore AC{\text{ }} = {\text{ }}BD\] (By CPCT)

Exercise (8.2)

2. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:

(i) \[SR{\text{ }}||{\text{ }}AC\] and \[SR = \dfrac{1}{2}\;AC\]

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Given: ABCD is a quadrilateral

To prove: (i) \[SR{\text{ }}||{\text{ }}AC\] and \[SR = \dfrac{1}{2}\;AC\]

     (ii) PQ = SR

     (iii) PQRS is a parallelogram.

(i) In \[\Delta ADC\] , S and R are the mid-points of sides AD and CD respectively.

In a triangle, the line segment connecting the midpoints of any two sides is parallel to and half of the third side.

\[\therefore SR{\text{ }}||{\text{ }}AC\] and \[SR{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\] ... (1)

(ii) In ∆ABC, P and Q are mid-points of sides AB and BC respectively. Therefore, by using midpoint theorem,

\[PQ{\text{ }}||{\text{ }}AC\] and \[PQ{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\] ... (2)

Using Equations (1) and (2), we obtain

\[PQ{\text{ }}||{\text{ }}SR\] and \[PQ{\text{ }} = {\text{ }}\dfrac{1}{2}SR\] ... (3)

\[\therefore PQ{\text{ }} = {\text{ }}SR\]

 (iii) From Equation (3), we obtained

\[PQ{\text{ }}||{\text{ }}SR\] and  \[PQ{\text{ }} = {\text{ }}SR\]

Clearly, one pair of quadrilateral PQRS opposing sides is parallel and equal.

PQRS is thus a parallelogram.

3. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Given: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

To find: Quadrilateral PQRS is a rectangle

In \[\Delta ABC\] , P and Q are the mid-points of sides AB and BC respectively.

\[PQ{\text{ }}||{\text{ }}AC{\text{ , }}PQ{\text{ }} = {\text{ }}\dfrac{1}{2}AC\] (Using mid-point theorem) ... (1)

In \[\Delta ADC\] ,

R and S are the mid-points of CD and AD respectively.

\[RS{\text{ }}||{\text{ }}AC{\text{ , }}RS{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\] (Using mid-point theorem) ... (2)

\[PQ{\text{ }}||{\text{ }}RS\] and \[PQ{\text{ }} = {\text{ }}RS\]

It is a parallelogram because one pair of opposing sides of quadrilateral PQRS is equal and parallel to each other. At position O, the diagonals of rhombus ABCD should cross.

In quadrilateral OMQN,

\[MQ{\text{ }}\left| {\left| {{\text{ }}ON{\text{ }}({\text{ }}PQ{\text{ }}} \right|} \right|{\text{ }}AC)\]

\[QN{\text{ }}\left| {\left| {{\text{ }}OM{\text{ }}({\text{ }}QR{\text{ }}} \right|} \right|{\text{ }}BD)\]

Hence , OMQN is a parallelogram.

\[\begin{array}{*{20}{l}} {\therefore \angle MQN{\text{ }} = \angle NOM} \\ {\therefore \angle PQR{\text{ }} = \angle NOM} \end{array}\]

Since,  \[\angle NOM{\text{ }} = {\text{ }}90^\circ \] (Diagonals of the rhombus are perpendicular to each other)

\[\therefore \angle PQR{\text{ }} = {\text{ }}90^\circ \]

Clearly, PQRS is a parallelogram having one of its interior angles as .

So , PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Given: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

To prove: The quadrilateral PQRS is a rhombus.

Let us join AC and BD.

P and Q are the mid-points of AB and BC respectively.

\[\therefore PQ{\text{ }}||{\text{ }}AC\] and \[PQ{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\] (Mid-point theorem) ... (1)

Similarly in \[\Delta ADC\] ,

\[SR{\text{ }}||{\text{ }}AC{\text{ , }}SR{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\] (Mid-point theorem) ... (2)

Clearly, \[PQ{\text{ }}||{\text{ }}SR\] and \[PQ{\text{ }} = {\text{ }}SR\]

It is a parallelogram because one pair of opposing sides of quadrilateral PQRS is equal and parallel to each other.

\[\therefore PS{\text{ }}||{\text{ }}QR{\text{ }},{\text{ }}PS{\text{ }} = {\text{ }}QR\] (Opposite sides of parallelogram) ... (3)

In \[\Delta BCD\] , Q and R are the mid-points of side BC and CD respectively.

\[\therefore QR{\text{ }}||{\text{ }}BD{\text{ , }}QR{\text{ }} = {\text{ }}\dfrac{1}{2}BD\] (Mid-point theorem) ... (4)

Also, the diagonals of a rectangle are equal.

\[\therefore AC{\text{ }} = {\text{ }}BD\] …(5)

By using Equations (1), (2), (3), (4), and (5), we obtain

\[PQ{\text{ }} = {\text{ }}QR{\text{ }} = {\text{ }}SR{\text{ }} = {\text{ }}PS\]

So , PQRS is a rhombus

4. ABCD is a trapezium in which \[AB{\text{ }}||{\text{ }}DC\], BD is a diagonal and E is the mid - point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC.

Given: ABCD is a trapezium in which \[AB{\text{ }}||{\text{ }}DC\] , BD is a diagonal and E is the mid - point of AD. A line is drawn through E parallel to AB intersecting BC at F.

To prove: F is the mid-point of BC.

Let EF intersect DB at G.

We know that a line traced through the mid-point of any side of a triangle and parallel to another side bisects the third side by the reverse of the mid-point theorem.

In \[\Delta ABD\] ,

\[EF{\text{ }}||{\text{ }}AB\] and E is the mid-point of AD.

Hence , G will be the mid-point of DB.

As \[EF{\text{ }}\left| {\left| {{\text{ }}AB{\text{ , }}AB{\text{ }}} \right|} \right|{\text{ }}CD\] ,

\[\therefore EF{\text{ }}||{\text{ }}CD\] (Two lines parallel to the same line are parallel)

In \[\Delta BCD\] , \[GF{\text{ }}||{\text{ }}CD\] and G is the mid-point of line BD. So , by using converse of mid-point

theorem, F is the mid-point of BC.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively To prove: The line segments AF and EC trisect the diagonal BD.

\[AB{\text{ }}||{\text{ }}CD\]

And hence, \[AE{\text{ }}||{\text{ }}FC\]

Again, AB = CD (Opposite sides of parallelogram ABCD)

\[\dfrac{1}{2}AB{\text{ }} = {\text{ }}\dfrac{1}{2}CD\]

\[AE{\text{ }} = {\text{ }}FC\] (E and F are mid-points of side AB and CD)

In quadrilateral AECF, one pair of the opposite sides (AE and CF) is parallel and same to each other. So , AECF is a parallelogram.

\[\therefore AF{\text{ }}||{\text{ }}EC\] (Opposite sides of a parallelogram)

In \[\Delta DQC\] , F is the mid-point of side DC and \[FP{\text{ }}||{\text{ }}CQ\] (as \[AF{\text{ }}||{\text{ }}EC\] ). So , by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.

\[\therefore DP{\text{ }} = {\text{ }}PQ\] ... (1)

Similarly, in \[\Delta APB\] , E is the mid-point of side AB and \[EQ{\text{ }}||{\text{ }}AP\] (as \[AF{\text{ }}||{\text{ }}EC\] ).

As a result, the reverse of the mid-point theorem may be used to say that Q is the mid-point of PB.

\[\therefore PQ{\text{ }} = {\text{ }}QB\] ... (2)

\[DP{\text{ }} = {\text{ }}PQ{\text{ }} = {\text{ }}BQ\]

Hence, the line segments AF and EC trisect the diagonal BD.

6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

To find: The line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other

Let ABCD be a quadrilateral in which the midpoints of sides AB, BC, CD, and DA are P, Q, R, and S, respectively. PQ, QR, RS, SP, and BD are all joined

S and P are the mid-points of AD and AB, respectively, in \[\Delta ABD\] . As a result, using the mid-point theorem, it can be shown that

\[SP{\text{ }}||{\text{ }}BD{\text{ , }}SP{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}BD\] ... (1)

Similarly in \[\Delta BCD\] ,

\[QR{\text{ }}||{\text{ }}BD{\text{ , }}QR{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}BD\] ... (2)

From the Equations (1) and (2), we obtain

\[SP{\text{ }}||{\text{ }}QR{\text{ , }}SP{\text{ }} = {\text{ }}QR\]

One set of opposing sides of quadrilateral SPQR is equal and parallel to each other. SPQR is a parallelogram as a result.

The diagonals of a parallelogram are known to bisect each other.

As a result, PR and QS cut each other in half.

7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC

(ii) MD $ \bot $ AC

(iii) \[CM{\text{ }} = {\text{ }}MA{\text{ }} = \dfrac{1}{2}AB\]

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To prove: (i) D is the mid-point of AC

(i) In \[\Delta ABC\] ,

It is given that M is the mid-point of AB and \[MD{\text{ }}||{\text{ }}BC\] .

Therefore, D is the mid-point of AC. (Converse of the mid-point theorem)

(ii) As \[DM{\text{ }}||{\text{ }}CB\] and AC is a transversal line for them, therefore,

 (Co-interior angles)

(iii) Join MC.

In \[\Delta AMD\] and \[\Delta CMD\] ,

\[AD{\text{ }} = {\text{ }}CD\] (D is the mid-point of side AC)

\[\angle ADM{\text{ }} = \angle CDM\] (Each )

DM = DM (Common)

\[\therefore \Delta AMD \cong \Delta CMD\] (By SAS congruence rule)

Therefore, \[AM{\text{ }} = {\text{ }}CM\] (By CPCT)

However, \[{\text{ }}AM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\] (M is mid-point of AB)

Therefore, it is said that

\[CM{\text{ }} = {\text{ }}AM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\]

NCERT Solutions for Maths Class 9 Chapter 8 Quadrilateral - PDF Download

NCERT Solutions Class 9 Chapter 8 provides students with tricks and techniques to solve the problem in abstract and easier methods, helping the students master the skill of critical thinking and problem-solving at their early stages. An established support system like Vedantu's Free Download NCERT Solutions Class 9 Maths at their early stage of education strengthens their reasoning and logical abilities paving the way to great heights in the field of Mathematics. 

For upcoming exams, you can choose Chapter 8 - Quadrilaterals NCERT Solutions for Class 9 Maths in PDF. Additionally, you can access solutions for all the math chapters below.

NCERT Solutions for Class 9 Maths

Chapter 1 - Number System

Chapter 2 - Polynomials  

Chapter 3 - Coordinate Geometry

Chapter 4 - Linear Equations in Two Variables

Chapter 5 - Introductions to Euclids Geometry

Chapter 6 - Lines and Angles

Chapter 7 - Triangles

Chapter 8 - Quadrilaterals

Chapter 9 - Areas of Parallelogram and Triangles

Chapter 10 - Circles

Chapter 11 - Constructions

Chapter 12 - Herons formula

Chapter 13 - Surface area and Volumes

Chapter 14 - Statistics

Chapter 15 - Probability

NCERT Class 9 Maths Chapter 8 Quadrilaterals

In NCERT Class 9 Maths Chapter 8, you'll find a concise explanation of special quadrilaterals like Parallelogram, Rhombus, Rectangle, Square, and Trapezium. Vedantu's solutions delve into crucial basics, covering angles and diagonal relations for each quadrilateral. Clear understanding from NCERT Solutions Class 9 Quadrilaterals aids students in mastering these topics. 

The Class 9 Maths Chapter 8 concentrates on sections of the Chapter Quadrilateral. Section 1 is an introduction to the quadrilateral, and some of its general properties are introduced. Section 2 is the introduction to types of quadrilateral and their properties. Section 3 covers the properties of a parallelogram whose significance is discussed elaborately in Maths NCERT Solutions Class 9 Chapter 8. Section 4 includes mainly the Mid-point theorem and its wide range of applications.

Exercise 8.1

A figure formed by joining four non-collinear points is called a quadrilateral. In other words, it is a four-sided polygon. It has four sides and four vertices. By joining the opposite vertices of the quadrilateral, we obtain a diagonal.

Angle Sum Property of Quadrilateral

The sum of the angles of a quadrilateral is 360 degrees. Chapter 8 Maths Class 9 has a crystal clear explanation of the angle sum property of a quadrilateral through a well-expounded problem. Chapter 8 Class 9 Maths NCERT Solutions registers these concepts strong in young minds.

Types of Quadrilaterals.

The parallelism of sides plays a vital role in the determination of the properties of the Quadrilaterals. As the knowledge application of concepts is the pillar of good scores in Maths, Maths NCERT Solutions Class 9 Chapter 8 is the best stop. If one pair of opposite sides are parallel, it is called a Trapezium. If both the pairs of opposite sides are parallel, then it is called a parallelogram. A parallelogram in which all the sides are equal is called Rhombus. A parallelogram in which all angles are equal and measure 90 degrees is called a rectangle. A Rectangle whose all sides are equal is called a square. A kite is a Quadrilateral whose pairs of adjacent sides are equal.

The properties of the parallelogram have great application when it comes to solving Polygon problems. NCERT Solutions Class 9 Chapter 8 explains the properties of parallelogram elaborately through the illustrated issues. 

Some properties of parallelograms are listed and are proficiently used in Chapter 8 Class 9 Maths NCERT Solutions. The opposite sides of the parallelogram are equal and parallel. The opposite angles are equal. The properties concerning the diagonals of the parallelogram have high significance that NCERT Class 9 Maths Chapter 8 applies them in most sums. The diagonals form congruent triangles with the sides of the parallelogram. The diagonals bisect each other. x The diagonals are angle bisectors of the vertices.

The Experts of Vedantu emphasize the importance of the properties of a parallelogram, as it being the fundamental shape of other quadrilaterals like Rectangle, Rhombus, and Square. Also, these shapes have higher usage in daily life than any other Quadrilateral. Maths NCERT Class 9 Chapter 8 also uses all the properties of a parallelogram in certain proofs and solutions of the problems in the most efficient manner.

Exercise 8.2

The line segment joining the mid-points of two sides of a triangle is parallel to the third side. Conversely, The line drawn through the midpoint of one side of a triangle, parallel to another side bisects the third side. NCERT Solutions for Class 9 Maths Chapter 8 provides the derailed solutions for all the problems based on mid-point theorems. The type of problems in which the line segments joining the midpoints of sides of quadrilaterals in various combinations is dealt with in detail. It also includes midpoints intersecting the diagonals of trapeziums and some of their extended versions. Added to it are right-angled triangles and the application of the mid-point theorem. NCERT Solutions for Class 9 Chapter 8 breaks down all these involuted problems simple to the ease of students.

Why are NCERT Solutions for Class 9 Math Chapter 8 - Quadrilaterals

NCERT Solutions for Math Class 9 Chapter 8 “Quadrilaterals” is the finest material to understand the topics in the best way.

The material carries all the information in detail and pointwise.

All the cases of the material are solved in the simplest way which explains the term clearly.

NCERT Solutions Class 9 Math Chapter 8  “Quadrilaterals” provides all the main topics underlined so that the student can focus on it.

Class 9 Maths Chapter 8 is an important chapter that lays the foundation for future mathematics. Vedantu's Class 9 Maths Chapter 8 Solutions is a comprehensive and informative resource that will help students to understand the concepts, solve problems, and improve their analytical skills.

Success in exams requires regular practice. Vedantu's Class 9 Maths Chapter 8 Solutions offers an extensive set of practice questions along with solutions, aiding students in thorough exam preparation.

Students can also download a free PDF of Vedantu's Class 9 Maths Chapter 8 Solutions for easy access and offline use.

FAQs on NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

1. What is the relation between square, Rectangle, and Rhombus?

A Square is a Rectangle and a Rhombus. But Rhombus and Rectangle are not Squares.

2. What is the relation between a trapezium and a parallelogram?

A parallelogram is a trapezium but a trapezium is not a parallelogram. 

3. How to avoid silly mistakes?

Paying attention to the theorems and their usage in the notes provided would minimize the silly mistakes. A clear understanding of the steps involved and practice is the key. Going through the  NCERT Solutions for Class 9 Chapter 8 would clear all your dangling doubts and would learn a lot of alternative steps.

4. How to increase our score in CBSE examinations?

CBSE test papers mainly test the understanding of the students. So a clear explanation for every step would help you score more. As the Quadrilateral chapter involves long steps of proof, the reason for every conclusion has to be stated neatly and clearly. Presentation in an organized manner can give a further push to your grades.

5. What is the theorem of Chapter 8 quadrilateral?

Theorem 8.2 can be stated as given below : If a quadrilateral is a parallelogram, then each pair of its opposite sides is equal. So its converse is : Theorem 8.3 : If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

6. What is the formula of quadrilateral ABCD?

The area of the quadrilateral ABCD = Sum of areas of ΔBCD and ΔABD. Thus, the area of the quadrilateral ABCD = (1/2) × d × h1 + (1/2) × d × h2 = (1/2) × d × (h2 + h2 ).

NCERT Solutions for Class 9

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CBSE Class 9 Mathematics Case Study Questions

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If you’re looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter. With over 10,000 study notes, solved sample papers and practice questions, it’s got everything you need to ace your exams. Plus, it’s updated regularly to keep you aligned with the latest CBSE syllabus . So why wait? Start your journey to success with myCBSEguide today!

Significance of Mathematics in Class 9

Mathematics is an important subject for students of all ages. It helps students to develop problem-solving and critical-thinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.

CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .

Case studies in Class 9 Mathematics

A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.

Example of Case study questions in Class 9 Mathematics

The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on real-life scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.

The following are some examples of case study questions from Class 9 Mathematics:

Class 9 Mathematics Case study question 1

There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of   XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

• Now he told Raju to draw another line CD as in the figure
• The teacher told Ajay to mark  ∠ AOD  as 2z
• Suraj was told to mark  ∠ AOC as 4y
• Clive Made and angle  ∠ COE = 60°
• Peter marked  ∠ BOE and  ∠ BOD as y and x respectively

Now answer the following questions:

• 2y + z = 90°
• 2y + z = 180°
• 4y + 2z = 120°
• (a) 2y + z = 90°

Class 9 Mathematics Case study question 3

• (a) 31.6 m²
• (c) 513.3 m³
• (b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.

Class 9 Mathematics Curriculum at Glance

At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve real-world problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.

The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.

CBSE Class 9 Mathematics (Code No. 041)

Class 9 Mathematics question paper design

The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problem-solving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.

QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)

Mycbseguide: blessing in disguise.

Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.

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14 thoughts on “CBSE Class 9 Mathematics Case Study Questions”

This method is not easy for me

aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10-?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2-?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3-?2 4+ ?2 2+ ?2 d) 2-?3 the identity applied to solve (?10-?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (a-b)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b

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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 8 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

Class 9 Maths Chapter 8 Quadrilaterals NCERT Solutions

Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.

NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.1

NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2

NCERT Solutions for Class 9 Maths Chapter 8 – Topic Discussion

Below we have listed the topics that have been discussed in this chapter.

• The Mid-point Theorem
• Another Condition for a Quadrilateral to be a Parallelogram
• Properties of a Parallelogram
• Angle Sum Property of a Quadrilateral
• Types of Quadrilaterals

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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8: Quadrilaterals are an educational aid for students that help them solve and learn simple and difficult tasks. It includes a complete set of questions organized with advanced level of difficulty, which provide students ample opportunity to apply combinations and skills. Get free NCERT Solutions for Class 9 Maths devised according to the latest update on term-wise CBSE Syllabus for 2021-22. These NCERT Solutions will help the students to understand the concept of Quadrilaterals mainly basics, properties and some important theorems. These solutions can not only help students to clear their doubts but also to prepare more efficiently for the second term examination.

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NCERT Solutions for Class 9 Maths Chapter 8 explains Angle Sum Property of a Quadrilateral, Types of Quadrilaterals and Mid-Point theorem.

Topics covered under this chapter help the students to understand the basics of a geometrical figure named as a quadrilateral, its properties and various important theorems. This chapter of NCERT Solutions for Class 9 Maths is extremely crucial as the formulas and theorem results are extensively used in several other maths concepts in higher grades.

Chapter 8 Quadrilaterals is included in the second term CBSE Syllabus 2021-22 and is a part of Unit-Geometry which holds 28 marks of weightage in the term exams of CBSE Class 9 Maths. Two or three questions are asked every year in the second term examination from this chapter.

NCERT Solutions For Class 9 Maths Chapter 8 Exercises:

Get detailed solutions for all the questions listed under the below exercises:

Exercise 8.1 Solutions (12 Questions)

Exercise 8.2 Solutions (7 Questions) NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals NCERT Solutions for Class 9 Maths Chapter 8 is about Theorems and properties on Quadrilaterals. They are accompanied with explanatory figures and solved examples, which are explained in a comprehensive way. The main topics covered in this chapter include:

Key Features of Class 9 Maths Chapter 8 Quadrilaterals

• NCERT solutions have been prepared in a logical and simple language.
• Pictorial presentation of all the questions.
• Emphasizes that learning should be activity-based and knowledge-driven.
• The solutions are explained in a well-organised way.
• Step by step approach used to solve all NCERT questions.

Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 8

What are the main topics covered in NCERT Solutions for Class 9 Maths Chapter 8?

The main topics covered in NCERT Solutions for Class 9 Maths Chapter 8 are given below:

8.1 Introduction of quadrilaterals

8.2 Angle Sum Property of a Quadrilateral

8.3 Types of Quadrilaterals

8.4 Properties of a Parallelogram

8.5 Another Condition for a Quadrilateral to be a Parallelogram

8.6 The Mid-point Theorem

8.7 Summary

Question: How many questions are there in NCERT Solutions for Class 9 Maths Chapter 8?

Answer: NCERT Solutions for Class 9 Maths Chapter 8 contains two exercises. The first exercise has 12 questions and the second exercise has 7 questions. Practising these exercises help you in scoring high in second term exams and also help to ease the subject. These solutions are explained by subject matter experts to help you in clearing all the doubts.

Question: What is the meaning of quadrilaterals according to NCERT Solutions for Class 9 Maths Chapter 8?

Answer: According to NCERT Solutions for Class 9 Maths Chapter 8 quadrilateral is a plane figure that has four sides or edges, and also has four corners or vertices. Quadrilaterals will typically be of standard shapes with four sides like rectangle, square, trapezoid, and kite or irregular and uncharacterized shapes.

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Test: Quadrilaterals- Case Based Type Questions- 1 - Class 9 MCQ

10 questions mcq test - test: quadrilaterals- case based type questions- 1, harish makes a poster in the shape of a parallelogram on the topic save electricity for an inter school competition as shown in the follow figure. q. if ab = (2y – 3) and cd = 5 cm then what is the value of y.

⇒ 2y – 3 = 5

Harish makes a poster in the shape of a parallelogram on the topic SAVE ELECTRICITY for an inter school competition as shown in the follow figure. Q. If ∠B = (2y)° and ∠D = (3y – 6)°, then find the value of y.

(opposite angles of a parallelogram are equal)

⇒ 2y = 3y – 6

⇒ 2y – 3y = – 6

⇒ – y = – 6

Harish makes a poster in the shape of a parallelogram on the topic SAVE ELECTRICITY for an inter school competition as shown in the follow figure. Q. If ∠ A = (4x + 3)° and ∠D = (5x – 3)°, then find the measure of ∠B

∠A + ∠D = 180°

(adjacent angles of a quadrilateral are equal)

(4x + 3)° + (5x – 3)° = 180°

∠D = (5x – 3)° = 97°

Thus, ∠B = 97°

Harish makes a poster in the shape of a parallelogram on the topic SAVE ELECTRICITY for an inter school competition as shown in the follow figure.

Q. If ∠A = (2x – 3)° and ∠C = (4y + 2)°, then find how x and y relate.

• A. x = 2y + 3

• D. x = y – 7

⇒ 2x – 3 = 4y + 2

⇒ 2x = 4y + 5

Q. Which mathematical concept is used here?

• A. Co-ordinate geometry
• B. Surface area and volume
• C. Properties of a parallelogram
• D. Probability

If one pair of opposite sides of a quadrilateral is equal and parallel, then the quadrilateral is a parallelogram.

During maths lab activity, teacher gives four sticks of lengths 6 cm, 6 cm, 4 cm and 4 cm to each student to make different types of quadrilateral.

She asks following questions from the students:

Q. A student formed a rectangle with these sticks. What is the length of the diagonal of the rectangle formed by the student?

6 2 + 4 2 = l 2

36 + 16 = l 2

Q. Write the name of quadrilateral that can be formed with these sticks.

• A. Kite, rectangle, rhombus
• B. Parallelogram, rectangle , trapezium
• C. Kite, rectangle, parallelogram
• D. Square, rectangle, kite

Q. How many types of quadrilaterals can be possible?

Q. Which statement is incorrect ?

• A. Opposite sides of a parallelogram are equal
• B. A kite is not a parallelogram
• C. Diagonals of a parallelogram bisect each other
• D. A trapezium is a parallelogram.

Q. A diagonal of a parallelogram divides it into two _______ triangles.

• B. Congruent
• C. Equilateral
• D. Right angled

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Important Questions for Quadrilaterals- Case Based Type Questions- 1

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• Quadrilaterals Class 9

Quadrilaterals Class 9 Notes - Chapter 8

Cbse class 9 maths quadrilaterals notes:- download pdf here.

Get the complete notes on quadrilaterals class 9 here. A quadrilateral is a shape which has four sides. In this article, we are going to discuss the different types of quadrilaterals such as square, rectangle, parallelogram properties with proofs.

To know more about Parallelogram, visit here .

Parallelogram: Opposite sides of a parallelogram are equal

In Δ A B C   a n d   Δ C D A

A C = A C [Common / transversal]

∠ B C A = ∠ D A C [alternate angles]

∠ B A C = ∠ D C A [alternate angles]

Opposite angles in a parallelogram are equal

In parallelogram A B C D

A B ‖ C D ; and A C is the transversal

Hence, ∠ 1 = ∠ 3 ….(1) (alternate interior angles)

B C ‖ D A ; and A C is the transversal

Hence, ∠ 2 = ∠ 4 ….(2) (alternate interior angles)

Adding (1) and (2)

∠ 1 + ∠ 2 = ∠ 3 + ∠ 4

∠ B A D = ∠ B C D

Similarly, ∠ A D C = ∠ A B C

Properties of diagonal of a parallelogram

Diagonals of a parallelogram bisect each other.

In  Δ A O B   a n d   Δ C O D ,

∠ 3 = ∠ 5 [alternate interior angles]

∠ 1 = ∠ 2   [vertically opposite angles]

A B = C D [opp. Sides of parallelogram]

Δ A O B ≅ Δ C O D [AAS rule]

Hence, proved

Conversely, If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Diagonal of a parallelogram divides it into two congruent triangles.

In Δ A B C and Δ C D A ,

A B = C D [Opposite sides of parallelogram]

B C = A D [Opposite sides of parallelogram]

A C = A C [Common side]

Hence, proved.

For more information on Properties Of Parallelogram, watch the below video

Diagonals of a rhombus bisect each other at right angles

Diagonals of a rhombus bisect each – other at right angles

In Δ A O D   a n d   Δ C O D ,

O A = O C [Diagonals of parallelogram bisect each other]

O D = O D [Common side]

A D = C D [Adjacent sides of a rhombus]

Δ A O D ≅ Δ C O D [SSS rule]

∠ A O D = ∠ D O C [C.P.C.T]

Hence, ∠ A O D = ∠ D O C = 90

Hence proved.

For more information on Properties Of Rhombus, watch the below video

Diagonals of a rectangle bisect each other and are equal

Rectangle ABCD

In Δ A B C   a n d   Δ B A D ,

A B = B A [Common side]

B C = A D [Opposite sides of a rectangle]

∠ A B C = ∠ B A D [Each = 90 0   ∵ ABCD is a Rectangle]

Δ A B C ≅ Δ B A D [SAS rule]

Consider  Δ O A D   a n d   Δ O C B ,

A D = C B   [Opposite sides of a rectangle]

∠ O A D = ∠ O C B   [ ∵ AD||BC and transversal AC intersects them]

∠ O D A = ∠ O B C   [ ∵ AD||BC and transversal BD intersects them]

Δ O A D ≅ Δ O C B [ASA rule]

Similarly we can prove O B = O D

For more information on Properties Of Rectangle, watch the below video

Diagonals of a square bisect each other at right angles and are equal

Square ABCD

B C = A D [Opposite sides of a Square]

∠ A B C = ∠ B A D [Each = 90 0   ∵ ABCD is a Square]

A D = C B   [Opposite sides of a Square]

In Δ O B A   a n d   Δ O D A ,

O B = O D    [ proved above]

B A = D A   [Sides of a Square]

O A = O A   [ Common side]

Δ O B A ≅ Δ O D A ,   [ SSS rule]

But,  ∠ A O B + ∠ A O D = 180 0   [ Linear pair]

∴ ∠ A O B = ∠ A O D = 90 0

Important results related to parallelograms

Parallelogram ABCD

Opposite sides of a parallelogram are parallel and equal .

A B | | C D , A D | | B C , A B = C D , A D = B C

Opposite angles of a parallelogram are equal adjacent angels are supplementary .

∠ A = ∠ C , ∠ B = ∠ D ,

∠ A + ∠ B = 180 0 , ∠ B + ∠ C = 180 0 , ∠ C + ∠ D = 180 0 , ∠ D + ∠ A = 180 0

A diagonal of parallelogram divides it into two congruent triangles .

Δ A B C ≅ Δ C D A [With respect to AC as diagonal]

The diagonals of a parallelogram bisect each other.

A E = C E , B E = D E

∠ 1 = ∠ 5 (alternate interior angles)

∠ 2 = ∠ 6 (alternate interior angles)

∠ 3 = ∠ 7 (alternate interior angles)

∠ 4 = ∠ 8 (alternate interior angles)

∠ 9 = ∠ 11 (vertically opp. angles)

∠ 10 = ∠ 12 (vertically opp. angles)

The Mid-Point Theorem

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of the third side

In Δ A B C ,  E – the midpoint of A B ;  F – the midpoint of A C

Construction : Produce E F to D such that E F = D F .

In Δ A E F and Δ C D F ,

A F = C F   [ F is the midpoint of AC]

E F = D F [ Construction]

∠ E A F = ∠ D C F ….(1)

D C = E A = E B   [ E is the midpoint of AB]

D C ‖ E A ‖ A B [Since, (1), alternate interior angles]

So E B C D is a parallelogram

Therefore, B C = E D and B C ‖ E D

We have, E F = 1 2 B C and E F | | B C

To know more about Mid-Point Theorem, visit here .

Introduction to Quadrilaterals

For more information on quadrilaterals, watch the below videos.

Quadrilaterals

Any four points in a plane, of which three are non-collinear are joined in order results into a four-sided closed figure called ‘quadrilateral’

Quadrilateral

For more information on Elementary Shapes – Quadrilaterals, watch the below video

To know more about Quadrilaterals, visit here .

Angle sum property of a quadrilateral

Angle sum property – Sum of angles in a quadrilateral is 360

In △ A D C ,

∠ 1 + ∠ 2 + ∠ 4 = 180 (Angle sum property of triangle)…………….(1)

In △ A B C ,

∠ 3 + ∠ 5 + ∠ 6 = 180 (Angle sum property of triangle)………………(2)

∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 = 360

I.e, ∠ A + ∠ B + ∠ C + ∠ D = 360

Hence proved

Types of Quadrilaterals

A trapezium is a quadrilateral with any one pair of opposite sides parallel .

P Q R S is a trapezium in which P Q | | R S

Parallelogram

A parallelogram is a quadrilateral, with both pair of opposite sides parallel and equal . In a parallelogram, diagonals bisect each other.

Parallelogram ABCD in which A B | | C D , B C | | A D   and A B = C D , B C = A D

A rhombus is a parallelogram with all sides equal.  In a rhombus, diagonals bisect each other perpendicularly

Rhombus ABCD

A rhombus A B C D in which A B = B C = C D = A D and   A C ⊥ B D

A rectangle is a parallelogram with all angles as right angles .

A rectangle A B C D in which, ∠ A = ∠ B = ∠ C = ∠ D = 90 0

A square is a special case of a parallelogram with all angles as right angles and all sides equal.

A   square A B C D in which ∠ A = ∠ B = ∠ C = ∠ D = 90 0 and A B = B C = C D = A D

A kite is a quadrilateral with adjacent sides equal .

A kite A B C D in which A B = B C and A D = C D

For more information on Types Of Quadrilaterals, watch the below videos

To know more about Different Types of Quadrilateral, visit here .

Venn diagram for different types of quadrilaterals

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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Unit 8: Quadrilaterals

Quadrilaterals 8.1.

• Proof: Opposite sides of a parallelogram (Opens a modal)
• Proof: Opposite angles of a parallelogram (Opens a modal)
• Proof: Diagonals of a parallelogram (Opens a modal)
• Side and angle properties of a parallelogram (level 1) Get 3 of 4 questions to level up!
• Quadrilaterals 8.1 Get 5 of 7 questions to level up!

Quadrilaterals 8.2

• Proof: Rhombus diagonals are perpendicular bisectors (Opens a modal)
• Proof: The diagonals of a kite are perpendicular (Opens a modal)
• Quadrilaterals 8.2 Get 5 of 7 questions to level up!

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Ncert solutions for class 9 maths chapter 8 quadrilaterals| pdf download.

• Exercise 8.1 Chapter 8 Class 9 Maths NCERT Solutions
• Exercise 8.2 Chapter 8 Class 9 Maths NCERT Solutions

NCERT Solutions for Class 9 Maths Chapters:

What are the benefits of ncert solutions for chapter 8 quadrilaterals class 9 ncert solutions, what is a rhombus, what is mid-point theorem, in a quadrilateral, ∠ a : ∠ b : ∠ c : ∠ d = 1 : 2 : 3 : 4, then find the measure of each angle of the quadrilateral., contact form.

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Quadrilaterals Class 9 Notes PDF (Short & Handwritten)

Revision notes make you aware of those topics that you might have missed during your regular classes. Thus, Quadrilaterals is an extremely important chapter of Class 9 Maths and so, all students who have opted for Maths in their intermediate should refer to the Quadrilaterals Class 9 Notes. The revision notes work as a reference that help students like you to revise the concepts and formulas which you have studied earlier from your Mathematics textbook.

Therefore, to help you better learn the Quadrilaterals concepts and techniques to solve questions, we have mentioned here the PDF of revision notes. The Quadrilaterals notes in the PDF file can be downloaded for free of cost from this page.

Quadrilaterals Notes PDF

The PDF file is easy to access and often free, it is popular among students as they can carry them anywhere they want into their device like smartphone, Laptop, etc. Keeping in mind the convenience of students, our subject matter experts have prepared the Quadrilaterals Notes PDF which is very illustrative and can help you recall all of your previous learning of Class 9 Maths Quadrilaterals.

The Quadrilaterals PDF notes can be a fun and time saving study resource that can help students to cover all the important and highlighted key points of the chapter.

Topic wise Class 9 Quadrilaterals Notes

Our Maths experts have organised the Class 9 Quadrilaterals Notes in a topic wise manner. Because of that students are able to find an organised study resource that enables them to focus on study and avoid clutter in learning.

The topic wise Quadrilaterals notes PDF is an ideal study resource that improves memory power and increases the attention. The notes are considered a great tool to develop a deeper understanding in the concepts explained. Keeping in mind this, our subject experts have prepared our own revision notes for Quadrilaterals.

How to Download Quadrilaterals Class 9 Notes in PDF for Free?

Follow the below-given steps to download Quadrilaterals Class 9 Notes in PDF for Free.

• Open Selfstudys.com: Selfstudys is an online platform that offers free study resources for K-11 students. Thus, to download Quadrilaterals Class 9 Notes in PDF for free you have to open Selfstudys.com on your browser.
• Navigate to NCERT Books & Solutions: The links to download Class 9 Quadrilaterals notes PDF are available within the NCERT Books & Solutions which you can find in the navigation menu/button.
• Enter into NCERT Notes Menu: In order to access the Notes of Class 9 Quadrilaterals, you have to enter into NCERT Notes Menu. To do this, click on NCERT Notes within the NCERT Books and Solutions menu

• Choose Class: There is a list of classes, but you have to choose Class 9 in order to download the Quadrilaterals Class 9 Notes PDF.

• Now, Select Maths and Access Quadrilaterals Notes for Free: At this moment, you are two steps away from downloading your revision notes. Click on Maths as mentioned in the given image and then the page will reload where you can download the Quadrilaterals Class 9 Notes for free of cost.

Quadrilaterals Class 9 Notes That Can Help You Score Better Marks

We all agree that students need to do lots of practice in order to master Maths; however, it isn’t an easy task because without a proper understanding of the concepts it wouldn’t be easy to solve even a single question from the chapter Quadrilaterals. But the Class 9 Quadrilaterals notes that we provide here are a very ideal study resource to develop a higher level of understanding in the topic.

A good grip on the concepts of Class 9 Quadrilaterals can assist candidates to easily score better marks due to two reasons:

• Revision notes are short yet precise that help students to cover more important concepts in a lesser time.
• Research suggests that repeatedly memorising something boosts depth of mental processing as a result; thus, it develops deeper levels of analysis that produce more elaborate, longer-lasting, and stronger memory.

In conclusion, we can say that Notes of Quadrilaterals along with a regular practice of questions can help students to score better marks as they can remember the topics for a longer period of time.

Other than Notes of Quadrilaterals You Can Use

In case, if you make your mind to not only focus on Quadrilaterals Class 9 notes then you can use -

• MCQ Test of Quadrilaterals: Similar to notes, solving the Mock test of Quadrilaterals can help you recall your previous learning from the chapter. If you want to level up your preparation level and want to challenge your conceptual understanding of the Quadrilaterals you can use the CBSE MCQ test.
• Chapter End Questions of Quadrilaterals: The Quadrilaterals notes PDF that we provide here are prepared referring to the NCERT Class 9 Maths Book so, those who want to use other study resources than revision notes of Quadrilaterals can use the NCERT Books. The book contains exercises that enable the learners to solve various questions to cross-check how well they understood the topic.
• Highlighted Points of Topic: Every single chapter contains the key points that grab attention to readers, even such highlighted points are considered important. Thus, if you want to take a pause for a while from using the Quadrilaterals PDF notes then you can go back to your preferred textbook to read those highlighted points of the topic.

The Right Time To Use Quadrilaterals Class 9 Notes

There is no right or wrong time to use Quadrilaterals Class 9 Maths Notes as long as you are sincere about your study; however, there are 3 most important times that we think all students should revise whatever they have studied in Quadrilaterals. 

• On a weekly basis after completing the chapter Quadrilaterals: There’s a few students who dedicate themselves in learning Maths during regular classes; however, experts suggests that revision just after completing the chapter Quadrilaterals can help students to deepen their understanding in the concepts that were explained in the lesson. Thus, the right time to use the Quadrilaterals Class 9 Notes can be on a weekly basis after finishing the chapter’s basics subtopics.
• During Board Exam Preparation: The importance of revision Notes of Quadrilaterals can be understood during the board exam preparation because it saves time and allows students to cover the complete chapter in a lesser time. Also, the board exam preparation time often becomes a mess but the notes are neat, clean and organised in a manner that helps students to be more productive.
• While preparing for competitive exams like JEE: Class 9 Quadrilaterals is not limited to board exams only, rather it plays a crucial role in national level competitive exam preparation. Therefore, the use of Quadrilaterals Class 9 Notes PDF while preparing for competitive exams like JEE can be a great time.

What You Definitely Need to Revise Class 9 Quadrilaterals?

The Quadrilaterals is an important part of higher level Maths, because in standard like 11, students have to study various topics at the same time, it becomes a challenge to handle so many topics and subjects at the same time. Thus, here we have listed what you definitely need to revise Class 9 Quadrilaterals for the ease of your study and exam preparation.

• A Short Notes of Quadrilaterals: You wouldn’t like to use the revision Notes of Quadrilaterals that look like bulky textbooks, cluttered with so much information in an unorganised manner, right? So, you need a short yet precise Short Notes of Quadrilaterals that can help you recall your learnings from the chapter within a few minutes. You can find such Notes of Quadrilaterals here on Selfstudys website.
• Easy To Understand Definitions: Unclear and lengthy definitions never work, and it becomes very challenging to retain such definitions for a longer time. Therefore, you need revision notes that can help you easily understand the definitions of Quadrilaterals. Easy to understand definitions help you retain them for a longer time and give you an ability to recall them wherever you need them to solve a particular question related to Quadrilaterals.
• Some Questions of Quadrilaterals to Practise: Without solving the questions of Quadrilaterals a revision is incomplete. If you really want to revise the topics, you must solve the questions based on Quadrilaterals. Such questions can be found from Notes of Quadrilaterals that we provide here or from the NCERT Class 9 Maths textbooks. Our Maths experts have mentioned the questions of Quadrilaterals especially in Class 9 Maths notes so that a learner can use them to practise.

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