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NCERT Solutions for Exercise 8.1 Class 9 Maths Chapter 8 - Quadrilaterals Exercise 8.1

Last updated date: 13th Jul 2024
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NCERT Chapter 8 of Ex 8.1 Class 9 Maths, titled "Quadrilaterals," introduces various types of quadrilaterals and their properties. Class 9 maths 8.1 Exercise focuses on definitions and basic properties, covering parallelograms, rectangles, rhombuses, and squares.

Table of Content
2. Glance of NCERT Solutions for Maths Chapter 8 Ex 8.1 Class 9 Quadrilaterals | Vedantu
3. Topics Covered in Class 9 Maths Chapter 8 Exercise 8.1
4. Access NCERT Solution for Class 9 Maths Chapter 8 – Quadrilateral Exercise 8.1
5. Class 9 Maths Chapter 8 : Exercises Breakdown
6. CBSE Class 9 Maths Chapter 8 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs

In Understanding the properties of each quadrilateral type in class 9 Quadrilateral 8.1 Exercise, such as side and angle relationships, is crucial. This exercise prepares students for more complex problems later in the chapter, making it essential for building a strong mathematical foundation.

Glance of NCERT Solutions for Maths Chapter 8 Ex 8.1 Class 9 Quadrilaterals | Vedantu

• The topics covered in this chapter are definitions and properties of quadrilaterals.

• This Chapter focuses on different types of quadrilaterals: parallelograms, rectangles, rhombuses, squares, and trapeziums.

• This even covers the angle sum property of quadrilaterals.

• The important formulas are Angle sum property, Properties of parallelogram, Properties of rectangle, Properties of rhombus.

• Here we enhance understanding of quadrilateral properties.

• Different types of quadrilaterals (a figure that is formed by joining four points in order is called a quadrilateral)

• Properties of the quadrilaterals (especially those of parallelograms, such as the diagonal of a parallelogram and its sides)

• This chapter of Class 9 Maths Exercise 8.1 helps students to understand the concepts of Quadrilaterals.

• There are links to video tutorials explaining class 9 chapter 8 Exercise 8.1  Quadrilaterals for better understanding.

• Class 9 math chapter 8 exercise 8.1 NCERT Solutions has over 7 Questions.

Topics Covered in Class 9 Maths Chapter 8 Exercise 8.1

1. Properties of Parallelograms:

• The opposite sides of a parallelogram are always equal in length (e.g., AB = CD and AD = BC).

• The opposite angles in a parallelogram are also equal (e.g., angle A = angle C and angle B = angle D).

• While diagonals (opposite corners connected by a line) of a parallelogram aren't necessarily equal, the exercise might explore the fact that they bisect each other (divide each other into two segments of equal length).

1. Identifying Parallelograms: The exercise will provide methods to identify quadrilaterals as parallelograms based on the properties mentioned above.

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Access NCERT Solution for Class 9 Maths Chapter 8 – Quadrilateral Exercise 8.1

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

Ans: Given that the diagonals of a parallelogram are equal. Let us assume a parallelogram $ABCD$ whose diagonals are equal as shown in the figure below.

Here $ABCD$ is a parallelogram, so we can write that $AB=DC$ since opposite sides of a parallelogram are equal.

Consider the triangles $ABC$ and $DCB$. In these two triangles we have $AB=DC$, $BC=BC$ and as per given data $AC=DB$. So, we can write that $\Delta ABC\cong \Delta DCB$ by SSS Congruence rule.

$\Rightarrow \angle ABC=\angle DCB$

We know that the sum of the angles on the same side of the transversal is $180{}^\circ$.

$\therefore \angle ABC+\angle DCB=180{}^\circ$

Substitute the value $\angle ABC=\angle DCB$ in the above equation.

$\Rightarrow \angle ABC+\angle ABC=180{}^\circ$

$\Rightarrow 2\angle ABC=180{}^\circ$

$\Rightarrow \angle ABC=\dfrac{180{}^\circ }{2}$

$\Rightarrow \angle ABC=90{}^\circ$

Here we have the angle $ABC$ and $DCB$ as $90{}^\circ$. When the angles of the parallelogram are equal to $90{}^\circ$, then it is called as Rectangle.

Hence, if the diagonals of the parallelogram are equal then it is called as Rectangle.

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

Ans: Consider a square $ABCD$ whose diagonals are $AC$ and $BD$. The intersecting point of the diagonals will be $O$ as shown in below figure.

According to the given statement we need to prove that $AC=BD$, $OA=OC$, $OB=OD$ and $\angle AOB=90{}^\circ$.

Consider $\Delta ABC$ and $\Delta DCB$.

In both the triangles we can write

$AB=DC$ since the sides of a square are equal to each other, $\angle ABC=\angle DCB=90{}^\circ$ and

$BC=CB$.

So, by using SAS congruency law we can say that $\Delta ABC\cong \Delta DCB$.

$\therefore AC=DB$ by CPCT.

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

Now consider $\Delta AOB$ and $\Delta COD$.

In both the triangles we can write

$\angle AOB=\angle COD$ since vertically opposite angles are equal,

$\angle ABO=\angle CDO$ since alternate interior angles are also equal and

$AB=CD$ since the sides of a square are always equal.

So, by using AAS congruence rule we can say that $\Delta AOB\cong \Delta COD$.

$\therefore AO=CO$ and $OB=OD$ by CPCT.

Now consider $\Delta AOB$ and $\Delta COB$,

As we have proved earlier $AO=CO$,

$AB=CB$ since sides of a square are equal,

$BO=BO$.

By using SSS congruency rule we can write that $\Delta AOB\cong \Delta COB$.

$\therefore \angle AOB=\angle COB$ by CPCT.

But the angles $AOB$ and $COB$ are linear pair.

$\therefore \angle AOB+\angle COB=180{}^\circ$

$2\angle AOB=180{}^\circ$

$\angle AOB=90{}^\circ$

Hence, the diagonals of a square bisect each other at right angles.

3. Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$ as shown in figure.

• Show that it bisects $\angle C$.

Ans: Given that $ABCD$ is a parallelogram.

$\therefore \angle DAC=\angle BCA$ and $\angle BAC=\angle DCA$ since alternative interior angles

But in the problem, we have given that $AC$ bisect $\angle A$.

$\therefore \angle DAC=\angle BAC$

From all the above equations we can write that

$\angle DAC=\angle BCA=\angle BAC=\angle DCA$

$\angle DCA=\angle BAC$

Hence, $AC$ bisects $\angle C$.

• Show that $ABCD$ is a rhombus.

Ans: From the equation $\angle DAC=\angle BCA=\angle BAC=\angle DCA$, we can write that $\angle DAC=\angle DCA$.

$DA=DC$ since side opposite to equal angles are equal.

But we have $DA=BC$ and $AB=CD$ since opposite sides of the parallelogram are equal.

$\therefore AB=BC=CD=DA$

4. $ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$.

• $ABCD$ is a square.

Ans: Let the diagram of the given $ABCD$ rectangle is given by

We have given that $ABCD$ is a rectangle. So, we can write that

$\angle A=\angle C$

$\Rightarrow \dfrac{1}{2}\angle A=\dfrac{1}{2}\angle C$

But they have mentioned that $AC$ bisects $\angle A$ as well as $\angle C$.

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

$\therefore CD=DA$ since sides opposite to equal angles are also equal.

But $DA=BC$ and $AB=CD$ since opposite sides of a rectangle are equal.

$\therefore AB=BC=CD=DA$

Here $ABCD$ is given to be a rectangle but all the sides are equal.

Hence $ABCD$ is a square.

• Diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Ans: Now consider $\Delta BCD$, in this triangle we can write that

$BC=CD$ since sides of a square are equal,

$\angle CDB=\angle CBD$ since angles opposite to equal sides are equal.

But we have $\angle CDB=\angle ADB$ since alternate interior angles for parallel lines $AB$ and $CD$.

$\therefore \angle CBD=\angle ABD$

Hence $BD$ bisects $\angle B$.

Also, $\angle CBD=\angle ADB$ which are alternate interior angles for parallel lines $BC$ and $AD$.

$\therefore \angle CDB=\angle ADB$

Finally, $BD$ bisects $\angle D$ and $\angle B$.

5. In parallelogram $ABCD$, two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP=BQ$ as shown in below figure.

• Show that $\Delta APD\cong \Delta CQB$

Ans: Consider $\Delta APD$ and $\Delta CQB$, in these two triangles we can write

$\angle ADP=\angle CBQ$ since alternate interior angles for the parallel lines $BC$ and $AD$,

$AD=CB$ since opposite sides of parallelogram are equal.

$DP=BQ$ which is in the given data.

By using SAS congruence rule, we can say that $\Delta APD\cong \Delta CQB$

• Show that $AP=CQ$

Ans: As we have $\Delta APD\cong \Delta CQB$. By using CPCT we can write that $AP=CQ$.

• Show that $\Delta AQB\cong \Delta CPD$

Ans: Consider $\Delta AQB$ and $\Delta CPD$, in these two triangles we can write

$\angle ABQ=\angle CDP$ since alternate interior angles for the parallel lines $AB$ and $CD$,

$AB=CD$ since opposite sides of parallelogram are equal.

$BQ=DP$ which is in the given data.

By using SAS congruence rule, we can say that $\Delta AQB\cong \Delta CPD$

• Show that $AQ=CP$

Ans: As we have $\Delta AQB\cong \Delta CPD$. By using CPCT we can write that $AQ=CP$.

• Show that $APCQ$ is a Parallelogram.

Ans: From the results obtained in (ii) and (iv), which are $AQ=CP$ and $AP=CQ$. That means the opposite sides of the quadrilateral $APCQ$ are equal, so $APCQ$ is a parallelogram.

6. $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ as shown in figure.

• Show that  $\Delta APB\cong \Delta CQD$

Ans: Consider $\Delta APB$ and $\Delta CQD$, in these two triangles we can write that

$\angle APB=\angle CQD=90{}^\circ$,

$AB=CD$ since opposite sides of parallelogram,

$\angle ABP=\angle CDQ$ since alternate interior angles for parallel lines $AB$ and $CD$.

By using AAS congruency rule we can say that $\Delta APB\cong \Delta CQD$.

• Show that $AP=CQ$

Ans: From the statement $\Delta APB\cong \Delta CQD$, by using CPCT we can write that $AP=CQ$.

7. $ABCD$ is a trapezium on which $AB\parallel CD$ and $AD=BC$ as shown in figure. (Hint: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ at point $E$).

• Show that $\angle A=\angle B$

Ans: We have $AD=CE$ opposite sides of parallelogram and $AD=BC$.

$\therefore BC=CE$ and $\angle CEB=\angle CBE$ since angle opposite to equal sides are also equal.

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

So,

$\angle A+\angle CEB=180{}^\circ$ since angles on the same side of transversal.

$\Rightarrow \angle A+\angle CBE=180{}^\circ$

Also, $\angle B+\angle CBE=180{}^\circ$ as they are linear pair of angles. So, from these two equals we can write that

$\angle A=\angle B$

• Show that $\angle C=\angle D$

Ans: Given $AB\parallel CD$, so

$\angle A+\angle D=180{}^\circ$ , $\angle C+\angle B=180{}^\circ$since angles on the same side of the transversal.

$\therefore \angle A+\angle D=\angle C+\angle B$

But we have $\angle A=\angle B$, so we will have

$\angle C=\angle D$

• Show that $\Delta ABC\cong \Delta BAD$

Ans: Consider $\Delta ABC$ and $\Delta BAD$, in these two triangles we can write that

$AB=BA$, $BC=AD$ and $\angle B=\angle A$.

By using SAS congruence rule, we can say that $\Delta ABC\cong \Delta BAD$.

• Show that diagonal $AC=\text{ diagonal }BD$

Ans: We have $\Delta ABC\cong \Delta BAD$, so by using CPCT we can write that $AC=BD$

Conclusion

NCERT Class 9 Maths Ex 8.1 is a crucial exercise that teaches students the properties and classifications of quadrilaterals, including squares, rectangles, rhombuses, and parallelograms. It builds foundational understanding for more complex geometrical concepts and enhances analytical skills by asking students to prove or disprove certain characteristics of these shapes. Mastering quadrilaterals class 9 exercise 8.1 solutions can help students secure marks and deepen their understanding of geometry, as questions from this chapter have appeared frequently in exams.

Class 9 Maths Chapter 8 : Exercises Breakdown

 Exercise Number of Questions Exercise 8.2 6 Questions & Solutions

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for class 9th maths chapter 8 exercise 8.1 Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Exercise 8.1 Class 9 Maths Chapter 8 - Quadrilaterals Exercise 8.1

1. What is a quadrilateral? Explain

Quadrilateral is a polygon that has four sides and four angles. These are available in different shapes. Rhombus, rectangle, parallelogram, Square, etc., all are examples of quadrilaterals. These can be seen everywhere, especially in the manufacturing of electronic gadgets, stationery items, architectural designs.

2. Explain a few properties of a parallelogram?

Parallelogram is one of the examples of a quadrilateral that has different properties. They are-

• The opposite sides and angles are equal.

• It has supplementary angles on adjacent sides.

• The diagonals always bisect each other

3. Where can I find NCERT Solutions for Exercise 8.1 of Chapter 8 of Class 9 Maths online?

It’s very important for the students to find accurate NCERT solutions on the internet. To make it easier, Students can find the NCERT Solutions for Exercise 8.1 of Chapter 8 of Class 9 Maths easily on the Vedantu website as well as Vedantu Mobile app. Stepwise solutions have been provided for all the questions present in the exercise. The solutions are available in PDF format which the students can access either online or download for free. All the solutions are created by our expert teachers as per the CBSE guidelines.

4. Do I need to practise all questions provided in Exercise 8.1 of Chapter 8 of Class 9 Maths?

Yes, all questions must be practised constantly as it is the key for the students to score high marks in exams. It is advised for the students to practice all the questions provided in Exercise 8.1 of Chapter 8 of Class 9 Maths as everything is important from an exam perspective. It will help the students to clear their doubts, increase speed and reduce making mistakes in the exam. Students must memorize formulae. Practising regularly will help the students to remember the problem-solving methods easily.

5. How CBSE Students can utilize NCERT Solutions for Exercise 8.1 of Chapter 8 of Class 9 Maths effectively?

Students can effectively use the NCERT Solutions for Exercise 8.1 of Chapter 8 of Class 9 Maths by understanding the derivations of all the formulae. Students should also revise all the concepts and understand the theories from the chapter. Once students get a complete idea of all the theories and concepts, they can start to solve the exercises to know how much they have understood. Finally, they should use NCERT solutions to check if their answers are right or wrong and understand the problem-solving methods.

6. Do NCERT Solutions for Exercise 8.1 of Chapter 8 of Class 9 Maths help you to score well in the exam?

yes, the NCERT solutions for exercise 8.1 chapter 8 class 9 helps the students to score well in the exam. With the practice of the NCERT solutions, the student will be able to figure out the areas that they find difficult and thereby give extra time on mastering them. Apart from this , every question from the NCERT solution is exam based. Thus, with the practice of the NCERT solutions, the student is preparing for the exams, which will ultimately help them answer any question asked in the examination.

7. Are NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.1 the best study material for the students during revision?

Yes, NCERT Solutions provided by Vedantu for Exercise 8.1 of Chapter 8 of Class 9 Maths are the most reliable study material for the CBSE students. It helps the students to learn and revise difficult concepts easily. Every solution is provided by an explanation to make learning easier for the students. The experts have designed stepwise solutions to make students understand the steps. During exam time, it will help them to save a lot of time by giving a quick revision.

8. Are there any theorems introduced in Class 9 Maths Ch 8 Ex 8.1 ?

In Exercise 8.1 of Class 9 Maths Chapter 8 on Quadrilaterals, several important theorems about parallelograms are introduced. These include theorems stating that in a parallelogram, opposite sides are equal, opposite angles are equal, and the diagonals bisect each other. Another key theorem states that if, in a quadrilateral, opposite sides are equal, then it must be a parallelogram. Understanding these theorems is crucial, as they form the foundation for proving other properties of parallelograms and are often used in solving various geometrical problems, not only in this exercise but also in more advanced studies of geometry.

9. What should I focus on in Class 9 Ex 8.1 to perform well in exams?

Understanding the principles and characteristics of parallelograms, using these properties to derive other facts, and improving problem-solving techniques are necessary for mastering quadrilateral exams. Solve past papers and examples from textbooks to improve your problem-solving skills. Build comprehensive proofs as a practice for expressing understanding clearly in exams. This will get you ready for both easy and difficult parallelogram questions.

10. Any tips for mastering Class 9 Maths Ch 8 Ex 8.1?

Regular practice and revisiting theorems and their proofs can significantly help. Also, solving additional problems from other resources like Vedantu can provide deeper insights and better preparation.

11. Are diagrams important in solving problems from Class 9 Maths Chapter 8 Exercise 8.1 ?

Yes, accurately drawing diagrams helps in better visualization and solving of problems related to quadrilaterals.