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

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

Edited By Vishal kumar | Updated on Oct 07, 2023 10:16 AM IST

NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1- This class 9 Maths chapter 8 exercise 8.1 focuses on quadrilaterals, four-sided shapes. Exercise 8.1 helps you understand their basics, properties, and types. Our free PDF solutions simplify complex ideas, providing clear explanations and step-by-step guidance. These resources aid your understanding, exam preparation, and math success.

In NCERT solutions for Class 9 Maths chapter 8 exercise 8.1, the types of quadrilaterals are determined by the angles and lengths of their sides. Because the word "quad" means "four," all of these quadrilaterals have four sides, and the sum of their angles is 360 degrees.

Types of quadrilaterals are Squares, Trapezium, Parallelogram, Rectangle, Rhombus, and Kite.

A square can likewise be considered a rectangle and furthermore, a rhombus, yet every rectangle or a rhombus is certainly not a square. A trapezium isn't a parallelogram (as just one set of inverse sides are equal in a trapezium and we require the two sets to be equal in a parallelogram)

Along with NCERT book Class 9 Maths, chapter 8 exercise 8.1 the following exercises are also present.

## Access Quadrilaterals Class 9 Chapter 8 Exercise: 8.1

Given : The angles of a quadrilateral are in the ratio $\small 3:5:9:13$ .
Let the angles of quadrilateral be $3x,5x,9x ,13x$ .

Sum of all angles is 360.

Thus, $3x+5x+9x +13x=360 \degree$

$\Rightarrow 30x=360 \degree$

$\Rightarrow x=12 \degree$

All four angles are : $3x=3\times 12=36\degree$

$5x=5\times 12=60 \degree$

$9x=9\times 12=108 \degree$

$13x=13\times 12=156 \degree$

Given: ABCD is a parallelogram with AC=BD.

To prove: ABCD is a rectangle.

Proof: In $\triangle$ ABC and $\triangle$ BAD,

BC= AD (Opposite sides of parallelogram)

AC=BD (Given )

AB=AB (common)

$\triangle$ ABC $\cong$ $\triangle$ BAD (By SSS)

$\angle ABC=\angle BAD$ (CPCT)

and $\angle ABC+\angle BAD=180 \degree$ (co - interior angles)

$2\angle BAD=180 \degree$

$\angle BAD= 90 \degree$

Hence, it is a rectangle.

Given: the diagonals of a quadrilateral bisect each other at right angles i.e.BO=OD.

To prove: ABCD is a rhombus.

Proof : In $\triangle$ AOB and $\triangle$ AOD,

$\angle AOB=\angle AOD$ (Each $90 \degree$ )

BO=OD (Given )

AO=AO (common)

$\triangle$ AOB $\cong$ $\triangle$ AOD (By SAS)

Similarly, AB=BC and BC=CD

Hence, it is a rhombus.

Given : ABCD is a square i.e. AB=BC=CD=DA.

To prove : the diagonals of a square are equal and bisect each other at right angles i.e. AC=BD,AO=CO,BO=DO and $\angle COD=90 \degree$

Proof : In $\triangle$ BAD and $\triangle$ ABC,

$\angle BAD = \angle ABC$ (Each $90 \degree$ )

AB=AB (common)

$\triangle$ BAD $\cong$ $\triangle$ ABC (By SAS)

BD=AC (CPCT)

In $\triangle$ AOB and $\triangle$ COD,

$\angle$ OAB= $\angle$ OCD (Alternate angles)

AB=CD (Given )

$\angle$ OBA= $\angle$ ODC (Alternate angles)

$\triangle$ AOB $\cong$ $\triangle$ COD (By AAS)

AO=OC ,BO=OD (CPCT)

In $\triangle$ AOB and $\triangle$ AOD,

OB=OD (proved above)

OA=OA (COMMON)

$\triangle$ AOB $\cong$ $\triangle$ AOD (By SSS)

$\angle$ AOB= $\angle$ AOD (CPCT)

$\angle$ AOB+ $\angle$ AOD = $180 \degree$

2. $\angle$ AOB = $180 \degree$

$\angle$ AOB = $90 \degree$

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

Given : ABCD is a quadrilateral with AC=BD,AO=CO,BO=DO, $\angle$ COD = $90 \degree$

To prove: ABCD is a square.

Proof: Since the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a rhombus.

Thus, AB=BC=CD=DA

In $\triangle$ BAD and $\triangle$ ABC,

AB=AB (common)

BD=AC

$\triangle$ BAD $\cong$ $\triangle$ ABC (By SSS)

$\angle BAD = \angle ABC$ (CPCT)

$\angle$ BAD+ $\angle$ ABC = $180 \degree$ (Co-interior angles)

2. $\angle$ ABC = $180 \degree$

$\angle$ ABC = $90 \degree$

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

it bisects $\small \angle C$ also.

Given: $\angle$ DAC= $\angle$ BAC ................1

$\angle$ DAC= $\angle$ BCA.................2 (Alternate angles)

$\angle$ BAC= $\angle$ ACD .................3 (Alternate angles)

From equation 1,2 and 3, we get

$\angle$ ACD= $\angle$ BCA...................4

Hence, diagonal AC bisect angle C also.

Given: $\angle$ DAC= $\angle$ BAC ................1

$\angle$ DAC= $\angle$ BCA.................2 (Alternate angles)

$\angle$ BAC= $\angle$ ACD .................3 (Alternate angles)

From equation 1,2 and 3, we get

$\angle$ ACD= $\angle$ BCA...................4

From 2 and 4, we get

$\angle$ ACD= $\angle$ DAC

In $\triangle$ ADC,

$\angle$ ACD= $\angle$ DAC (proved above )

AD=DC (In a triangle,sides opposite to equal angle are equal)

A parallelogram whose adjacent sides are equal , is a rhombus.

Thus, ABCD is a rhombus.

In $\triangle$ ADC,

AD = CD (ABCD is a rhombus)

$\angle$ 3= $\angle$ 1.................1(angles opposite to equal sides are equal )

$\angle$ 3= $\angle$ 2.................2 (alternate angles)

From 1 and 2, we have

$\angle$ 1= $\angle$ 2.................3

and $\angle$ 1= $\angle$ 4.................4 (alternate angles)

From 1 and 4, we get

$\angle$ 3= $\angle$ 4.................5

Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.

In $\triangle$ ADB,

AD = AB (ABCD is a rhombus)

$\angle$ 5= $\angle$ 7.................6(angles opposite to equal sides are equal )

$\angle$ 7= $\angle$ 6.................7 (alternate angles)

From 6 and 7, we have

$\angle$ 5= $\angle$ 6.................8

and $\angle$ 5= $\angle$ 8.................9(alternate angles)

From 6 and 9, we get

$\angle$ 7= $\angle$ 8.................10

Hence, from 8 and 10, diagonal BD bisects angles B as well as angle D.

Given: ABCD is a rectangle with AB=CD and BC=AD $\angle$ 1= $\angle$ 2 and $\angle$ 3= $\angle$ 4.

To prove: ABCD is a square.

Proof : $\angle$ 1= $\angle$ 4 .............1(alternate angles)

$\angle$ 3= $\angle$ 4 ................2(given )

From 1 and 2, $\angle$ 1= $\angle$ 3.....................................3

In $\triangle$ ADC,

$\angle$ 1= $\angle$ 3 (from 3 )

DC=AD (In a triangle, sides opposite to equal angle are equal )

A rectangle whose adjacent sides are equal is a square.

Hence, ABCD is a square.

diagonal BD bisects $\small \angle B$ as well as $\small \angle D$ .

In $\triangle$ ADB,

AD = AB (ABCD is a square)

$\angle$ 5= $\angle$ 7.................1(angles opposite to equal sides are equal )

$\angle$ 5= $\angle$ 8.................2 (alternate angles)

From 1 and 2, we have

$\angle$ 7= $\angle$ 8.................3

and $\angle$ 7 = $\angle$ 6.................4(alternate angles)

From 1 and 4, we get

$\angle$ 5= $\angle$ 6.................5

Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ .

To prove : $\small \Delta APD\cong \Delta CQB$

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ (Given )

$\angle$ ADP= $\angle$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$ (By SAS)

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ .

To prove : $\small AP=CQ$

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ (Given )

$\angle$ ADP= $\angle$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$ (By SAS)

$\small AP=CQ$ (CPCT)

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ .

To prove : $\small \Delta AQB\cong \Delta CPD$

Proof :

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ (Given )

$\angle$ ABQ= $\angle$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$ (By SAS)

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ .

To prove : $\small AQ=CP$

Proof :

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ (Given )

$\angle$ ABQ= $\angle$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$ (By SAS)

$\small AQ=CP$ (CPCT)

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ .

To prove: APCQ is a parallelogram

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ (Given )

$\angle$ ADP= $\angle$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$ (By SAS)

$\small AP=CQ$ (CPCT)...............................................................1

Also,

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ (Given )

$\angle$ ABQ= $\angle$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$ (By SAS)

$\small AQ=CP$ (CPCT)........................................2

From equation 1 and 2, we get

$\small AP=CQ$

$\small AQ=CP$

Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To prove : $\small \Delta APB\cong \Delta CQD$

Proof: In $\small \Delta APB\, \, and\, \, \Delta CQD$ ,

$\angle$ APB= $\angle$ CQD (Each $90 \degree$ )

$\angle$ ABP= $\angle$ CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, $\small \Delta APB\cong \Delta CQD$ (By SAS)

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To prove : $\small AP=CQ$

Proof: In $\small \Delta APB\, \, and\, \, \Delta CQD$ ,

$\angle$ APB= $\angle$ CQD (Each $90 \degree$ )

$\angle$ ABP= $\angle$ CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, $\small \Delta APB\cong \Delta CQD$ (By SAS)

$\small AP=CQ$ (CPCT)

Given : In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF$ .

To prove : quadrilateral ABED is a parallelogram

Proof : In ABED,

AB=DE (Given)

AB||DE (Given )

Hence, quadrilateral ABED is a parallelogram.

Given: In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF$ .

To prove: quadrilateral BEFC is a parallelogram

Proof: In BEFC,

BC=EF (Given)

BC||EF (Given )

Hence, quadrilateral BEFC is a parallelogram.

To prove: $\small AD\parallel CF$ and $\small AD=CF$

Proof :

In ABED,

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

From 1 and 3, we get

To prove : quadrilateral ACFD is a parallelogram

Proof :

In ABED,

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

From 1 and 3, we get

From 5 and 6, we get

Thus, quadrilateral ACFD is a parallelogram

In $\small \Delta ABC$ and $\small \Delta DEF$ ,

AB=DE (Given )

BC=EF (Given )

AC=DF ( proved in (v) part)

$\small \Delta ABC\cong \Delta DEF$ (By SSS rule)

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

Given: ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$

To prove : $\small \angle A=\angle B$

Proof: Let $\angle$ A be $\angle$ 1, $\angle$ ABC be $\angle$ 2, $\angle$ EBC be $\angle$ 3, $\angle$ BEC be $\angle$ 4.

In AECD,

AE||DC (Given)

Hence, AECD is a parallelogram.

From 1 and 2, we get

CE=BC

In $\triangle$ BCE,

$\angle 3=\angle 4$ .................3 (opposite angles of equal sides)

$\angle 2+\angle 3=180 \degree$ ...................4(linear pairs)

$\angle 1+\angle 4=180 \degree$ .....................5(Co-interior angles)

From 4 and 5, we get

$\angle 2+\angle 3=\angle 1+\angle 4$

$\therefore \angle 2=\angle 1 \rightarrow \angle B=\angle A$ (Since, $\angle 3=\angle 4$ )

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

Given: ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$

To prove : $\small \angle C=\angle D$

Proof: Let $\angle$ A be $\angle$ 1, $\angle$ ABC be $\angle$ 2, $\angle$ EBC be $\angle$ 3, $\angle$ BEC be $\angle$ 4.

$\angle 1+\angle D= 180 \degree$ (Co-interior angles)

$\angle 2+\angle C= 180 \degree$ (Co-interior angles)

$\therefore \angle 1+\angle D=\angle 2+\angle C$

Thus, $\small \angle C=\angle D$ (Since , $\small \angle 1=\angle 2$ )

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

Given: ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$

To prove : $\small \Delta ABC\cong \Delta BAD$

Proof: In $\small \Delta ABC\, \, and\, \, \, \Delta BAD$ ,

AB=AB (Common )

$\angle ABC=\angle BAD$ (proved in (i) )

Thus, $\small \Delta ABC\cong \Delta BAD$ (By SAS rule)

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

Given: ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$

To prove: diagonal AC $\small =$ diagonal BD

Proof: In $\small \Delta ABC\, \, and\, \, \, \Delta BAD$ ,

AB=AB (Common )

$\angle ABC=\angle BAD$ (proved in (i) )

Thus, $\small \Delta ABC\cong \Delta BAD$ (By SAS rule)

diagonal AC $\small =$ diagonal BD (CPCT)

## More About NCERT Solutions for Class 9 Maths Exercise 8.1

NCERT syllabus Class 9 Maths exercise 8.1 includes some important theorems that are listed below:

A parallelogram diagonal divides it into two congruent triangles.

The opposite sides of a parallelogram are equal.

A parallelogram is formed when each pair of opposite sides of a quadrilateral is equal.

The opposite angles in a parallelogram are equal.

A parallelogram is formed when each pair of opposite angles in a quadrilateral is equal.

A parallelogram's diagonals cross each other.

A parallelogram is formed when the diagonals of a quadrilateral cross each other.

If two opposite sides of a quadrilateral are equal and parallel, the quadrilateral is a parallelogram.

## Benefits of NCERT Solutions for Class 9 Maths Exercise 8.1

• Exercise 8.1 Class 9 Maths, is based on the concepts of QUADRILATERALS.

• From Class 9 Maths chapter 8 exercise 8.1 we learn about the angle property of quadrilaterals and the different types of quadrilaterals.

• Understanding the concepts from Class 9 Maths chapter 8 exercise 8.1 will easily help us to comprehend the ideas identified with the various sorts of QUADRILATERALS like a parallelogram, square shape.

## Key Features of 9th Class Maths Exercise 8.1 Answers

1. Basic Understanding: The answers in this class 9 maths ex 8.1 provide a fundamental understanding of quadrilaterals, their properties, and classifications.

2. Clear Explanations: Class 9 ex 8.1 offer clear and concise explanations for each question, making it easier for students to comprehend the concepts.

3. Step-by-Step Solutions: The answers provide step-by-step solutions, helping students follow the problem-solving process easily.

4. Varied Problem Types: Ex 8.1 class 9 cover a variety of problems related to quadrilaterals, ensuring that students are well-prepared for different question types.

5. Practice Material: Exercise 8.1 answers serve as valuable practice material for students to reinforce their knowledge of quadrilaterals.

6. Preparation for Assessments: These 9th class maths exercise 8.1 answers help students prepare for exams by offering a range of problems that may be encountered in tests and assessments.

Also, See

## Subject Wise NCERT Exemplar Solutions

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 8.1?

A quadrilateral is a closed four-sided shape formed by joining four points by straight lines.

3. What is the sum total of all angles of a quadrilateral?

The sum total of all angles of a quadrilateral is 360 degrees.

4. The minimum number of triangles that can be formed from a quadrilateral are?

We can get a minimum of two triangles to form a quadrilateral.

5. What are the different types of quadrilaterals?

1. Square

2. Rectangle

3. Rhombus

4. Trapezium

5. Parallelogram

6. Kite

6. What are the similarities between square and rhombus?

The similarities between square and rhombus are:

1. All the sides are equal to each other

2. Their diagonal intersects at an angle of 90 degrees

7. The diagonals of a parallelogram separates it into how many congruent triangles?

The diagonals of a parallelogram separate it into two congruent triangles.

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