VMC VIQ Scholarship Test
ApplyRegister for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.
Search Colleges, Exams, Schools & more
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- Download Free PDF
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.
Scholarship Test: Vidyamandir Intellect Quest (VIQ) | ALLEN TALLENTEX
Don't Miss: JEE Main 2027: Narayana Scholarship Test Preparation Kit for Class 9
This Story also Contains
- NCERT Solutions for class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1- Download Free PDF
- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1
- Access Quadrilaterals Class 9 Chapter 8 Exercise: 8.1
- More About NCERT Solutions for Class 9 Maths Exercise 8.1
- Benefits of NCERT Solutions for Class 9 Maths Exercise 8.1
- Key Features of 9th Class Maths Exercise 8.1 Answers
- NCERT Solutions of Class 10 Subject Wise
- Subject Wise NCERT Exemplar Solutions
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.
Some points to be noted about the types of quadrilaterals are:
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.
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1
Access Quadrilaterals Class 9 Chapter 8 Exercise: 8.1
Q1 The angles of quadrilateral are in the ratio . Find all the angles of the quadrilateral.
Answer:
Given : The angles of a quadrilateral are in the ratio .
Let the angles of quadrilateral be .
Sum of all angles is 360.
Thus,
All four angles are :
Q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Answer:
Given: ABCD is a parallelogram with AC=BD.
To prove: ABCD is a rectangle.
Proof: In ABC and
BAD,
BC= AD (Opposite sides of parallelogram)
AC=BD (Given )
AB=AB (common)
ABC
BAD (By SSS)
(CPCT)
and (co - interior angles)
Hence, it is a rectangle.
Answer:
Given: the diagonals of a quadrilateral bisect each other at right angles i.e.BO=OD.
To prove: ABCD is a rhombus.
Proof : In AOB and
AOD,
(Each
)
BO=OD (Given )
AO=AO (common)
AOB
AOD (By SAS)
AB=AD (CPCT)
Similarly, AB=BC and BC=CD
Hence, it is a rhombus.
Q4 Show that the diagonals of a square are equal and bisect each other at right angles.
Answer:
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
Proof : In BAD and
ABC,
(Each
)
AD=BC (Given )
AB=AB (common)
BAD
ABC (By SAS)
BD=AC (CPCT)
In AOB and
COD,
OAB=
OCD (Alternate angles)
AB=CD (Given )
OBA=
ODC (Alternate angles)
AOB
COD (By AAS)
AO=OC ,BO=OD (CPCT)
In AOB and
AOD,
OB=OD (proved above)
AB=AD (Given )
OA=OA (COMMON)
AOB
AOD (By SSS)
AOB=
AOD (CPCT)
AOB+
AOD =
2. AOB =
AOB =
Hence, the diagonals of a square are equal and bisect each other at right angles.
Answer:
Given : ABCD is a quadrilateral with AC=BD,AO=CO,BO=DO, COD =
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 BAD and
ABC,
AD=BC (proved above )
AB=AB (common)
BD=AC
BAD
ABC (By SSS)
(CPCT)
BAD+
ABC =
(Co-interior angles)
2. ABC =
ABC =
Hence, the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Q6 (i) Diagonal AC of a parallelogram ABCD bisects (see Fig.
). Show that
it bisects also.
Answer:
Given: DAC=
BAC ................1
DAC=
BCA.................2 (Alternate angles)
BAC=
ACD .................3 (Alternate angles)
From equation 1,2 and 3, we get
ACD=
BCA...................4
Hence, diagonal AC bisect angle C also.
Q6 (ii) Diagonal AC of a parallelogram ABCD bisects (see Fig.
). Show that
Answer:
Given: DAC=
BAC ................1
DAC=
BCA.................2 (Alternate angles)
BAC=
ACD .................3 (Alternate angles)
From equation 1,2 and 3, we get
ACD=
BCA...................4
From 2 and 4, we get
ACD=
DAC
In ADC,
ACD=
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.
Q7 ABCD is a rhombus. Show that diagonal AC bisects as well as
and diagonal BD bisects
as well as
.
Answer:
In ADC,
AD = CD (ABCD is a rhombus)
3=
1.................1(angles opposite to equal sides are equal )
3=
2.................2 (alternate angles)
From 1 and 2, we have
1=
2.................3
and 1=
4.................4 (alternate angles)
From 1 and 4, we get
3=
4.................5
Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.
In ADB,
AD = AB (ABCD is a rhombus)
5=
7.................6(angles opposite to equal sides are equal )
7=
6.................7 (alternate angles)
From 6 and 7, we have
5=
6.................8
and 5=
8.................9(alternate angles)
From 6 and 9, we get
7=
8.................10
Hence, from 8 and 10, diagonal BD bisects angles B as well as angle D.
Q8 (i) ABCD is a rectangle in which diagonal AC bisects as well as
. Show that:
Answer:
Given: ABCD is a rectangle with AB=CD and BC=AD 1=
2 and
3=
4.
To prove: ABCD is a square.
Proof : 1=
4 .............1(alternate angles)
3=
4 ................2(given )
From 1 and 2, 1=
3.....................................3
In ADC,
1=
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.
Q8 (ii) ABCD is a rectangle in which diagonal AC bisects as well as
. Show that:
diagonal BD bisects as well as
.
Answer:
In ADB,
AD = AB (ABCD is a square)
5=
7.................1(angles opposite to equal sides are equal )
5=
8.................2 (alternate angles)
From 1 and 2, we have
7=
8.................3
and 7 =
6.................4(alternate angles)
From 1 and 4, we get
5=
6.................5
Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.
Q9 (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig.
). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ADP=
CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
Q9 (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig.
). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ADP=
CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
(CPCT)
Q9 (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig.
). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ABQ=
CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
Q9 (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig.
). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ABQ=
CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
(CPCT)
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove: APCQ is a parallelogram
Proof :
In
DP=BQ (Given )
ADP=
CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
(CPCT)...............................................................1
Also,
In
DP=BQ (Given )
ABQ=
CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
(CPCT)........................................2
From equation 1 and 2, we get
Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.
Answer:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove :
Proof: In ,
APB=
CQD (Each
)
ABP=
CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, (By SAS)
Answer:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove :
Proof: In ,
APB=
CQD (Each
)
ABP=
CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, (By SAS)
(CPCT)
Answer:
Given : In and
,
and
.
To prove : quadrilateral ABED is a parallelogram
Proof : In ABED,
AB=DE (Given)
AB||DE (Given )
Hence, quadrilateral ABED is a parallelogram.
Answer:
Given: In and
,
and
.
To prove: quadrilateral BEFC is a parallelogram
Proof: In BEFC,
BC=EF (Given)
BC||EF (Given )
Hence, quadrilateral BEFC is a parallelogram.
Answer:
To prove: and
Proof :
In ABED,
AD=BE.................1(ABED is a parallelogram)
AD||BE .................2(ABED is a parallelogram)
In BEFC,
BE=CF.................3(BEFC is a parallelogram)
BE||CF .................4(BEFC is a parallelogram)
From 2 and 4 , we get
AD||CF
From 1 and 3, we get
AD=CF
Answer:
To prove : quadrilateral ACFD is a parallelogram
Proof :
In ABED,
AD=BE.................1(ABED is a parallelogram)
AD||BE .................2(ABED is a parallelogram)
In BEFC,
BE=CF.................3(BEFC is a parallelogram)
BE||CF .................4(BEFC is a parallelogram)
From 2 and 4 , we get
AD||CF...........................5
From 1 and 3, we get
AD=CF...........................6
From 5 and 6, we get
AD||CF and AD=CF
Thus, quadrilateral ACFD is a parallelogram
Answer:
In ACFD,
AC=DF (Since, ACFD is a parallelogram which is prooved in part (iv) of the question)
Answer:
In and
,
AB=DE (Given )
BC=EF (Given )
AC=DF ( proved in (v) part)
(By SSS rule)
Q12 (i) ABCD is a trapezium in which and
(see Fig.
). Show that
[ Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which and
To prove :
Proof: Let A be
1,
ABC be
2,
EBC be
3,
BEC be
4.
In AECD,
AE||DC (Given)
AD||CE (By cnstruction)
Hence, AECD is a parallelogram.
AD=CE...............1(opposite sides of a parallelogram)
AD=BC.................2(Given)
From 1 and 2, we get
CE=BC
In BCE,
.................3 (opposite angles of equal sides)
...................4(linear pairs)
.....................5(Co-interior angles)
From 4 and 5, we get
(Since,
)
Q12 (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that
[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which and
To prove :
Proof: Let A be
1,
ABC be
2,
EBC be
3,
BEC be
4.
(Co-interior angles)
(Co-interior angles)
Thus, (Since ,
)
Q12 (iii) ABCD is a trapezium in which and
(see Fig.
). Show that
[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which and
To prove :
Proof: In ,
BC=AD (Given )
AB=AB (Common )
(proved in (i) )
Thus, (By SAS rule)
Q12 (iv) ABCD is a trapezium in which and
(see Fig.
). Show that diagonal AC
diagonal BD
[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which and
To prove: diagonal AC diagonal BD
Proof: In ,
BC=AD (Given )
AB=AB (Common )
(proved in (i) )
Thus, (By SAS rule)
diagonal AC 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.
Also Read| Quadrilaterals Class 9 Notes
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
Basic Understanding: The answers in this class 9 maths ex 8.1 provide a fundamental understanding of quadrilaterals, their properties, and classifications.
Clear Explanations: Class 9 ex 8.1 offer clear and concise explanations for each question, making it easier for students to comprehend the concepts.
Step-by-Step Solutions: The answers provide step-by-step solutions, helping students follow the problem-solving process easily.
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.
Practice Material: Exercise 8.1 answers serve as valuable practice material for students to reinforce their knowledge of quadrilaterals.
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
NCERT Solutions of Class 10 Subject Wise
Subject Wise NCERT Exemplar Solutions
Frequently Asked Questions (FAQs)
1. What is the main concept of NCERT solutions for Class 9 Maths exercise 8.1?
Quadrilateral, types of quadrilaterals and their properties and different theorem.
2. Define quadrilateral?
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?
Different types of quadrilaterals are:
Square
Rectangle
Rhombus
Trapezium
Parallelogram
Kite
6. What are the similarities between square and rhombus?
The similarities between square and rhombus are:
All the sides are equal to each other
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.
Also Read
Articles
Sep 17, 2024
Upcoming School Exams
Application Date:09 September,2024 - 14 November,2024
Application Date:09 September,2024 - 14 November,2024
Admit Card Date:13 September,2024 - 07 October,2024
Admit Card Date:13 September,2024 - 07 October,2024
Get answers from students and experts
Popular Questions
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
In the reaction,
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
| Option 1)
0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
Colleges After 12th
Popular Course After 12th
Applications for Admissions are open.
Tallentex 2025 - ALLEN's Talent Encouragement Exam
ApplyRegister for Tallentex '25 - One of The Biggest Talent Encouragement Exam
JEE Main Important Physics formulas
ApplyAs per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
JEE Main Important Chemistry formulas
ApplyAs per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
TOEFL ® Registrations 2024
ApplyAccepted by more than 11,000 universities in over 150 countries worldwide
Pearson | PTE
ApplyRegister now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Limited Period Offer! Trusted by 3,500+ universities globally
News and Notifications
Download Careers360 App's
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile
300M+
Students
36,000+
Colleges
550+
Exams
1500+
E-Books
16000+
Certifications
We Appeared in
Back to top