NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
In the world of geometry, quadrilaterals teach us that shape and structure can come in many forms. Have you noticed that the things surrounding you, like windows, doors, your notebook or tablet, all have a similar four-sided shape? Well, these four-sided shapes are called quadrilaterals. The beauty of quadrilaterals lies in their diversity—trapeziums, parallelograms, rectangles, squares, and rhombuses all under one family. From NCERT Class 9 Maths, the chapter Quadrilaterals, students will learn the concepts of Properties of a Parallelogram, theorems related to angles and sides, the Mid-point Theorem, etc. These NCERT Solutions for Class 9 will help the students grasp more advanced geometry concepts easily and enhance their problem-solving ability in real-world applications.
This Story also Contains
- Quadrilaterals Class 9 Questions And Answers PDF Free Download
- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals: Exercise Questions
- Quadrilaterals Class 9 NCERT Solutions: Exercise-wise
- Class 11 Maths NCERT Chapter 8: Extra Question
- Quadrilaterals Class 9 Chapter 8: Topics
- Quadrilaterals Class 9 NCERT Solutions - Important Formulae And Points
- Approach to Solve Questions of Quadrilaterals Class 9
- NCERT Solutions For Class 9 Maths - Chapter Wise
Understanding the properties of sides, angles, and diagonals is key to mastering Quadrilaterals. This article on NCERT Solutions for class 9 Maths provides easy-to-follow step-by-step solutions for all the exercise problems. Students who are searching for Quadrilaterals class 9 solutions will find this article beneficial. The article covers all the important Class 9 Maths Chapter 8 question answers. These NCERT Solutions are trustworthy and reliable, as they are created by subject matter experts at Careers360, making them an essential resource for exam preparation. For full syllabus coverage and solved exercises as well as a downloadable PDF, please visit this NCERT article.
Quadrilaterals Class 9 Questions And Answers PDF Free Download
To make maths learning easier, Careers360 experts have created these NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals. Students can also access these solutions in PDF form.
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals: Exercise Questions
| Class 9 Maths Chapter 8 NCERT Solutions - Exercise: 8.1 Page number: 110-111 Total Questions: 7 |
Question 1: 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 a parallelogram)
AC=BD (Given)
AB=AB (common)
△ ABC $\cong$ △ BAD (By SSS)
$\angle$ABC=$\angle$BAD (CPCT)
and $\angle$ABC+$\angle$BAD=180° (co - interior angles)
2$\angle$BAD=180°
$\angle$BAD=90°
Hence, it is a rectangle.
Question 2: 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 $\angle$COD=90°
Proof : In △ BAD and △ ABC,
$\angle$BAD=$\angle$ABC (Each 90° )
AD=BC (Given )
AB=AB (common)
△ BAD $\cong$ △ ABC (By SAS)
BD=AC (CPCT)
In △ AOB and △ COD,
$\angle$ OAB= $\angle$ OCD (Alternate angles)
AB=CD (Given )
$\angle$ OBA= $\angle$ ODC (Alternate angles)
△ AOB $\cong$ △ COD (By AAS)
AO=OC ,BO=OD (CPCT)
In △ AOB and △ AOD,
OB=OD (proved above)
AB=AD (Given )
OA=OA (COMMON)
△ AOB $\cong$ △ AOD (By SSS)
$\angle$ AOB= $\angle$ AOD (CPCT)
$\angle$ AOB+ $\angle$ AOD = 180°
2. $\angle$ AOB = 180°
$\angle$ AOB = 90°
Hence, the diagonals of a square are equal and bisect each other at right angles.
Question 3: (i) Diagonal AC of a parallelogram ABCD bisects $\angle$A (see Fig. 8.19 ). Show that it bisects $\angle$C also.
Answer:
Given: $\angle$ DAC= $\angle$ BAC ................1
$\angle$ DAC= $\angle$ BCA.................2 (Alternate angles)
$\angle$ BAC= $\angle$ ACD .................3 (Alternate angles)
From equations 1,2 and 3, we get
$\angle$ ACD= $\angle$ BCA...................4
Hence, diagonal AC bisect angle C also.
Question 3: (ii) Diagonal AC of a parallelogram ABCD bisects $\angle$A (see Fig. 8.19 ). Show that ABCD is a rhombus.
Answer:
Given: $\angle$ DAC= $\angle$ BAC ................1
$\angle$ DAC= $\angle$ BCA.................2 (Alternate angles)
$\angle$ BAC= $\angle$ ACD .................3 (Alternate angles)
From equations 1,2, and 3, we get
$\angle$ ACD= $\angle$ BCA...................4
From 2 and 4, we get
$\angle$ ACD= $\angle$ DAC
In △ 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.
Question 4: (i) ABCD is a rectangle in which diagonal AC bisects $\angle$A as well as $\angle$C . Show that: ABCD is a square.
Answer:
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 △ 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.
Question 4: (ii) ABCD is a rectangle in which diagonal AC bisects $\angle$A as well as $\angle$C . Show that: Diagonal BD bisects $\angle$B as well as $\angle$D.
Answer:
In △ 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.
Question 5: (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see Fig. 8.20 ). Show that: ΔAPD$\cong$ΔCQB
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ.
To prove : ΔAPD$\cong$ΔCQB
Proof :
In ΔAPDandΔCQB,
DP=BQ (Given )
$\angle$ ADP= $\angle$ CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
ΔAPD$\cong$ΔCQB (By SAS)
Question 5: (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see Fig. 8.20 ). Show that: AP=CQ
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ .
To prove : AP=CQ
Proof :
In ΔAPDandΔCQB,
DP=BQ (Given )
$\angle$ ADP= $\angle$ CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
ΔAPD$\cong$ΔCQB (By SAS)
AP=CQ (CPCT)
Question 5: (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see Fig. 8.20 ). Show that: ΔAQB$\cong$ΔCPD
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ .
To prove : ΔAQB$\cong$ΔCPD
Proof :
In ΔAQBandΔCPD,
DP=BQ (Given )
$\angle$ ABQ= $\angle$ CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
ΔAQB$\cong$ΔCPD (By SAS)
Question 5: (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see Fig. 8.20 ). Show that: AQ=CP
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ.
To prove : AQ=CP
Proof :
In ΔAQBandΔCPD,
DP=BQ (Given )
$\angle$ ABQ= $\angle$ CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
ΔAQB$\cong$ΔCPD (By SAS)
AQ=CP (CPCT)
Question 5: (v) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see Fig. 8.20 ). Show that: APCQ is a parallelogram
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ.
To prove: APCQ is a parallelogram
Proof :
In ΔAPDandΔCQB,
DP=BQ (Given )
$\angle$ ADP= $\angle$ CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
ΔAPD$\cong$ΔCQB (By SAS)
AP=CQ (CPCT)...............................................................(1)
Also,
In ΔAQBandΔCPD,
DP=BQ (Given )
$\angle$ ABQ= $\angle$ CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
ΔAQB$\cong$ΔCPD (By SAS)
AQ=CP (CPCT)........................................(2)
From equations 1 and 2, we get
AP=CQ
AQ=CP
Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.
Question 6: (i) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21 ). Show that ΔAPB$\cong$ΔCQD
Answer:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove : ΔAPB$\cong$ΔCQD
Proof: In ΔAPBandΔCQD ,
$\angle$ APB= $\angle$ CQD (Each 90° )
$\angle$ ABP= $\angle$ CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, ΔAPB$\cong$ΔCQD (By SAS)
Question 6: (ii) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21 ). Show that AP=CQ
Answer:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove : AP=CQ
Proof: In ΔAPBandΔCQD ,
$\angle$ APB= $\angle$ CQD (Each 90° )
$\angle$ ABP= $\angle$ CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, ΔAPB$\cong$ΔCQD (By SAS)
AP=CQ (CPCT)
Question 7: (i) ABCD is a trapezium in which AB∥CD and AD=BC (see Fig. 8.23 ). Show that $\angle$A=$\angle$B
[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 AB∥CD and AD=BC
To prove : $\angle$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)
AD||CE (By construction)
Hence, AECD is a parallelogram.
AD=CE...............(1)(opposite sides of a parallelogram)
AD=BC.................(2)(Given)
From equations 1 and 2, we get,
CE=BC
In △ BCE,
$\angle$3=$\angle$4 .................3 (opposite angles of equal sides)
$\angle$2+$\angle$3=180° ...................4(linear pairs)
$\angle$1+$\angle$4=180° .....................5(Co-interior angles)
From 4 and 5, we get
$\angle$2+$\angle$3=$\angle$1+$\angle$4
∴$\angle$2=$\angle$1→$\angle$B=$\angle$A (Since, $\angle$3=$\angle$4 )
Question 7: (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that $\angle$C=$\angle$D
[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 AB∥CD and AD=BC
To prove : $\angle$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° (Co-interior angles)
$\angle$2+$\angle$C=180° (Co-interior angles)
∴$\angle$1+$\angle$D=$\angle$2+$\angle$C
Thus, $\angle$C=$\angle$D (Since , $\angle$1=$\angle$2 )
Question 7: (iii) ABCD is a trapezium in which AB∥CD and AD=BC (see Fig. 8.23 ). Show that ΔABC$\cong$ΔBAD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which AB∥CD and AD=BC
To prove : ΔABC$\cong$ΔBAD
Proof: In ΔABCandΔBAD ,
BC=AD (Given )
AB=AB (Common )
$\angle$ABC=$\angle$BAD (proved in (i) )
Thus, ΔABC$\cong$ΔBAD (By SAS rule)
Question 7: (iv) ABCD is a trapezium in which AB∥CD and AD=BC (see Fig. 8.23 ). 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 AB∥CD and AD=BC
To prove: diagonal AC = diagonal BD
Proof: In ΔABCandΔBAD ,
BC=AD (Given )
AB=AB (Common )
$\angle$ABC=$\angle$BAD (proved in (i) )
Thus, ΔABC$\cong$ΔBAD (By SAS rule)
diagonal AC = diagonal BD (CPCT)
| Class 9 Maths Chapter 8 NCERT Solutions - Exercise: 8.2 Page Number: 113-114 Total Questions: 6 |
Answer:
Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig. 8.29 ). AC is a diagonal.
To prove :
SR∥AC and SR=12AC
Proof: In △ ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By the midpoint theorem,
SR∥AC and SR=12AC
Question 1: (ii) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29 ). AC is a diagonal. Show that : PQ=SR
Answer:
Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig. 8.29 ). AC is a diagonal.
To prove : PQ=SR
Proof : In △ ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By the mid-point theorem,
SR∥AC and SR=12AC ...................................(1)
In △ ABC,
P is mid point of AB. (Given)
Q is mid point of BC. (Given)
By mid-point theorem,
PQ∥AC and PQ=12AC .................................(2)
From equations 1 and 2, we get
PQ∥SR and PQ=SR=12AC
Thus, PQ=SR
Answer:
Given : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29 ). AC is a diagonal.
To prove : PQRS is a parallelogram.
Proof : In PQRS,
Since,
PQ∥SR and PQ=SR .
So, PQRS is a parallelogram.
Answer:
Given: ABCD is a rhombus in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA . AC, BD are diagonals.
To prove: the quadrilateral PQRS is a rectangle.
Proof: In △ ACD,
S is midpoint of DA. (Given)
R is midpoint of DC. (Given)
By the midpoint theorem,
SR∥AC and SR=12AC ...................................(1)
In △ ABC,
P is the midpoint of AB. (Given)
Q is mid point of BC. (Given)
By the mid-point theorem,
PQ∥AC and PQ=12AC .................................(2)
From 1 and 2, we get
PQ∥SR and PQ=SR=12AC
Thus, PQ=SR and PQ∥SR
So, the quadrilateral PQRS is a parallelogram.
Similarly, in △ BCD,
Q is mid point of BC. (Given)
R is mid point of DC. (Given)
By mid point theorem,
QR∥BD
So, QN || LM ...........(5)
LQ || MN ..........(6) (Since, PQ || AC)
From 5 and 6, we get
LMPQ is a parallelogram.
Hence, $\angle$ LMN= $\angle$ LQN (opposite angles of the parallelogram)
But, $\angle$ LMN= 90 (Diagonals of a rhombus are perpendicular)
so, $\angle$ LQN=90
Thus, a parallelogram whose one angle is a right angle is a rectangle.
Hence, PQRS is a rectangle.
Answer:
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.
Proof :
In △ ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By the midpoint theorem,
SR∥AC and SR=12AC ...................................(1)
In △ ABC,
P is the midpoint of AB. (Given)
Q is the midpoint of BC. (Given)
By the midpoint theorem,
PQ∥AC and PQ=12AC .................................(2)
From 1 and 2, we get
PQ∥SR and PQ=SR=12AC
Thus, PQ=SR and PQ∥SR
So, the quadrilateral PQRS is a parallelogram.
Similarly, in △ BCD,
Q is the midpoint of BC. (Given)
R is the midpoint of DC. (Given)
By the midpoint theorem,
QR∥BD and QR=12BD ...................(5)
AC = BD.......................(6)(diagonals )
From 2, 5 and 6, we get
PQ=QR
Thus, a parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.
Answer:
Given: ABCD is a trapezium in which AB∥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 Fig. 8.30 ).
To prove: F is the mid-point of BC.
In △ ABD,
E is the midpoint of AD. (Given)
EG || AB (Given)
By converse of midpoint theorem,
G is the midpoint of BD.
In △ BCD,
G is mid point of BD. (Proved above)
FG || DC (Given)
By converse of midpoint theorem,
F is the midpoint of BC.
Answer:
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.
Proof : In quadrilatweral ABCD,
AB=CD (Given)
12AB=12CD
⇒AE=CF (E and F are midpoints of AB and CD)
In quadrilateral AECF,
AE=CF (Given)
AE || CF (Opposite sides of a parallelogram)
Hence, AECF is a parallelogram.
In △ DCQ,
F is the midpoint of DC. (given )
FP || CQ (AECF is a parallelogram)
By the converse of the midpoint theorem,
P is the midpoint of DQ.
DP= PQ....................(1)
Similarly,
In △ ABP,
E is the midpoint of AB. (given )
EQ || AP (AECF is a parallelogram)
By converse of midpoint theorem,
Q is the midpoint of PB.
OQ= QB....................(2)
From 1 and 2, we have
DP = PQ = QB.
Hence, the line segments AF and EC trisect the diagonal BD.
Answer:
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 :D is mid point of AC.
Proof: In △ ABC,
M is mid point of AB. (Given)
DM || BC (Given)
By converse of mid point theorem,
D is the mid point of AC.
Question 6: (ii) 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 MD⊥AC
Answer:
Given: ABC is a triangle right angled at C. A line through the mid-point M of the hypotenuse AB and parallel to BC intersects AC at D.
To prove : MD⊥AC
Proof : $\angle$ ADM = $\angle$ ACB (Corresponding angles)
$\angle$ ADM= 90° . ( $\angle$ ACB = 90° )
Hence, MD⊥AC .
Question 6: (iii) 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 CM=MA=12AB
Answer:
Given: ABC is a triangle right angled at C. A line through the mid-point M of the hypotenuse AB and parallel to BC intersects AC at D.
To prove : CM=MA=12AB
Proof : In △ ABC,
M is the midpoint of AB. (Given)
DM || BC (Given)
By the converse of the midpoint theorem,
D is the midpoint of AC i.e. AD = DC.
In △ AMD and △ CMD,
AD = DC (proved above)
$\angle$ ADM = $\angle$ CDM (Each right angle)
DM = DM (Common)
△ AMD $\cong$ △ CMD (By SAS)
AM = CM (CPCT)
But , AM=12AB
Hence, CM=MA=12AB.
Quadrilaterals Class 9 NCERT Solutions: Exercise-wise
Exercise-wise NCERT Solutions of Quadrilaterals Class 9 Maths Chapter 8 are provided in the link below.
Class 11 Maths NCERT Chapter 8: Extra Question
Question: In quadrilateral ABCD, AB parallel CD, and $\angle$A = 70°, $\angle$D = 110°. Find $\angle$B and $\angle$C.
Answer:
We know that, in a quadrilateral with one pair of opposite sides parallel, adjacent angles on the same side of a transversal are supplementary.
Now, by the angle sum property of a quadrilateral:
$\angle$A+$\angle$B+$\angle$C+$\angle$D = 360°
70°+\angle B+\angle C$ + 110° = 360°
$\angle$B + $\angle$C = 180°
As we are given that AB is parallel to CD
So, $\angle$B=$\angle$C (alternate interior angles)
Therefore, $\angle$B=$\angle$C = 90°
Quadrilaterals Class 9 Chapter 8: Topics
Students will explore the following topics in NCERT Class 9 Maths Chapter 8 Quadrilaterals:
- Properties of a Parallelogram
- The Mid-point Theorem
- Summary
Quadrilaterals Class 9 NCERT Solutions - Important Formulae And Points
Quadrilateral:
-
Sum of all angles = 360°
Parallelogram:
-
A diagonal of a parallelogram divides it into two congruent triangles.
-
In a parallelogram, the diagonals bisect each other.
-
In a parallelogram, opposite angles are equal.
-
In a parallelogram, opposite sides are equal.
Square:
-
Diagonals of a square bisect each other at right angles and are equal, and vice versa.
Triangle:
-
A line through the midpoint of a side of a triangle parallel to another side bisects the third side (Midpoint theorem).
-
The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half the third side.
Parallelogram Angle Bisectors:
-
In a parallelogram, the bisectors of any two consecutive angles intersect at a right angle.
-
If a diagonal of a parallelogram bisects one of the angles of a parallelogram, it also bisects the second angle.
Rectangle:
-
The angle bisectors of a parallelogram form a rectangle.
-
Each of the four angles of a rectangle is a right angle.
Rhombus:
-
The diagonals of a rhombus are perpendicular to each other.
Approach to Solve Questions of Quadrilaterals Class 9
Using these approaches, students can tackle the Quadrilaterals Class 9 Chapter 8 Question Answers with greater confidence.
1. Learn basic properties of quadrilaterals: Any shape that comprises four sides proves to be a quadrilateral, and its total internal angles create a 360-degree circle.
2. Identify types of quadrilaterals: The study of special parallelograms and rhombuses and rectangles, and trapeziums, along with their specific properties, should be studied.
3. Use angle sum property: Solving for unknown angles requires using the formula, which calculates that $\angle$A + $\angle$B + $\angle$C + $\angle$D = 360.
4. Apply properties of parallelograms: The shape contains equal and parallel sides, which have identical opposite angles.
5. Use diagonal properties: The diagonals of rectangles and rhombuses hold either equal measurement lengths or bisect each other, providing valuable tools for question verification.
6. Solve reasoning-based and proof questions: Develop proof statements together with supporting reasons to show whether a particular quadrilateral qualifies as a parallelogram.
NCERT Solutions For Class 9 Maths - Chapter Wise
For students' preparation, Careers360 has gathered all Class 9 Maths NCERT solutions here for quick and convenient access.
Also, read,
- NCERT Notes Class 9 Maths Chapter 8 Quadrilaterals
- NCERT Exemplar Solutions for Class 9 Maths Chapter 8 Quadrilaterals
NCERT Class 9 Books and Syllabus
Given below are some useful links for NCERT books and the NCERT syllabus for class 9:
Frequently Asked Questions (FAQs)
Q: Where can I download NCERT Solutions for Quadrilaterals?
A:
NCERT Quadrilaterals Class 9 solutions PDFs are freely available on various trusted educational platforms such as Careers360.
Q: What are the important topics in maths chapter 8 class 9 Quadrilaterals?
A:
Angle sum property of a quadrilateral, properties of the parallelogram, another condition for a quadrilateral to be a parallelogram, and the mid-point theorem are the important topics covered in NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals.
Q: How does the NCERT solutions are helpful ?
A:
NCERT solutions are not only helpful for students if they get stuck while solving NCERT problems, but also it will provide them with new ways to solve the problems. We at Careers360 provide these solutions in a very detailed manner, giving students conceptual clarity.
Q: Which theorems are important to deal with class 9 maths chapter 8 question answer?
A:
Here is the list of theorems that are important for this chapter.
A diagonal of a parallelogram divides it into two congruent triangles.
Opposite sides of a parallelogram are equal.
Opposite angles of a parallelogram are equal.
Diagonals of a parallelogram bisect each other.
A quadrilateral is a parallelogram if opposite sides/angles are equal or if its diagonals bisect each other.
Q: Why are NCERT Solutions for Chapter 8 useful?
A:
They provide step-by-step proofs of all theorems, solved examples, and exercises with logical reasoning, making it easier for students to understand geometry.
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| 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 |
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