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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
Edited By Komal Miglani | Updated on Apr 30, 2025 07:41 PM IST
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. Quadrilaterals can be of various types, such as a Parallelogram, a Rhombus, a Trapezium, etc. From NCERT Class 9 Maths, the chapter Quadrilaterals contains the concepts of Properties of a Parallelogram, theorems related to angles and sides, the Mid-point Theorem, etc. These concepts will help the students grasp more advanced geometry concepts easily and enhance their problem-solving ability in real-world applications through the NCERT Solutions.
This Story also Contains
- Quadrilaterals Class 9 Questions And Answers PDF Free Download
- Quadrilaterals Class 9 NCERT Solutions - Important Formulae And Points
- NCERT Solutions for Class 9 Maths Chapter 8: Exercise Questions
- Class 11 Maths NCERT Chapter 5: Extra Question
- Approach to Solve Questions of Quadrilaterals Class 9
- Quadrilaterals Class 9 Solutions - Exercise Wise
- NCERT Solutions For Class 9 Maths - Chapter Wise
- NCERT Solutions For Class 9 - Subject Wise
- NCERT Class 9 Books and Syllabus
This article on NCERT solutions for class 9 Maths Chapter 8 Quadrilaterals offers clear and step-by-step solutions for the exercise problems. Students who are in need of Quadrilaterals class 9 solutions will find this article very useful. It covers all the important Class 9 Maths Chapter 8 question answers. These Quadrilaterals class 9 ncert solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT Solutions for Class 9 and NCERT Solutions for other subjects and classes can be downloaded from the website.
Quadrilaterals Class 9 Questions And Answers PDF Free Download
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.
NCERT Solutions for Class 9 Maths Chapter 8: 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 $\triangle$ 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 $\angle ABC+\angle BAD=180 ^\circ$ (co - interior angles)
$2\angle BAD=180 ^\circ$
$\angle BAD= 90 ^\circ$
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 $\angle COD=90 ^\circ$
Proof : In $\triangle$ BAD and $\triangle$ ABC,
$\angle BAD = \angle ABC$ (Each $90 ^\circ$ )
AD=BC (Given )
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)
AB=AD (Given )
OA=OA (COMMON)
$\triangle$ AOB $\cong$ $\triangle$ AOD (By SSS)
$\angle$ AOB= $\angle$ AOD (CPCT)
$\angle$ AOB+ $\angle$ AOD = $180 ^\circ$
2. $\angle$ AOB = $180 ^\circ$
$\angle$ AOB = $90 ^\circ$
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 $\small \angle A$ (see Fig. $\small 8.19$ ). Show that it bisects $\small\angle C$ also.
Answer:
Given: $\angle$ DAC= $\angle$ BAC ................1
$\angle$ DAC= $\angle$ BCA.................2 (Alternate angles)
$\angle$ BAC= $\angle$ ACD .................3 (Alternate angles)
From equation 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 $\small \angle A$ (see Fig. $\small 8.19$ ). 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 $\triangle$ 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.
Question 4: (i) ABCD is a rectangle in which diagonal AC bisects $\small \angle A$ as well as $\small \angle C$ . 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 $\triangle$ 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.
Answer:
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.
Answer:
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)
Question 5: (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ (see Fig. $\small 8.20$ ). Show that: $\small AP=CQ$
Answer:
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)
Answer:
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)
Question 5: (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ (see Fig. $\small 8.20$ ). Show that: $\small AQ=CP$
Answer:
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)
Answer:
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.
Answer:
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 $90 ^\circ$ )
$\angle$ ABP= $\angle$ CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, $\small \Delta APB\cong \Delta CQD$ (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. $\small 8.21$ ). Show that $\small AP=CQ$
Answer:
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 $90 ^\circ$ )
$\angle$ 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)
Question 7: (i) ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$ (see Fig. $\small 8.23$ ). Show that $\small \angle A=\angle B$
[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 $\small AB\parallel CD$ 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)
AD||CE (By construction)
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 $\triangle$ BCE,
$\angle 3=\angle 4$ .................3 (opposite angles of equal sides)
$\angle 2+\angle 3=180 ^\circ$ ...................4(linear pairs)
$\angle 1+\angle 4=180 ^\circ$ .....................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, $\angle 3=\angle 4$ )
Question 7: (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that $\small \angle C=\angle D$
[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 $\small AB\parallel CD$ 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.
$\angle 1+\angle D= 180 ^\circ$ (Co-interior angles)
$\angle 2+\angle C= 180 ^\circ$ (Co-interior angles)
$\therefore \angle 1+\angle D=\angle 2+\angle C$
Thus, $\small \angle C=\angle D$ (Since , $\small \angle 1=\angle 2$ )
Question 7: (iii) ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$ (see Fig. $\small 8.23$ ). Show that $\small \Delta ABC\cong \Delta BAD$
[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 $\small AB\parallel CD$ and $\small AD=BC$
To prove : $\small \Delta ABC\cong \Delta BAD$
Proof: In $\small \Delta ABC\, \, and\, \, \, \Delta BAD$ ,
BC=AD (Given )
AB=AB (Common )
$\angle ABC=\angle BAD$ (proved in (i) )
Thus, $\small \Delta ABC\cong \Delta BAD$ (By SAS rule)
Question 7: (iv) ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$ (see Fig. $\small 8.23$ ). Show that diagonal AC $\small =$ 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 $\small AB\parallel CD$ and $\small AD=BC$
To prove: diagonal AC $\small =$ diagonal BD
Proof: In $\small \Delta ABC\, \, and\, \, \, \Delta BAD$ ,
BC=AD (Given )
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)
Class 9 maths chapter 8 question answer - 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. $\small 8.29$ ). AC is a diagonal.
To prove :
$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$
Proof: In $\triangle$ ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$
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. $\small 8.29$ ). AC is a diagonal.
To prove : $\small PQ=SR$
Proof : In $\triangle$ ACD,
S is mid point of DA. (Given)
R is mid point of DC. (Given)
By mid-point theorem,
$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1
In $\triangle$ ABC,
P is mid point of AB. (Given)
Q is mid point of BC. (Given)
By mid-point theorem,
$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2
From 1 and 2,we get
$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$
Thus, $\small PQ=SR$
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 $\small 8.29$ ). AC is a diagonal.
To prove : PQRS is a parallelogram.
Proof : In PQRS,
Since,
$\small PQ\parallel SR$ and $\small PQ=SR$ .
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 $\triangle$ ACD,
S is midpoint of DA. (Given)
R is midpoint of DC. (Given)
By midpoint theorem,
$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1
In $\triangle$ ABC,
P is midpoint of AB. (Given)
Q is mid point of BC. (Given)
By mid point theorem,
$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2
From 1 and 2,we get
$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$
Thus, $\small PQ=SR$ and $\small PQ\parallel SR$
So,the quadrilateral PQRS is a parallelogram.
Similarly, in $\triangle$ BCD,
Q is mid point of BC. (Given)
R is mid point of DC. (Given)
By mid point theorem,
$\small QR\parallel BD$
So, QN || LM ...........5
LQ || MN ..........6 (Since, PQ || AC)
From 5 and 6, we get
LMPQ is a parallelogram.
Hence, $\small \angle$ LMN= $\small \angle$ LQN (opposite angles of the parallelogram)
But, $\small \angle$ LMN= 90 (Diagonals of a rhombus are perpendicular)
so, $\small \angle$ 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 $\triangle$ ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1
In $\triangle$ ABC,
P is the midpoint of AB. (Given)
Q is the midpoint of BC. (Given)
By midpoint theorem,
$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2
From 1 and 2, we get
$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$
Thus, $\small PQ=SR$ and $\small PQ\parallel SR$
So, the quadrilateral PQRS is a parallelogram.
Similarly, in $\triangle$ BCD,
Q is the midpoint of BC. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
$\small QR\parallel BD$ and $\small QR=\frac{1}{2}BD$ ...................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 $\small AB\parallel DC$ , 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. $\small 8.30$ ).
To prove: F is the mid-point of BC.
In $\triangle$ ABD,
E is the midpoint of AD. (Given)
EG || AB (Given)
By converse of midpoint theorem,
G is the midpoint of BD.
In $\triangle$ 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)
$\frac{1}{2}AB=\frac{1}{2}CD$
$\Rightarrow AE=CF$ (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 $\triangle$ DCQ,
F is the midpoint of DC. (given )
FP || CQ (AECF is a parallelogram)
By converse of midpoint theorem,
P is the mid point of DQ.
DP= PQ....................1
Similarly,
In $\triangle$ 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 $\triangle$ 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 $\small MD\perp AC$
Answer:
Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersects AC at D.
To prove : $\small MD\perp AC$
Proof : $\angle$ ADM = $\angle$ ACB (Corresponding angles)
$\angle$ ADM= $90 ^\circ$ . ( $\angle$ ACB = $90 ^\circ$ )
Hence, $\small MD\perp AC$ .
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 $\small CM=MA=\frac{1}{2}AB$
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 : $\small CM=MA=\frac{1}{2}AB$
Proof : In $\triangle$ 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 $\triangle$ AMD and $\triangle$ CMD,
AD = DC (proved above)
$\angle$ ADM = $\angle$ CDM (Each right angle)
DM = DM (Common)
$\triangle$ AMD $\cong$ $\triangle$ CMD (By SAS)
AM = CM (CPCT)
But , $\small AM=\frac{1}{2}AB$
Hence, $\small CM=MA=\frac{1}{2}AB$.
Class 11 Maths NCERT Chapter 5: 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°
Approach to Solve Questions of Quadrilaterals Class 9
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 ∠A + ∠B + ∠C + ∠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.
Quadrilaterals Class 9 Solutions - Exercise Wise
The NCERT Solutions for Class 9 provide detailed steps to solve both numerical and theorem-based quadrilateral problems. Students can practice Class 9 Maths Chapter 8 question answers using the exercise link given below.
NCERT Solutions For Class 9 Maths - Chapter Wise
NCERT Solutions For Class 9 - Subject Wise
Here are the subject-wise links for the NCERT solutions of class 9:
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)
1. What are the important topics in Quadrilaterals ch 8 maths class 9?
Angle sum property of a quadrilateral, properties of the parallelogram, another condition for a quadrilateral to be a parallelogram, and mid-point theorem are the important topics covered in class 9th quadrilateral solution. students can practice these NCERT solutions for class 9 to command the concepts.
2. How does the NCERT solutions are helpful ?
NCERT solutions are not only helpful for the students if they stuck while solving NCERT problems but also it will provide them new ways to solve the problems. These solutions are provided in a very detailed manner which will give them conceptual clarity.
3. In NCERT Solutions for Class 9 Maths Chapter 8, how does the concept of "quadrilaterals" get defined?
Quadrilateral class 9 solutions define a quadrilateral as a two-dimensional shape that has four sides or edges and four corners or vertices. Quadrilaterals are commonly recognized by their standard shapes, such as rectangle, square, trapezoid, and kite, but they can also have irregular and undefined shapes.
4. Where can I find the complete solutions of class 9 chapter 8 maths ?
Here you will get the detailed NCERT solutions for class 9 by clicking on the link. for ease students can study quadrilateral class 9 pdf both online and offline mode.
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| 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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