In NCERT Class 9 Maths Chapter 8, you will find a concise and comprehensive explanation of the properties of special quadrilaterals, such as Parallelogram, Rhombus, Rectangle, Square, and Trapezium. The solutions provided by the experts at GeeksforGeeks cover the essential principles regarding angles, diagonals, and relations specific to each quadrilateral. These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals offer conceptual clarity, enabling students to grasp the topics effectively.
The Class 9 Maths Chapter 8 focuses on the Quadrilateral section. The significance of the properties of parallelograms is thoroughly discussed in the NCERT Solutions for Class 9 Maths Chapter 8. Section 4 of this chapter primarily covers the Mid-point theorem and its wide range of applications. By studying these topics, students can gain a better understanding and strengthen their mathematical skills.
NCERT Class 9 Maths Chapter 8 Quadrilaterals covers variety of topics such as:
NCERT Answers for Class 9 Mathematics Chapter 8: Exercise 8.1
Question 1. The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.
Solution:
As we know that, the sum of the angles of a quadrilateral is 360° (Angle sum property of quadrilateral)
As they are in ratio 3 : 5 : 9 : 13, so we can assume angles be as 3x, 5x, 9x and 13x.
So,
3x + 5x + 9x + 13x = 360°
30x = 360°
x = 360/30 = 12°
So the angles will be as follows:
3x = 3×12 = 36°
5x = 5×12 = 60°
9x = 9×12 = 108°
13x = 13×12 = 156°
Hence, the angles of quadrilateral are 36°, 60°, 108° and 156°.
Question 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
As mentioned parallelogram, so let PQRS be a parallelogram
where, given
PR = QS
In ∆PQR and ∆QRS,
PR = QS …………………[Given]
PQ = RS …………………[Opposite sides of a parallelogram]
QR = RQ …………………[Common side]
∴ ∆PQR ≅ ∆QRS [By SSS congruency]
so, ∠PQR = ∠QRS [By C.P.C.T.] ………………………………………….(1)
Now, PQ || RS and QR is a transversal. …………………….[PQRS is a parallelogram]
∴ ∠PQR + ∠QRS = 180° [Co-interior angles of parallelogram]…………………………………… (2)
From (1) and (2), we have
∠PQR = ∠QRS = 90°
i.e., PQRS is a parallelogram having an angle equal to 90°.
Hence, PQRS is a rectangle. (having all angles equal to 90° and opposite sides are equal)
Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
As mentioned quadrilateral, so let PQRS be a quadrilateral
where, given
PO = RO
SO = QO
In ∆POQ and ∆POS, we have
PO = PO [Common]
OQ = OS [O is the mid-point of QS]
∠POQ= ∠POS [Each 90°]
∴ ∆POQ ≅ ∆POS [By, SAS congruency]
∴ PQ = PS [By C.P.C.T.] …….. (1)
Similarly, PQ = QR …………………..(2)
QR = RS ……………………………………..(3)
RS = SP ………………………………………(4)
∴ From (1), (2), (3) and (4), we have
PQ = QR = RS = SP
Thus, the quadrilateral PQRS is a rhombus.
Alternative Solution:
So as it is given that diagonals of a quadrilateral PQRS bisect each other
According to Theorem 8.7 NCERT it is a parallelogram
PQRS can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.
In ∆POQ and ∆POS, we have
PO = PO [Common]
OQ = OS [O is the mid-point of QS]
∠POQ= ∠POS [Each 90°]
∴ ∆POQ ≅ ∆POS [By,SAS congruency]
∴ PQ = PS [By C.P.C.T.] …….. (1)
Similarly, PQ = QR …………………..(2)
QR = RS ……………………………………..(3)
RS = SP ………………………………………(4)
From (1), (2), (3) and (4), we have
PQ = QR = RS = SP
Hence, as a parallelogram has all sides equal then it is called a rhombus.
Question 4. Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
As it is mentioned that it is a square so,
all sides are equal. (PQ = QR = RS = SP)
all angles at four corners = 90°
So, to prove that the diagonals are equal, we need to prove PR = QS.
In ∆PQR and ∆QPS, we have
PQ = QP ……………….[Common]
PS = PQ ………………..[Sides of a square PQRS]
∠PQR = ∠QPS …….[Each angle is 90°]
∆PQR ≅ ∆QPS [By SAS congruency]
AC = BD [By C.P.C.T.] …(1)
Now as we know square is also a parallelogram
In ∆POQ and ∆ROS, we have
∠POQ = ∠ROS ……………….[Opposite angles of two intersecting lines]
PQ = RS ………………..[Sides of a square PQRS]
∠PQO = ∠RSO …….[Alternate interior angles are equal]
∆PQR ≅ ∆QPS [By ASA congruency]
OP = OQ = OR = OS (Hence the diagonals are equal and bisect each other)
Now , In ∆OQP and ∆OSP, we have
OQ = OS [Proved]
QP = SP [Sides of a square PQRS]
OP = OP [Common]
∆OQP ≅ ∆OSP [By SSS congruency]
∠POQ = ∠POS [By C.P.C.T.] …(3)
∠POQ + ∠POS = 180° (∵ ∠POQ and ∠POS form a linear pair)
∠POQ = ∠POS = 90° [By(3)]
PR ⊥ QS …(4)
From (1), (2) and (4), we get PR and QS are equal and bisect each other at right angles (90°).
Question 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:
As it is mentioned Quadrilateral, so let PQRS be a quadrilateral
where,
PR = QS
OP = OR = OQ = OS
∠POQ = ∠QOR = ∠ROS = ∠SOP = 90°
Now, in ∆POS and ∆POQ, We have
∠POS = ∠POQ [Each 90°]
PO = PO [Common]
OS= OQ [ ∵ O is the midpoint of BD]
∆POS ≅ ∆POQ [By SAS congruency]
PS = PQ [By C.P.C.T.] …(1)
Similarly, we have
PQ = QR … (2)
QR = RS …(3)
RS = SP…(4)
From (1), (2), (3) and (4), we have
PQ = QR = RS = SP
Hence, Quadrilateral PQRS have all sides equal.
Now, In ∆POS and ∆ROQ, we have
PO = RO [Given]
OS = OQ [Given]
∠POS = ∠ROQ [Vertically opposite angles]
So, ∆POS ≅ ∆ROQ [By SAS congruency]
∠PSO = ∠RQO [By C.P.C.T.]
So as they form a pair of alternate interior angles.
PS || QR
Similarly, PQ || RS
PQRS is a parallelogram.
Parallelogram having all its sides equal is a rhombus.
PQRS is a rhombus.
Now, in ∆PQR and ∆QPS, we have
PR = QS [Given]
QR = SP [Proved]
PQ = QP [Common]
∆PQR ≅ ∆QPS [By SSS congruency]
∠PQR = ∠QPS [By C.P.C.T.] ……(5)
Since, PS || QR and PQ is a transversal.
∠PQR + ∠QPS = 180° .. .(6) [ Co – interior angles]
∠PQR = ∠QPS = 90° [By(5) & (6)]
So, rhombus PQRS is having corners angle equal to 90°.
Hence, PQRS is a square.
Question 6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Solution:
(i) As, ABCD is a parallelogram.
∠BAC = ∠DCA ……………………………(1) [Alternate interior angles are equal]
∠CAD = ∠BCA ………………………….(2) [Alternate interior angles are equal]
Also, ∠CAD = ∠CAB …………………….(3) [ (Given) as AC bisects ∠A]
From (1), (2) and (3), we have
∠DCA = ∠BCA
Hence, AC bisects ∠C.
(ii) In ∆ABC,
∠BAC = ∠DCA ………………………….. [Alternate interior angles are equal]
BC = AB …………………………….(4) [ Sides opposite to equal angles of a ∆ are equal]
Similarly, AD = DC ……..(5)
But, ABCD is a parallelogram. [Given]
AB = DC ………………….(6) (opposite sides of parallelogram)
From (4), (5) and (6), we have
AB = BC = CD = DA
As, ABCD is a parallelogram having all sides equal then it is a rhombus.
Question 7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Solution:
As, ABCD is a rhombus, so
AB = BC = CD = DA
Now, As CD = AD
∠ADB= ∠ABD……………………….(1) [Angles opposite to equal sides of a triangle are equal]
∠ADB = ∠CBD …………………..(2) [ ∵ Alternate interior angles are equal] (Rhombus is a parallelogram)
From (1) and (2), we have
∠CBD = ∠ABD …………………..(3)
∠ABD= ∠CDB ……………………..(4) [ ∵ Alternate interior angles are equal]
From (1) and (4),
we have ∠ADB = ∠CDB
Hence, BD bisects ∠B as well as ∠D.
Similarly, we can prove that AC bisects ∠C as well as ∠A..
Question 8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.
Solution:
There is rectangle ABCD such that AC bisects ∠A as well as ∠C, so
∠BAC = ∠DAC and,
∠DCA = ∠BCA ………………………………(1)
(i) As we know that every rectangle is a parallelogram.
ABCD is a parallelogram.
∠BCA = ∠DAC …………………….(2) [ Alternate interior angles are equal]
From (1) and (2), we have
∠DCA= ∠DAC……………………….(3)
In ∆ABC, ∠DCA= ∠DAC then,
CD = DA [Sides opposite to equal angles of a ∆ are equal]
Similarly, AB = BC
So, ABCD is a rectangle having adjacent sides equal.
ABCD is a square.
(ii) Since, ABCD is a square
AB = BC = CD = DA
so, In ∆ABD, as AB = AD
∠ABD = ∠ADB [Angles opposite to equal sides of a ∆ are equal]……………………..(1)
Similarly, ∠CBD = ∠CDB…………………..(2)
∠CBD = ∠ADB [Alternate interior angles are equal]………………(3)
From (1) and (3)
∠CBD = ∠ABD
From (2) and (3)
∠ADB = ∠CBD
So, BD bisects ∠B as well as ∠D.
Question 9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that :
(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Solution:
ABCD is a parallelogram
DP = BQ
(i) As ABCD is a parallelogram
∠ADB = ∠CBD [Alternate interior angles are equal]……………….(1)
∠ABD = ∠CDB [Alternate interior angles are equal]…………………(2)
Now, in ∆APD and ∆CQB, we have
AD = CB [Opposite sides of a parallelogram ABCD are equal]
PD = QB [Given]
∠ADP = ∠CBQ [Proved]
Hence, ∆APD ≅ ∆CQB [By SAS congruency]
(ii) As, ∆APD ≅ ∆CQB [Proved]
AP = CQ [By C.P.C.T.]…………………(3)
(iii) Now, in ∆AQB and ∆CPD, we have
QB = PD [Given]
∠ABQ = ∠CDP [Proved]
AB = CD [ Opposite sides of a parallelogram ABCD are equal]
Hence, ∆AQB ≅ ∆CPD [By SAS congruency]
(iv) As, ∆AQB ≅ ∆CPD [Proved]
AQ = CP [By C.P.C.T.] …………………………..(4)
(v) In a quadrilateral APCQ,
Opposite sides are equal. [From (3) and (4)]
Hence, APCQ is a parallelogram. (NCERT Theorem 8.3)
Question 10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) ∆APB ≅ ∆CQD
(ii) AP = CQ
Solution:
ABCD is a parallelogram
DP = BQ
(i) In ∆APB and ∆CQD, we have
∠APB = ∠CQD [Each 90°]
AB = CD [ Opposite sides of a parallelogram ABCD are equal]
∠ABP = ∠CDQ [Alternate angles are equal as AB || CD and BD is a transversal]
Hence, ∆APB ≅ ∆CQD [By AAS congruency]
(ii) As, ∆APB ≅ ∆CQD [Proved]
AP = CQ [By C.P.C.T.]
Question 11. In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF.
Solution:
AB = DE, and AB || DE,
BC = EF and BC || EF
(i) We have AB = DE and AB || DE [Given]
so here., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.
Hence, ABED is a parallelogram.
(ii) BC = EF and BC || EF [Given]
i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.
Hence, BEFC is a parallelogram.
(iii) as, ABED is a parallelogram [Proved]
∴ AD || BE and AD = BE …(1) [Opposite sides of a parallelogram are equal and parallel]
Also, BEFC is a parallelogram. [Proved]
BE || CF and BE = CF …(2) [Opposite sides of a parallelogram are equal and parallel]
From (1) and (2), we have
AD || CF and AD = CF
(iv) Since, AD || CF and AD = CF [Proved]
i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.
Hence, Quadrilateral ACFD is a parallelogram.
(v) Since, ACFD is a parallelogram. [Proved]
So, AC =DF [Opposite sides of a parallelogram are equal]
(vi) In ∆ABC and ∆DEF, we have
AB = DE [Given]
BC = EF [Given]
AC = DE [Proved in (v) part]
∆ABC ≅ ∆DEF [By SSS congruency]
Question 12. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD
(iv) diagonal AC = diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Solution:
so basically here is a trapezium ABCD in which AB || CD and AD = BC.
Extended AB and draw a line through C parallel to DA intersecting AB produced at E
(i) AB || DC , AE || DC Also AD || CE
then, AECD is a parallelogram.
AD = CE …………………………(1) [Opposite sides of the parallelogram are equal]
But AD = BC …(2) [Given]
From (1) and (2),
BC = CE
Now, in ∆BCF, we have BC = CF
∠CEB = ∠CBE …(3) [Angles opposite to equal sides of a triangle are equal]
Also, ∠ABC + ∠CBE = 180° … (4) [Linear pair]
and ∠A + ∠CEB = 180° …(5) [Co-interior angles of a parallelogram ADCE]
From (4) and (5), we get
∠ABC + ∠CBE = ∠A + ∠CEB
∠ABC = ∠A [From (3)]
∠B = ∠A …(6)
(ii) AB || CD and AD is a transversal.
∠A + ∠D = 180° …(7) [Co-interior angles in parallelogram]
Similarly, ∠B + ∠C = 180° … (8)
From (7) and (8), we get
∠A + ∠D = ∠B + ∠C
∠C = ∠D [From (6)]
(iii) In ∆ABC and ∆BAD, we have
AB = BA [Common]
BC = AD [Given]
∠ABC = ∠BAD [Proved]
Hence, ∆ABC ≅ ∆BAD [By SAS congruency]
(iv) Since, ∆ABC ≅ ∆BAD [Proved]
AC = BD [By C.P.C.T.]
NCERT Answers for Class 9 Mathematics Chapter 8: Exercise 8.2
Question 1. 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:
(i) SR || AC and SR = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram
Solution:
Given that, P, Q, R and S are the mid points of quadrilateral ABCD
Theorem 8.9: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
(i) So here, taking ∆ACD
we can see S and R are the mid points of side AD and DC respectively. [Given]
Hence, SR || AC and SR = ½ AC (NCERT Theorem 8.9)……………………….(1)
(ii) So here, taking ∆ACB
we can see P and Q are the mid points of side AB and BC respectively. [Given]
Hence, PQ || AC and PQ = ½ AC (NCERT Theorem 8.9)…………………………(2)
From (1) and (2) we can say,
PQ = SR
(iii) so from (i) and (ii) we can say that
PQ || AC and SR || AC
so, PQ || SR and PQ = SR
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Hence, PQRS is a parallelogram.
Question 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Solution:
Given that, P, Q, R and S are the mid points of Rhombus ABCD.
Theorem 8.9: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Construction: Join AC and BD
So here, taking ∆ABD
We can see P and S are the mid points of side AB and AD respectively. [Given]
Hence, PS || BD and PS = ½ BD (NCERT Theorem 8.9)……………………….(1)
Similarly, is we take ∆CBD
We can see R and Q are the mid points of side CD and CB respectively. [Given]
Hence, RQ || BD and RQ = ½ BD (NCERT Theorem 8.9)……………………….(2)
So from (1) and (2), we conclude that
PS || RQ and PS = RQ
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Hence, PQRS is a parallelogram.
Now in ∆ACD
We can see S and R are the mid points of side AD and CD respectively. [Given]
Hence, SR || AC and RS = ½ AC (NCERT Theorem 8.9)
from (2) RQ || BD and RQ = ½ BD (NCERT Theorem 8.9)
Hence, OGSH is a parallelogram.
∠HOG = 90° (Diagonal of rhombus intersect at 90°)
So ∠HSG = 90° (opposite angle of a parallelogram are equal)
As, PQRS is a parallelogram having vertices angles equal to 90°.
Hence, PQRS is a Rectangle.
Question 3. ABCD is a rectangle and P, Q, R, and S are mid-points of the sides AB, BC, CD, and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Solution:
Given that, P, Q, R and S are the mid points of Rectangle ABCD.
Construction: Join AC
Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
So here, taking ∆ACD
we can see S and R are the mid points of side AD and DC respectively. [Given]
Hence, SR || AC and SR = ½ AC (NCERT Theorem 8.9)……………………….(1)
Now, taking ∆ACB
we can see P and Q are the mid points of side AB and BC respectively. [Given]
Hence, PQ || AC and PQ = ½ AC (NCERT Theorem 8.9)…………………………(2)
So from (1) and (2), we conclude that
SR || PQ and SR = PQ
Hence, PQRS is a parallelogram. (If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram)
Now in ∆QBP and ∆QCR
QC = QB (Q is the mid point of BC)
RC = PB (opposite sides are equal, hence half length is also equal)
∠QCR = ∠QBP (Each 90°)
∆QBP ≅ ∆QCR (By SAS congruency)
QR = QP (By C.P.C.T.)
As PQRS is a parallelogram and having adjacent sides equal
Hence, PQRS is a rhombus
Question 4. 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). Show that F is the mid-point of BC.
Solution:
Let O be the point of intersection of lines BD and EF
Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
So here, taking ∆ADB
we can see S is the mid point of side AD and ED || AB [Given]
Hence, OD = ½ BD ………..(NCERT Theorem 8.10)
Now, taking ∆BCD
we can see O is the mid point of side BD and OF || AB [Proved and Given]
Hence, CF = ½ BC…….. (NCERT Theorem 8.10)
Hence proved!!
Question 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.
Solution:
Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
Given, E and F are the mid points of side AB and CD of parallelogram ABCD.
As, AB || CD and AB = CD (opposite sides of parallelogram)………..(1)
AE = CF (halves of opposite sides of parallelogram)………………………(2)
from (1) and (2)
AECF is a parallelogram
Hence, AF || EC
Now taking ∆APB
we can see E is the mid point of side AB and EF || AP [Given and proved]
Hence, BQ = PQ………..(NCERT Theorem 8.10)………………….(1)
Now taking ∆CQD
we can see F is the mid point of side CD and CQ || FP [Given and proved]
Hence, DP = PQ………..(NCERT Theorem 8.10)………………..(2)
From (1) and (2) we conclude that,
BQ = PQ = DQ
Hence, we can say that line segments AF and EC trisect the diagonal BD
Question 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution:
Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
So here, taking ∆ACD
we can see S and R are the mid points of side AD and DC respectively. [Given]
Hence, SR || AC and SR = ½ AC (NCERT Theorem 8.9)……………………….(1)
Now, taking ∆ACB
we can see P and Q are the mid points of side AB and BC respectively. [Given]
Hence, PQ || AC and PQ = ½ AC (NCERT Theorem 8.9)…………………………(2)
So from (1) and (2), we conclude that
SR || PQ and SR = PQ
Hence, PQRS is a parallelogram. (If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram)
And since the diagonal of parallelogram bisects each other
so, QS and PR bisects each other.
Question 7. 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
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB
Solution:
Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
(i) while taking ∆ABC
we can see M is the mid point of side AB and DM || BC [Given]
This implies, DC= AD ……….. (NCERT Theorem 8.10)
Hence, D is the mid-point of AC.
(ii) As we know MD || BC and AC is transversal
This implies, ∠ACB = ∠ADM = 90°
Hence, MD ⊥AC
(iii) Considering ∆ADM and ∆CDM
AD = CD (D is the mid point of AC (Proved))
∠CDM = ∠ADM (proved, MD ⊥AC)
DM = DM (common)
∆ADM ≅ ∆CDM (By SAS congruency)
CM = AM (By C.P.C.T.)
CM = AM = ½ AB (M is the mid point of AB)
Important Topics under NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
The NCERT Class 9 Maths Chapter 8 syllabus focuses on Quadrilaterals, which is a crucial topic covered in class 9. The chapter is divided into two exercise sections and covers five main topics. To help students fully grasp the concepts and benefit from the solutions provided by GeeksforGeeks, it is important to pay close attention to the following essential topics:
- Introduction to Quadrilaterals
- The Angle Sum Property of Quadrilaterals
- Types of Quadrilaterals
- Properties of Parallelograms
- The Mid-point Theorem
By thoroughly understanding these topics, students will be well-prepared to excel in this chapter and build a solid foundation for further mathematical concepts.
Important Points to Remember:
- NCERT Solutions for Class 9 will help students to learn the solution for all the NCERT Problems.
- All the solutions here are comprehensive and step-by-step so that students can solve all the problems with ease.
- All the solutions provided are in a step-by-step format for better understanding.
Also Check:
FAQs on NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
Q1: Why is it important to learn about Quadrilaterals?
Learning about quadrilaterals is important because it helps develop geometric understanding and reasoning. Quadrilaterals are fundamental shapes in geometry, and studying them deepens your knowledge of angles, sides, diagonals, and symmetry. By understanding quadrilaterals, you can analyze shapes and structures, make informed decisions, and solve practical problems.
Q2: What topics are covered in NCERT Solutions for Chapter 8 Quadrilaterals?
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals covers several quadrilateral kinds (rectangle, square, parallelogram, trapezium, etc.) and their various features, quadrilateral’s angle sum property and midpoint theorem and its converse.
Q3: How can NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals help me?
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals can help you solve the NCERT exercise without any limitations. If you are stuck on a problem, you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.
Q4: How many exercises are there in Class 9 Maths Chapter 8 Quadrilaterals?
There are 2 exercises in the Class 9 Maths Chapter 8 Quadrilaterals which covers all the important topics and sub-topics.
Q5: Where can I find NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals?
You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.
Share your thoughts in the comments
Please Login to comment...