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NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

Updated on Apr 23, 2025 11:16 AM IST

A quadrilateral is a geometrical shape that has four sides and four corners. The sides of quadrilaterals are also called edges, and their corners are called vertices. It is a closed shape with four angles; quadrilaterals are also called polygons. In our daily life, many objects are similar to quadrilaterals. Quadrilaterals are used in graphic design, logos, architecture, like house blueprints, windows, floors, walls, and in everyday objects like paper, electronic devices, picture frames, stationery items, etc.

This Story also Contains
  1. NCERT Class 9 Maths Chapter 8 Quadrilaterals: Notes
  2. Class 9 Chapter Wise Notes
  3. NCERT Solutions for Class 9
  4. NCERT Exemplar Solutions for Class 9
  5. NCERT Books and Syllabus
NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF
NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

These notes cover different types of quadrilaterals and theorems related to opposite sides of parallelograms, properties of diagonals, diagonals of the rhombus, etc. CBSE Class 9 chapter includes examples for the given topics as required. NCERT class 9th maths notes include all the definitions, formulas, and examples and all related theorems. Students can download the NCERT notes according to their subjects and standard, which will help to understand all the required concepts in a single place.

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NCERT Class 9 Maths Chapter 8 Quadrilaterals: Notes

Quadrilateral: The quadrilateral combines two Latin words, "quadri" and "latus"; quadri means a variant of four and latus means sides. So, the word quadrilateral implies that a shape has four sides.

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Types of Quadrilaterals:

There are the following types of quadrilaterals -
1. Parallelogram
2. Trapezium
3. Rectangle
4. Square
5. Rhombus
6. Kite

Parallelogram: A parallelogram is a quadrilateral in which opposite sides are parallel and equal to each other.

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Trapezium: A quadrilateral in which any two sides are parallel is called a trapezium.
Theorem: The opposite sides of a parallelogram are equal.

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Rectangle: A quadrilateral in which two opposite sides are equal as well as well and all the angles are right angles.

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Square: It is a type of parallelogram in which all the angles are right angles and all sides are equal.

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Rhombus: A rhombus is a parallelogram in which all the sides are equal and the diagonals perpendicularly bisect each other.

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Kite: A quadrilateral in which adjacent sides are equal is called a kite.

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Theorem: In a parallelogram, opposite sides are equal.

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Let there be a diagonal that joins Q and S.In ΔPQS and ΔQRS
∠PQS = ∠QSR (Alternate angles)
QS = QS (Common)
∠PSQ = ∠RQS (Alternate angles)
ΔPQS≅ΔQRS (ASA rule)
Hence, PQ = RS and PS = QR.

Theorem: In a parallelogram, opposite angles are equal.

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In parallelogram PQRS∠1 = ∠4 (PS || QR) .... (1)
∠2 = ∠3 (PQ || SR) ..... (2)
After adding equations (1) and (2), we get,
∠1 + ∠2 = ∠4 + ∠3
∠PSR = ∠PQR
Similarly, ∠QPS = ∠QRS

Theorem: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

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In ΔQOR and ΔPOS,
∠SPO = ∠ORQ (Alternate angles)
∠PSO = ∠OQR (Alternate angles)
PS = QR (Opposite sides of parallelogram)
ΔQOR ≅ ΔPOS (AAS rule)
Therefore, PO = OR and SO = OQ

Theorem: The diagonal of a parallelogram divides it into two congruent triangles.

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In ΔPQS and ΔQRS,
PQ = RS (Opposite sides of parallelogram)
PS = QR (Opposite sides of a parallelogram)
QS = QS (Common side in both triangles)
ΔPQS ≅ ΔQRS (By SSS rule)

Theorem: The diagonals of a rhombus bisect each other at right angles.

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In ΔSOP and ΔPOQ,
PS = PQ (Adjacent sides of a rhombus)
PO = PO (Common side)
SO = OQ (Diagonals of a parallelogram that bisect each other)
ΔSOP ≅ΔPOQ (By SSS rule)
Therefore, ∠SOP =∠POQ ... (1)
∠SOP +∠POQ =180° (∵ Angle in a straight line is 180°)
From equation (1),
∠SOP +∠SOP =180°
∠SOP = 90°
Similarly, ∠SOP = ∠POQ = 90°

Theorem: The diagonals of a rectangle bisect each other and are equal.

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In ΔPSQ and ΔRSQ,
PS = QR (Opposite sides of a rectangle)
PQ = SR (Opposite sides of a rectangle)
∠SPQ = ∠SRQ (Angles of rectangle are 90°)
ΔPSQ ≅ΔRSQ (By SAS rule)
∴SQ = PR (Corresponding parts of congruent triangles)
Now, in ΔPOS and ΔQOR,
∠OSP = ∠OQR (∵ PS||QR)
∠SPO=∠ORQ (∵ AD||BC)
PS = QR (Opposite sides of a rectangle)
ΔPOS ≅ΔQOR (By ASA rule)
Therefore, SO = OQ (Corresponding parts of congruent triangles)
Similarly, we can prove PO = OR.

Theorem: The diagonals of a square bisect each other at right angles and are equal.

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In ΔPRS and ΔPQR,
PS = QR (Opposite sides of a Square)
RS = PQ (Opposite sides of a Square)
PR = PR (Common side)
∠PQR =∠PSR (Each angle of the square is 90°)
ΔPRS ≅ΔPQR (By SAS rule)
Therefore, PR = SQ (Corresponding parts of congruent triangles)
Now in ΔPOS and ΔPOQ,
PS = RQ (Opposite sides of a square)
∠PSO =∠OQR (∵ PS || RQ)
∠SPO =∠ORQ (∵ SR || PQ)
ΔPOS ≅ΔPOQ (By ASA rule)
∴PO = OR (Corresponding parts of congruent triangles)
Similarly, we can prove SO = OQ
In ΔPSO and ΔPOQ,
OP = OP (Common side)
SO = OQ (Already proved)
PS = PQ (Side of square)
ΔPSO ≅ΔPOQ (By SSS rule)
∴∠POQ = ∠POS (Corresponding parts of congruent triangles)
Now, ∠POS +∠POQ =180° (Angle in a line is equal to 180°)
∴∠POS = ∠POQ = 90°

The Mid-Point Theorem

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of the third side.

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In ΔPRS, Q is the midpoint of U; U is the midpoint of PS.
Draw a line QU to T such that QU = UT
In ΔPQU and ΔUST,
∠PUQ = ∠TUS (Vertically opposite angles)
PU = US (U is the midpoint of PS)
QU = UT (From construction)
∴ΔPQU ≅ΔUST (By SAS rule)
Therefore,
∠QPU =∠TSU….(1)
PQ = TS = QR (Q is the midpoint of PR)
TS || PQ || PR (From equation (1))
TS || RQ
Therefore, QRST is a parallelogram.
Therefore, RS = QT and RS || QT
Therefore QU || RS

Class 9 Chapter Wise Notes

Students must download the notes below for each chapter to ace the topics.

NCERT Solutions for Class 9

Students must check the NCERT solutions for Class 10 Maths and Science given below:

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NCERT Exemplar Solutions for Class 9

Students must check the NCERT exemplar solutions for Class 10 Maths and Science given below:

NCERT Books and Syllabus

To learn about the NCERT books and syllabus, read the following articles and get a direct link to download them.


Articles

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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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