NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

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.

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.

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

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

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

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

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

Theorem: In a parallelogram, opposite sides are equal.

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.

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.

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.

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.

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.

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.

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.

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

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