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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.
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.
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.
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.
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.
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
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
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)
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°
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.
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 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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