Quadrilaterals, Class 9 Mathematics Detailed Chapter Notes PDF Download

INTRODUCTION

We have studied in detail about the properties of a triangle. We also know that the triangle is a figure obtained by joining three non collinear points in pair. In this chapter we shall discuss about four non-collinear points such that no three of them are collinear.

1. QUADRILATERALS

We know that the figure obtained on joining three non-collinear points in pairs is a triangle. If we mark four points and join them in some order, then there are three possibilities for the figure obtained:
(i) If all the points are collinear (in the same line), we obtain a line segment.

(ii) If three out of four points are collinear, we get a triangle.

(iii) If no three points out of four are collinear, we obtain a closed figure with four sides.

Each of the figure obtained by joining four points in order is called a quadrilateral. (quad means four and lateral for sides).

2. CONSTITUENTS OF A QUADRILATERAL

A quadrilateral has four sides, four angles and four vertices.

In quadrilateral ABCD, AB, BC, CD and DA are the four sides; A, B, C and D are the four vertices and A, B, C and D are the four angles formed at the vertices. If we join the opposite vertices A to C and B to D, then AC and BD are the two diagonals of the quadrilateral ABCD.

3. QUADRILATERALS IN PRACTICAL LIFE

We find so many objects around us which are of the shape of a quadrilateral the floor, walls, ceiling, windows of our classroom, the blackboard, each face of the duster, each page of our mathematics book, the top of our study table, etc. Some of these are given below.

4. SOME RELATED TERMS TO QUADRILATERALS
In a quadrilateral ABCD, we have

(i) VERTICES :- The points A, B, C and D are called the vertices of quadrilateral ABCD.

(ii) SIDES :- The line segments AB, BC, CD and DA are called the sides of quadrilateral ABCD.

(iii) DIAGONALS :- The line segments AC and BD are called the diagonals of quadrilateral ABCD.

(iv) ADJACENT SIDES :- The sides of a quadrilateral are said to be adjacent sides if they have a common end point. Here, in the above figure, (AB,BC),(BC,CD),(CD,DA)and(DA,AB) are four pairs of adjacent sides or consecutive sides of quadrilateral ABCD.

(v) OPPOSITE SIDES :- Two sides of a quadrilateral are said to be opposite sides if they have no common end point. Here, in the above figure, (AB, DC) and (BC, AD) are two pairs of opposite sides of quadrilateral ABCD.

(vi) CONSECUTIVE ANGLES :- Two angles of a quadrilateral are said to be consecutive angles if they have a common arm. Here, in the above figure, (A,B), (B,C), (C, D) and (D, A) are four pairs of consecutive angles.

(vii) OPPOSITE ANGLES :- Two angles of a quadrilateral are said to be opposite angles if they have no common arm. Here, in the given figure, (A, C) and (B, D) are two pairs of opposite angles of quadrilateral ABCD.

5. TYPES OF QUADRILATERALS

(i) PARALLELOGRAM :- A quadrilateral in which Dboth pair of opposite Csides are parallel is called a parallelogram.

In figure, ABCD is a quadrilateral in which ABDC, BCAD.
quadrilateral ABCD is a parallelogram.

(ii) RHOMBUS :- A parallelogram whose all sides are equal is called rhombus.

In figure, ABCD is a parallelogram in which AB = BC = CD = DA, ABDC and BCAD .
parallelogram ABCD is a rhombus.

(iii) RECTANGLE :- A parallelogram whose each angle is equal to 90°, is called a rectangle. In figure, ABCD is a parallelogram in which
A = B = C = D = 90° ,
ABDC and BCAD.

Parallelogram ABCD is a rectangle. A B

(iv) SQUARE :- A rectangle in which a pair of adjacent sides are equal is said to be a square.
In figure, ABCD is a rectangle in which A = B = C = D = 90° ,
AB = BC, BC = CD, CD = DA, DA= AB.
i.e., AB = BC = CD = DA .

rectangle ABCD is a square. A B

(v) TRAPEZIUM :- A quadrilateral in which exactly one pair of opposite sides is parallel, is called a trapezium. D

A In figure, ABCD is a quadrilateral in which ABDC.
ABCD is trapezium.

(vi) ISOSCALES TRAPEZIUM :- A trapezium whose non-parallel sides are equal is called an isosceles trapezium. D

A In figure, ABCD is a trapezium in which ABDC and BC = AD .
trapezium ABCD is isosceles trapezium.

(vii) KITE :- A quadrilateral in which two pairs of adjacent sides are equal is called a kite. In figure, ABCD is a quadrilateral in which AB = AD and BC = CD.

quadrilateral ABCD is a kite.

6. ANGLE SUM PROPERTY OF A QUADRILATERAL

THEOREM-I : The sum of the four angles of a quadrilateral is 360°.
Given : A quadrilateral ABCD.
To Prove : A + B + C + D = 360°

Construction : Join AC. 2 1
Proof :

Hence, proved.

Ex.1 Three angles of a quadrilateral measure 56°, 100° and 88°. Find the measure of the fourth angle.
Sol. Let the measure of the fourth angle be x°.
56° + 100° + 88° + x° = 360° [Sum of all the angles of quadrileteral is 360°]
244+x=360
x = 360 – 244 = 116
Hence, the measure of the fourth angle is 116°.

Ex.2 The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Sol. Let the four angles of the quadrilateral be 3x, 5x, 9x and 13x . [NCERT]
3x + 5x + 9x + 13x = 360° [Sum of all the angles of quadrileteral is 360°]
30x = 360°
x =12°
Hence, the angles of the quadrilateral are 3 × 12° = 36°, 5 × 12° = 60°, 9 × 12° = 108° and 13 × 12° = 156°.

Ex.3 In figure, ABCD is a trapezium in which AB || CD. If D = 45° and C = 75°, find A and B.
Sol. We have, AB || CD and AD is a transversal.
so, A +D = 180° [Interior angles on the same side of the transversal]

A + 45° = 180° [D = 45°]
A = 180° – 45° = 135°
Similarly, AB || CD and BC is a transversal.
so, B + C = 180°
B + 75° = 180° [C = 75°]
B = 105°
Hence A = 135° and B = 105°

7. PROPERTIES OF A PARALLELOGRAM

THEOREM-1 A diagonal of a parallelogram divides it into two congruent triangles.
Given : ABCD is a parallelogram and AC is a diagonal which forms two triangles CAB and ACD.
To Prove : ΔACD ΔCAB

Proof :

Hence, proved.

THEOREM-2 In a parallelogram, opposite sides are equal.

Given : ABCD is a parallelogram.
To Prove : AB = CD and BC = DA

Construction : Join AC.
Proof :

Hence, proved.

THEOREM-3 In a parallelogram, opposite angles are equal.

Given : ABCD is a parallelogram.
To Prove : A = C and B = D
Proof :

Hence proved.

THEOREM-4 : The diagonals of a parallelogram bisect each other.

Given : ABCD is a parallelogram, diagonals AC and BD intersect at O. O
To Prove : OA = OC and OB = OD
Proof :

Hence, proved.

CONDITION FOR A QUADRILATERAL TO BE PARALLELOGRAM

THEOREM-5: If each pair of opposite sides of a quadrilateral is equal, then it is parallelogram.

Given : A quadrilateral ABCD in which AB = CD and CB = AD . c

To Prove : ABCD is a parallelogram.
Construction : Join AC.
Proof :

Hence Proved.

THEOREM-6 If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram:

Given : A quadrilateral ABCD in which A = C and B = D

To Prove : ABCD is parallelogram
Construction : Join AC & BD
Proof :

THEOREM-7 : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Given : ABCD is a quadrilateral whose diagonals AC and BD intersect at point O such that OA = OC and OB = OD.
To Prove : ABCD is a parallelogram .

Proof :

Hence proved.

THEOREM-8 : A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
Given : ABCD is a quadrilateral in which AB = CD and ABDC
To Prove : ABCD is a parallelogram.

Construction : Join AC. B
Proof :

Ex. In figure, ABCD is a parallelogram in which D = 72°.
Find A, B and C.

Sol. We have D = 72°
But B =D [Opposite angles of the parallelogram]
B = 72°

Now, AB || CD and AD and BC are two transversals.
So, A + D = 180° [Interior angles on the same side of the transversal AD]
A + 72° = 180°
A = 180° – 72° = 108°
C = A = 108° [Opposite angles of the parallelogram]
Hence, A = 108°, B = 72° and C = 108°.

Ex. In the given figure, ABCD is a parallelogram. Find the values of x and y.

MID- POINT THEOREM

THEOREM 1 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
GIVEN : E and F are the mid-points of the sides AB and AC respectively of the ΔABC.

TO PROVE : EF || BC.
CONSTRUCTION : Throught the vertex C, CG is drawn parallel to AB and it meets EF (produced) in G.
PROOF :

Converse of Theorem 1 : The line drawn through the mid-point of one side of a triangle and parallel to another side of the triangle.
GIVEN : ABC is a triangle in which D is mid-point of AB and DE || BC.

TO PROVE : E is the mid-point of AC.
CONSTRUCTION : Let E is not the mid-point of AC. If possible, let F is the mid-point of AC. Join DF.
PROOF :

Hence proved.

THEOREM 2 : Length of the line segment joining the mid points of two sides of a triangle is equal to half the length of the third side.

GIVEN : In ΔABC, EF is the line segement joining the mid-points of the sides AB and AC of ΔABC.
TO PROVE :

EF =

CONSTRUCTION : Through C, draw CG || BA. CG meets EF (produced) at G.
PROOF :

Ex. In the following figure, D, E and F are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that ΔDEF is also an equilateral triangle.

Sol. Given : D, E and F are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC.
To prove : ΔDEF is also an equilateral triangle.
Proof : since the segment joining the mid points of two sides of a triangle is half of the third side. Therefore D and E are the mid point of BC and AC respectively.

E and F are the mid point of AC and AB respectively

F and D are the mid point of AB and BC respectively

Ex. In figure, D and E are the mid-point of the sides AB and AC respectively of ΔABC. If BC = 5.6 cm, find DE.

Sol. D is mid-point of AB and E is mid-point of AC.

Ex. In figure, E and F are mid-points of the sides AB and AC respectively of the ABC, G and H are mid-points of the sides AE and AF respectively of the AEF. If GH = 1.8 cm, find BC.