Class 9 Exam  >  Class 9 Notes  >  Detailed Chapter Notes - Quadrilaterals, Class 9 Mathematics

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.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
(ii) If three out of four points are collinear, we get a triangle.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
(iii) If no three points out of four are collinear, we obtain a closed figure with four sides.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
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.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
In quadrilateral ABCD, AB, BC, CD and DA are the four sides; A, B, C and D are the four vertices and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D 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.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
(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, (NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A,NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B), (NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B,NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C), (NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D) and (NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A) 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, (NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C) and (NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D) 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.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
In figure, ABCD is a quadrilateral in which ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC, BCNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AD.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 quadrilateral ABCD is a parallelogram.

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

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

In figure, ABCD is a parallelogram in which AB = BC = CD = DA, ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC and BCNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AD .
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 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
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 90° ,
ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC and BCNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AD.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 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 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 90° ,
AB = BC, BC = CD, CD = DA, DA= AB.
i.e., AB = BC = CD = DA .
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 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
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
A In figure, ABCD is a quadrilateral in which ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ABCD is trapezium.

(vi) ISOSCALES TRAPEZIUM :- A trapezium whose non-parallel sides are equal is called an isosceles trapezium. D
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
A In figure, ABCD is a trapezium in which ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC and BC = AD .
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 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.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 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 : NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 360°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Construction : Join AC. 2 1
Proof :

STATEMENT REASON

1. In ΔABC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 91+ NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 94+ 6 = 180°

2. In ΔADC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 92 + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 93 + 5 =

3. (NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 91 +NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 92) + (NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 93 +NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 94) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 95 +NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 96 = 180° + 180°

4. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D + B = 360°

5. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C +NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 360°

Sum of the all angles of triangle is equal to 180°

 

180° Sum of the all angles of triangle is equal to 180°

 

Adding (1) & (2), we get

 

 

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°.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 56° + 100° + 88° + x° = 360° NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 [Sum of all the angles of quadrileteral is 360°]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 244+x=360
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 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]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 3x + 5x + 9x + 13x = 360° NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9[Sum of all the angles of quadrileteral is 360°]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 30x = 360°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 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 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 45° and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 75°, find NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B.
Sol.
We have, AB || CD and AD is a transversal.
so, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A +NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 180° [Interior angles on the same side of the transversal]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + 45° = 180° [NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 45°]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = 180° – 45° = 135°
Similarly, AB || CD and BC is a transversal.
so, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 180°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B + 75° = 180° [NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 75°]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 105°
Hence NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = 135° and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 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 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCAB
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Proof :

STATEMENT REASON

1. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC and ADNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC

2. (i)NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC and AC is a transversal
             NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CAB

(ii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ADNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC and AC is a transversal
    NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB

3. In ΔACD and ΔCAB,
      NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CAB
        AC = AC
         NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB
Therefore, ΔACD NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ΔCAB

ABCD is a parallelogram

Alternate angles

 

Alternate angles

 

From (2)

Common

From (3)

By ASA congruence rule

Hence, proved.

THEOREM-2 In a parallelogram, opposite sides are equal.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Given : ABCD is a parallelogram.
To Prove : AB = CD and BC = DA

Construction : Join AC.
Proof :

STATEMENT REASON

1. ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 DC and ADNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC

2. In ΔABC and ΔCDA

     NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAC = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCA
      AC = AC
    NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CAD
     ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ΔCDA
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AB = CD and BC = DA

Since ABCD is a parallelogram

Alternate angles

Common

Alternate angles

By ASA congruence rule

C.P.C.T.

Hence, proved.

THEOREM-3 In a parallelogram, opposite angles are equal.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Given : ABCD is a parallelogram.
To Prove : NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D
Proof :

STATEMENT REASON

1. ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 DC and ADNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC
2. ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 DC and AD is a transversal
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 180°
3. ADNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC and DC is a transversal
 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 180°
4. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C
           NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C
5. Similarly, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D
              NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D

Since ABCD is a parallelogram

Sum of consecutive interior angles is 180°

 

Sum of consecutive interior angles is 180°

From (2) & (3)

 

 

Hence proved.

THEOREM-4 : The diagonals of a parallelogram bisect each other.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Given : ABCD is a parallelogram, diagonals AC and BD intersect at O. O
To Prove : OA = OC and OB = OD
Proof :

STATEMENT REASON

1. ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 DC and ADNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC

2. In ΔAOB and ΔCOD ,
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAO = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCO
AB = CD
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABO = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CDO
3. ΔAOB NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCOD
4. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OA = OC and OB = OD

ABCD is a parallelogram

Alternate angles

Opposite sides of a parallelogram°

Alternate angles

By ASA congruence rule

[C.P.C.T.]

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
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
To Prove : ABCD is a parallelogram.
Construction : Join AC.
Proof :

STATEMENT REASON

1. In ΔABC and ΔCDA
      AB = CD
      CB = AD
      AC = AC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCDA
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAC = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCA C.P.C.T.
       AB NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 DC

2. Similarly NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCA = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DAC
BC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AD

3. ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 DC and BC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AD

4. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ABCD is a parallelogram

Given

Given

Common

By SSS congruence rule

alternate interior angles

C.P.C.T.

alternate interior angles

From (1) & (2)

 

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 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
To Prove : ABCD is parallelogram
Construction : Join AC & BD
Proof :

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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 .
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Proof :

STATEMENT REASON

1. In ΔAOB and ΔCOD ,
       OA = OC
    NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AOB =NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9COD
      OB = OD
       ΔAOB NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCOD
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAO = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCO
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 DC

2. Similarly, ADNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC

3. ABCD is a parallelogram.

Given

Vertically opposite angles

Given

By SAS congruence rule

C.P.C.T.

But these are Alternate interior angles

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 ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC
To Prove : ABCD is a parallelogram.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Construction : Join AC. B
Proof :

STATEMENT REASON

1. In ΔABC and ΔCDA
       AB = CD
       AC = AC
      NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAC = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCA
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCDA By SAS
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCA = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DAC C.P.C.T.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AD NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 BC

2. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CD and ADNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC
3. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ABCD is a parallelogram.

Given

Common

Alternate interior angles [NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ABNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DC]
and AC intersects them

But these are Alternate interior angles.

Adding (1) & (2), we get

 

Ex. In figure, ABCD is a parallelogram in which D = 72°.
Find NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Sol. We have NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 72°
But NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B =NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D [Opposite angles of the parallelogram]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 72°

Now, AB || CD and AD and BC are two transversals.
So, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = 180° [Interior angles on the same side of the transversal AD]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + 72° = 180°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = 180° – 72° = 108°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = 108° [Opposite angles of the parallelogram]
Hence, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = 108°, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 72° and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 108°.

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

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

 

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.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
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 :

STATEMENT REASON

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 D is the mid-point of AB  and F is the mid-point of AC.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 DF || BC
But, it is given that DE || BC

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 E and F coincide.
Hence, E is the mid-point of AC.

[Given]

[By construction]

[By mid-point theorem]

This is not possible that two lines parallel to the same line intersect each other.

[NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DE and DF intersect each other at D]

So, our supposition is wrong.

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.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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

EF = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
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.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
E and F are the mid point of AC and AB respectively
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
F and D are the mid point of AB and BC respectively
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Sol. D is mid-point of AB and E is mid-point of AC.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

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.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

The document Quadrilaterals, Class 9 Mathematics Detailed Chapter Notes is a part of Class 9 category.
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FAQs on Quadrilaterals, Class 9 Mathematics Detailed Chapter Notes

1. What are the properties of a quadrilateral?
Ans. A quadrilateral is a polygon with four sides and four vertices. The properties of a quadrilateral include: - The sum of all interior angles of a quadrilateral is equal to 360 degrees. - The opposite sides of a quadrilateral are parallel. - The opposite angles of a quadrilateral are equal. - The diagonals of a quadrilateral bisect each other.
2. What is a parallelogram?
Ans. A parallelogram is a special type of quadrilateral where opposite sides are parallel and equal in length. The opposite angles of a parallelogram are also equal. Examples of parallelograms include rectangles, squares, and rhombuses.
3. How can we determine if a quadrilateral is a rectangle?
Ans. To determine if a quadrilateral is a rectangle, we need to check if its opposite sides are parallel and equal in length, and if its angles are right angles (90 degrees). If these conditions are satisfied, then the quadrilateral is a rectangle.
4. What is the difference between a square and a rhombus?
Ans. A square is a special type of rhombus where all four sides are equal in length and all angles are right angles (90 degrees). On the other hand, a rhombus is a quadrilateral with all four sides equal in length, but its angles are not necessarily right angles. In a rhombus, opposite angles are equal, but they are not 90 degrees.
5. How can we find the area of a quadrilateral?
Ans. The area of a quadrilateral can be found using different formulas depending on the type of quadrilateral. - For a parallelogram, the area is given by the product of the base and the height. - For a rectangle, the area is given by the product of its length and width. - For a rhombus, the area is given by the product of its diagonals divided by 2. - For a square, the area is given by the product of its side length squared. - For a general quadrilateral, if we know its side lengths and the measure of an included angle, we can use the formula: Area = (1/2) × a × b × sin(C), where a and b are the lengths of two sides and C is the measure of the included angle.
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