Parallelograms encompass a diverse set of quadrilaterals, including rectangles, rhombuses, and squares. Each of these special parallelograms has distinct properties and characteristics.
Rectangle:
A rectangle is a parallelogram with all interior angles measuring 90 degrees, making it a right angle. Consequently, opposite sides of a rectangle are equal in length.
Rhombus:
A rhombus is a parallelogram with all sides of equal length. This means that opposite sides are equal and parallel. However, the angles of a rhombus are not necessarily 90 degrees, except in the case of a square.
Square:
A square is a special case of both a rectangle and a rhombus. It possesses all the properties of a rectangle, including right angles, and all the sides are equal in length like a rhombus.
Relationships Between Special Parallelograms:
In terms of relationships, every rectangle and rhombus is inherently a parallelogram. Therefore, they are depicted as subsets of a parallelogram. Furthermore, because a square possesses the characteristics of both a rectangle and a rhombus, it is represented by the overlapping shaded region in the diagram.
Understanding the distinctions and relationships between these special parallelograms is crucial for geometry and problem-solving applications.
Rectangle
A rectangle is a parallelogram with one of its angles as a right angle.
In the above figure,
Let,
A
ˆ
=
90
∘
𝐴^=90∘
Since,
A
D
∥
B
C
𝐴𝐷∥𝐵𝐶
,
A
ˆ
+
B
ˆ
=
180
∘
𝐴^+𝐵^=180∘
(Sum of interior angles on the same side of transversal
A
B
𝐴𝐵
)
Therefore,
B
ˆ
=
90
∘
𝐵^=90∘
Here,
A
B
∥
C
D
𝐴𝐵∥𝐶𝐷
and
A
ˆ
=
90
∘
𝐴^=90∘
(Given)
Therefore,
A
ˆ
+
D
ˆ
=
180
∘
𝐴^+𝐷^=180∘
∴
D
ˆ
=
90
∘
∴𝐷^=90∘
∴
C
ˆ
=
90
∘
∴𝐶^=90∘
Corollary: Each of the four angles of a rectangle is a right angle.
Rhombus
A rhombus is a parallelogram with a pair of its consecutive sides equal.
A
B
C
D
𝐴𝐵𝐶𝐷
is a rhombus in which
A
B
=
B
C
𝐴𝐵=𝐵𝐶
.
Since a rhombus is a parallelogram,
A
B
=
D
C
𝐴𝐵=𝐷𝐶
and
B
C
=
A
D
𝐵𝐶=𝐴𝐷
Thus,
A
B
=
B
C
=
C
D
=
A
D
𝐴𝐵=𝐵𝐶=𝐶𝐷=𝐴𝐷
Corollary: All the four sides of a rhombus are equal (congruent).
Square
A square is a rectangle with a pair of its consecutive sides equal.
Since square is a rectangle, each angle of a rectangle is a right angle and
A
B
=
D
C
𝐴𝐵=𝐷𝐶
,
B
C
=
C
D
𝐵𝐶=𝐶𝐷
.
Thus,
A
B
=
B
C
=
C
D
=
A
D
𝐴𝐵=𝐵𝐶=𝐶𝐷=𝐴𝐷
Each of the four angles of a square is a right angle and each of the four sides is of the same length.
Theorem
4
4
Statement:
The diagonals of a rectangle are equal in length.
Given:
A
B
C
D
𝐴𝐵𝐶𝐷
is a rectangle.
A
C
𝐴𝐶
and
B
D
𝐵𝐷
are diagonals.
To prove:
A
C
=
B
D
𝐴𝐶=𝐵𝐷
Proof:
Let,
A
ˆ
=
90
∘
𝐴^=90∘
(By definition of rectangle)
A
ˆ
+
B
ˆ
=
180
∘
𝐴^+𝐵^=180∘
(Consecutive interior angle)
A
ˆ
=
B
ˆ
=
90
∘
𝐴^=𝐵^=90∘
Now in triangles,
A
B
D
𝐴𝐵𝐷
and
A
B
C
𝐴𝐵𝐶
,
A
B
=
A
B
𝐴𝐵=𝐴𝐵
(Common side)
A
ˆ
=
B
ˆ
=
90
∘
𝐴^=𝐵^=90∘
(Each angle is a right angle)
A
D
=
B
C
𝐴𝐷=𝐵𝐶
(Opposite sides of parallelogram)
Therefore,
Δ
A
B
D
≅
Δ
B
A
C
Δ𝐴𝐵𝐷 ≅ Δ𝐵𝐴𝐶
Therefore,
B
D
=
A
C
𝐵𝐷=𝐴𝐶
(Corresponding parts of corresponding triangles)
Hence the theorem is proved.
Converse of Theorem
4
4
:
Statement:
If two diagonals of a parallelogram are equal, it is a rectangle.
Given:
A
B
C
D
𝐴𝐵𝐶𝐷
is a parallelogram in which
A
C
=
B
D
𝐴𝐶=𝐵𝐷
.
To prove:
Parallelogram
A
B
C
D
𝐴𝐵𝐶𝐷
is a rectangle.
Proof:
In triangles
A
B
C
𝐴𝐵𝐶
and
D
B
C
𝐷𝐵𝐶
,
A
B
=
D
C
𝐴𝐵=𝐷𝐶
(Opposite sides of parallelogram)
B
C
=
B
C
𝐵𝐶=𝐵𝐶
(Common side)
A
C
=
B
D
𝐴𝐶=𝐵𝐷
(Given)
Therefore,
Δ
A
B
C
≅
Δ
D
C
B
Δ𝐴𝐵𝐶 ≅ Δ𝐷𝐶𝐵
(
S
S
S
𝑆𝑆𝑆
congruency condition)
Therefore,
A
B
ˆ
C
=
D
C
ˆ
B
𝐴𝐵^𝐶=𝐷𝐶^𝐵
(Corresponding parts of corresponding triangles)
But these angles are consecutive interior angles on the same side of transversal
B
C
𝐵𝐶
and
A
B
∥
D
C
𝐴𝐵∥𝐷𝐶
.
Therefore,
A
B
ˆ
C
+
D
C
ˆ
B
=
180
∘
𝐴𝐵^𝐶+𝐷𝐶^𝐵=180∘
But,
A
B
ˆ
C
=
D
C
ˆ
B
𝐴𝐵^𝐶=𝐷𝐶^𝐵
Therefore,
A
B
ˆ
C
=
D
C
ˆ
B
=
90
∘
𝐴𝐵^𝐶=𝐷𝐶^𝐵=90∘
Therefore, by definition of rectangle, parallelogram
A
B
C
D
𝐴𝐵𝐶𝐷
is a rectangle.
Hence the theorem is proved.
Theorem
5
5
:
Statement:
The diagonals of a rhombus are perpendicular to each other.
Given:
A
B
C
D
𝐴𝐵𝐶𝐷
is a rhombus.
Diagonal
A
C
𝐴𝐶
and
B
D
𝐵𝐷
intersect at
O
𝑂
.
To prove:
A
C
𝐴𝐶
and
B
D
𝐵𝐷
bisect each other at right angles.
Proof:
A rhombus is a parallelogram such that
AB = DC = AD = BC
.
.
.
.
.
.
(
i
)
AB = DC = AD = BC ......(i)
Also the diagonals of a parallelogram bisect each other.
Hence,
B
O
=
D
O
𝐵𝑂=𝐷𝑂
and
A
O
=
OC
.
.
.
.
.
.
(
ii
)
𝐴𝑂=OC ......(ii)
Now, compare triangles
A
O
B
𝐴𝑂𝐵
and
A
O
D
𝐴𝑂𝐷
,
A
B
=
A
D
𝐴𝐵=𝐴𝐷
(From
(
i
)
(i)
above)
B
O
=
D
O
𝐵𝑂=𝐷𝑂
(From
(
ii
)
(ii)
above)
AO = AO
AO = AO
(Common side)
Therefore,
Δ
A
O
B
≅
Δ
A
O
D
Δ𝐴𝑂𝐵 ≅ Δ𝐴𝑂𝐷
(
S
S
S
𝑆𝑆𝑆
congruency condition)
Therefore,
A
O
ˆ
B
=
A
O
ˆ
D
𝐴𝑂^𝐵=𝐴𝑂^𝐷
(Corresponding parts of corresponding parts)
B
D
𝐵𝐷
is a straight line segment.
Therefore,
A
O
ˆ
B
+
A
O
ˆ
D
=
180
∘
𝐴𝑂^𝐵+𝐴𝑂^𝐷=180∘
But,
A
O
ˆ
B
=
A
O
ˆ
D
𝐴𝑂^𝐵=𝐴𝑂^𝐷
(Proved)
Therefore,
A
O
ˆ
B
=
A
O
ˆ
D
=
180
∘
2
𝐴𝑂^𝐵=𝐴𝑂^𝐷=180∘2
A
O
ˆ
B
=
A
O
ˆ
D
=
90
∘
𝐴𝑂^𝐵=𝐴𝑂^𝐷=90∘
That is, the diagonals bisect at right angles.
Hence the theorem is proved.
Converse of Theorem
5
5
:
Statement:
If the diagonals of a parallelogram are perpendicular then it is a rhombus.
Given:
A
B
C
D
𝐴𝐵𝐶𝐷
is a parallelogram in which
A
C
𝐴𝐶
and
B
D
𝐵𝐷
are perpendicular to each other.
To prove:
A
B
C
D
𝐴𝐵𝐶𝐷
is a rhombus.
Proof:
Let
A
C
𝐴𝐶
and
B
D
𝐵𝐷
intersect at right angles at
O
𝑂
.
A
O
ˆ
B
=
90
∘
𝐴𝑂^𝐵=90∘
In triangles
A
O
D
𝐴𝑂𝐷
and
C
O
D
𝐶𝑂𝐷
,
A
O
=
O
C
𝐴𝑂=𝑂𝐶
(Diagonals bisect each other)
O
D
=
O
D
𝑂𝐷=𝑂𝐷
(Common side)
A
O
ˆ
D
=
C
O
ˆ
D
=
90
∘
𝐴𝑂^𝐷=𝐶𝑂^𝐷=90∘
(Given)
Therefore,
Δ
A
O
D
≅
Δ
C
O
D
Δ𝐴𝑂𝐷 ≅ Δ𝐶𝑂𝐷
(
S
A
S
𝑆𝐴𝑆
congruency condition)
A
D
=
D
C
𝐴𝐷=𝐷𝐶
That is, the adjacent sides are equal.
Therefore, by definition,
A
B
C
D
𝐴𝐵𝐶𝐷
is a rhombus.
Hence the theorem is proved.
Benefits of CBSE Class 9 Maths Notes Chapter 8 Quadrilaterals
-
Conceptual Understanding:
These notes provide a comprehensive explanation of the properties and characteristics of quadrilaterals, helping students build a strong conceptual foundation in geometry.
-
Clarity in Definitions:
By clearly defining terms such as parallelograms, rectangles, rhombuses, and squares, the notes ensure that students understand the distinctions between different types of quadrilaterals.
-
Problem-Solving Skills:
Through worked examples and exercises, students can enhance their problem-solving abilities in geometry. Practice questions included in the notes enable students to apply the concepts they've learned to solve a variety of problems.
-
Preparation for Exams:
CBSE Class 9 Maths exams often include questions related to quadrilaterals. These notes serve as a valuable resource for exam preparation, ensuring that students are well-equipped to tackle questions on this topic.
