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
╦Ж
=
90
тИШ
ЁЭР┤^=90тИШ
Since,
A
D
тИе
B
C
ЁЭР┤ЁЭР╖тИеЁЭР╡ЁЭР╢
,
A
╦Ж
+
B
╦Ж
=
180
тИШ
ЁЭР┤^+ЁЭР╡^=180тИШ
(Sum of interior angles on the same side of transversal
A
B
ЁЭР┤ЁЭР╡
)
Therefore,
B
╦Ж
=
90
тИШ
ЁЭР╡^=90тИШ
Here,
A
B
тИе
C
D
ЁЭР┤ЁЭР╡тИеЁЭР╢ЁЭР╖
and
A
╦Ж
=
90
тИШ
ЁЭР┤^=90тИШ
(Given)
Therefore,
A
╦Ж
+
D
╦Ж
=
180
тИШ
ЁЭР┤^+ЁЭР╖^=180тИШ
тИ┤
D
╦Ж
=
90
тИШ
тИ┤ЁЭР╖^=90тИШ
тИ┤
C
╦Ж
=
90
тИШ
тИ┤ЁЭР╢^=90тИШ
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
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
is a rhombus in which
A
B
=
B
C
ЁЭР┤ЁЭР╡=ЁЭР╡ЁЭР╢
.
Since a rhombus is a parallelogram,
A
B
=
D
C
ЁЭР┤ЁЭР╡=ЁЭР╖ЁЭР╢
and
B
C
=
A
D
ЁЭР╡ЁЭР╢=ЁЭР┤ЁЭР╖
Thus,
A
B
=
B
C
=
C
D
=
A
D
ЁЭР┤ЁЭР╡=ЁЭР╡ЁЭР╢=ЁЭР╢ЁЭР╖=ЁЭР┤ЁЭР╖
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
ЁЭР┤ЁЭР╡=ЁЭР╖ЁЭР╢
,
B
C
=
C
D
ЁЭР╡ЁЭР╢=ЁЭР╢ЁЭР╖
.
Thus,
A
B
=
B
C
=
C
D
=
A
D
ЁЭР┤ЁЭР╡=ЁЭР╡ЁЭР╢=ЁЭР╢ЁЭР╖=ЁЭР┤ЁЭР╖
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
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
is a rectangle.
A
C
ЁЭР┤ЁЭР╢
and
B
D
ЁЭР╡ЁЭР╖
are diagonals.
To prove:
A
C
=
B
D
ЁЭР┤ЁЭР╢=ЁЭР╡ЁЭР╖
Proof:
Let,
A
╦Ж
=
90
тИШ
ЁЭР┤^=90тИШ
(By definition of rectangle)
A
╦Ж
+
B
╦Ж
=
180
тИШ
ЁЭР┤^+ЁЭР╡^=180тИШ
(Consecutive interior angle)
A
╦Ж
=
B
╦Ж
=
90
тИШ
ЁЭР┤^=ЁЭР╡^=90тИШ
Now in triangles,
A
B
D
ЁЭР┤ЁЭР╡ЁЭР╖
and
A
B
C
ЁЭР┤ЁЭР╡ЁЭР╢
,
A
B
=
A
B
ЁЭР┤ЁЭР╡=ЁЭР┤ЁЭР╡
(Common side)
A
╦Ж
=
B
╦Ж
=
90
тИШ
ЁЭР┤^=ЁЭР╡^=90тИШ
(Each angle is a right angle)
A
D
=
B
C
ЁЭР┤ЁЭР╖=ЁЭР╡ЁЭР╢
(Opposite sides of parallelogram)
Therefore,
╬Ф
A
B
D
тЙЕ
╬Ф
B
A
C
╬ФЁЭР┤ЁЭР╡ЁЭР╖┬атЙЕ┬а╬ФЁЭР╡ЁЭР┤ЁЭР╢
Therefore,
B
D
=
A
C
ЁЭР╡ЁЭР╖=ЁЭР┤ЁЭР╢
(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
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
is a parallelogram in which
A
C
=
B
D
ЁЭР┤ЁЭР╢=ЁЭР╡ЁЭР╖
.
To prove:
Parallelogram
A
B
C
D
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
is a rectangle.
Proof:
In triangles
A
B
C
ЁЭР┤ЁЭР╡ЁЭР╢
and
D
B
C
ЁЭР╖ЁЭР╡ЁЭР╢
,
A
B
=
D
C
ЁЭР┤ЁЭР╡=ЁЭР╖ЁЭР╢
(Opposite sides of parallelogram)
B
C
=
B
C
ЁЭР╡ЁЭР╢=ЁЭР╡ЁЭР╢
(Common side)
A
C
=
B
D
ЁЭР┤ЁЭР╢=ЁЭР╡ЁЭР╖
(Given)
Therefore,
╬Ф
A
B
C
тЙЕ
╬Ф
D
C
B
╬ФЁЭР┤ЁЭР╡ЁЭР╢┬атЙЕ┬а╬ФЁЭР╖ЁЭР╢ЁЭР╡
(
S
S
S
ЁЭСЖЁЭСЖЁЭСЖ
congruency condition)
Therefore,
A
B
╦Ж
C
=
D
C
╦Ж
B
ЁЭР┤ЁЭР╡^ЁЭР╢=ЁЭР╖ЁЭР╢^ЁЭР╡
(Corresponding parts of corresponding triangles)
But these angles are consecutive interior angles on the same side of transversal
B
C
ЁЭР╡ЁЭР╢
and
A
B
тИе
D
C
ЁЭР┤ЁЭР╡тИеЁЭР╖ЁЭР╢
.
Therefore,
A
B
╦Ж
C
+
D
C
╦Ж
B
=
180
тИШ
ЁЭР┤ЁЭР╡^ЁЭР╢+ЁЭР╖ЁЭР╢^ЁЭР╡=180тИШ
But,
A
B
╦Ж
C
=
D
C
╦Ж
B
ЁЭР┤ЁЭР╡^ЁЭР╢=ЁЭР╖ЁЭР╢^ЁЭР╡
Therefore,
A
B
╦Ж
C
=
D
C
╦Ж
B
=
90
тИШ
ЁЭР┤ЁЭР╡^ЁЭР╢=ЁЭР╖ЁЭР╢^ЁЭР╡=90тИШ
Therefore, by definition of rectangle, parallelogram
A
B
C
D
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
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
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
is a rhombus.
Diagonal
A
C
ЁЭР┤ЁЭР╢
and
B
D
ЁЭР╡ЁЭР╖
intersect at
O
ЁЭСВ
.
To prove:
A
C
ЁЭР┤ЁЭР╢
and
B
D
ЁЭР╡ЁЭР╖
bisect each other at right angles.
Proof:
A rhombus is a parallelogram such that
AB = DC = AD = BC
.
.
.
.
.
.
(
i
)
AB = DC = AD = BC┬а......(i)
Also the diagonals of a parallelogram bisect each other.
Hence,
B
O
=
D
O
ЁЭР╡ЁЭСВ=ЁЭР╖ЁЭСВ
and
A
O
=
OC
.
.
.
.
.
.
(
ii
)
ЁЭР┤ЁЭСВ=OC┬а......(ii)
Now, compare triangles
A
O
B
ЁЭР┤ЁЭСВЁЭР╡
and
A
O
D
ЁЭР┤ЁЭСВЁЭР╖
,
A
B
=
A
D
ЁЭР┤ЁЭР╡=ЁЭР┤ЁЭР╖
(From
(
i
)
(i)
above)
B
O
=
D
O
ЁЭР╡ЁЭСВ=ЁЭР╖ЁЭСВ
(From
(
ii
)
(ii)
above)
AO = AO
AO = AO
(Common side)
Therefore,
╬Ф
A
O
B
тЙЕ
╬Ф
A
O
D
╬ФЁЭР┤ЁЭСВЁЭР╡┬атЙЕ┬а╬ФЁЭР┤ЁЭСВЁЭР╖
(
S
S
S
ЁЭСЖЁЭСЖЁЭСЖ
congruency condition)
Therefore,
A
O
╦Ж
B
=
A
O
╦Ж
D
ЁЭР┤ЁЭСВ^ЁЭР╡=ЁЭР┤ЁЭСВ^ЁЭР╖
(Corresponding parts of corresponding parts)
B
D
ЁЭР╡ЁЭР╖
is a straight line segment.
Therefore,
A
O
╦Ж
B
+
A
O
╦Ж
D
=
180
тИШ
ЁЭР┤ЁЭСВ^ЁЭР╡+ЁЭР┤ЁЭСВ^ЁЭР╖=180тИШ
But,
A
O
╦Ж
B
=
A
O
╦Ж
D
ЁЭР┤ЁЭСВ^ЁЭР╡=ЁЭР┤ЁЭСВ^ЁЭР╖
(Proved)
Therefore,
A
O
╦Ж
B
=
A
O
╦Ж
D
=
180
тИШ
2
ЁЭР┤ЁЭСВ^ЁЭР╡=ЁЭР┤ЁЭСВ^ЁЭР╖=180тИШ2
A
O
╦Ж
B
=
A
O
╦Ж
D
=
90
тИШ
ЁЭР┤ЁЭСВ^ЁЭР╡=ЁЭР┤ЁЭСВ^ЁЭР╖=90тИШ
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
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
is a parallelogram in which
A
C
ЁЭР┤ЁЭР╢
and
B
D
ЁЭР╡ЁЭР╖
are perpendicular to each other.
To prove:
A
B
C
D
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
is a rhombus.
Proof:
Let
A
C
ЁЭР┤ЁЭР╢
and
B
D
ЁЭР╡ЁЭР╖
intersect at right angles at
O
ЁЭСВ
.
A
O
╦Ж
B
=
90
тИШ
ЁЭР┤ЁЭСВ^ЁЭР╡=90тИШ
In triangles
A
O
D
ЁЭР┤ЁЭСВЁЭР╖
and
C
O
D
ЁЭР╢ЁЭСВЁЭР╖
,
A
O
=
O
C
ЁЭР┤ЁЭСВ=ЁЭСВЁЭР╢
(Diagonals bisect each other)
O
D
=
O
D
ЁЭСВЁЭР╖=ЁЭСВЁЭР╖
(Common side)
A
O
╦Ж
D
=
C
O
╦Ж
D
=
90
тИШ
ЁЭР┤ЁЭСВ^ЁЭР╖=ЁЭР╢ЁЭСВ^ЁЭР╖=90тИШ
(Given)
Therefore,
╬Ф
A
O
D
тЙЕ
╬Ф
C
O
D
╬ФЁЭР┤ЁЭСВЁЭР╖┬атЙЕ┬а╬ФЁЭР╢ЁЭСВЁЭР╖
(
S
A
S
ЁЭСЖЁЭР┤ЁЭСЖ
congruency condition)
A
D
=
D
C
ЁЭР┤ЁЭР╖=ЁЭР╖ЁЭР╢
That is, the adjacent sides are equal.
Therefore, by definition,
A
B
C
D
ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖
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.