What is a Parallelogram?
A ‘parallelogram’ is a simple closed figure formed by line segments. As the name ‘parallelogram’ suggests, it is identified by the parallel line segments it is made of. Clearly, the parallel lines have to be opposite to each other, to form a quadrilateral. Therefore, we can define a parallelogram as a quadrilateral in which the opposite sides are parallel. The parallel sides of a parallelogram are also found to be equal in length. Also, the opposite pair of angles of a parallelogram are equal.
Here, AD = BC and AB = DC
∠A = ∠C and ∠B = ∠D
What is a Rhombus?
A rhombus is a special case of a parallelogram that has four sides that are equal. It is often called a diamond as it resembles the shape of a diamond in a 2-dimensional representation. A rhombus is also special because it also matches all the properties of a kite. A kite is a quadrilateral in which the opposite pairs of adjacent sides have equal length. As the rhombus is equilateral (all the sides are of equal length), it follows this property of a kite. We must remember that although a rhombus is equilateral, it need not be equiangular. It is only equilateral and equiangular in case of a square. The diagonals of a rhombus are not equal in length, but are perpendicular bisectors of each other. And each diagonal of a rhombus bisects a pair of opposite angles.
AB = BC = CD = DA
∠A = ∠C and ∠B = ∠D
AO = OC and DO = OB
∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Difference between Parallelogram and Rhombus
Parallelogram | Rhombus |
|---|---|
A parallelogram is a simple closed figure, in which the opposite sides are equal in length and parallel to each other. | A rhombus is a simple closed figure, in which all the four sides are equal in length, and the opposite sides are parallel. |
The diagonals of the parallelogram bisect each other but are not perpendicular to each other. | The diagonals of the rhombus are perpendicular bisectors of each other. |
If one of the sides of a parallelogram is ‘b’ units and its corresponding height is ‘h’ units then its area is equal to \(b~\times~h \) units squared. | If the diagonals of a rhombus are \(d_1 \) and \(d_2 \) units, then its area is \(\frac{1}{2}~\times~d_1~\times~d_2 \) square units. |
The formula for perimeter of parallelogram is \(2~(a~+~b) \), where a and b are the lengths of the adjacent sides of a parallelogram | The formula for perimeter of a Rhombus is 4a, where a is the length of each side of the rhombus. |
Solved Examples
Example 1: Find the measure of ∠1, ∠2, ∠3, and ∠4 in the rhombus ABCD given below.
Solution:
The diagonals of a rhombus are perpendicular bisectors of each other. Therefore,
\(\angle\)AOD = 90°
\(\angle\)1 = 90° \(\angle\)AOD = \(\angle\) 1
\(\angle\)BAC = 61° They form a pair of alternate interior angles.
\(\angle\)2 = 61° \(\angle\)BAC = 2
As each diagonal of a rhombus bisect a pair of opposite angles, therefore,
\(\angle\) DAC = \(\angle\) BAC = 61°
\(\angle\)3 = 61° \(\angle\)DAC = 3
Now in △AOD
\(\angle\)AOD + \(\angle\)DAO + \(\angle\)ADO = 180° Angle sum property for a triangle.
Or,
\(\angle\)1 + \(\angle\)3 + \(\angle\)4 = 180°
90° + 61° + \(\angle\)ADO = 180° Substitute the values found.
151° + \(\angle\)ADO – 151°= 180° – 151° Subtract 151° from both sides.
\(\angle\)ADO = 29°
\(\angle\)4 = 29° \(\angle\)ADO = \(\angle\)4
The unknown measurements are \(\angle\)1 = 90°, \(\angle\)2 = 61°, \(\angle\)3 = 61°, and \(\angle\)4 = 29°
Example 2: The extended arm of a desk lamp shown below is in the shape of a parallelogram. The angles of the parallelogram change as the lamp is raised and lowered. Find ∠BCD when \(\angle~ADC~=~120^\circ\)
Solution:
The adjacent angles of a parallelogram are supplementary.
So,
∠ADC + ∠BCD = 180°
120° + ∠BCD = 180° [ Substitute the value. ]
∠BCD = 180° – 120° [ Simplify ]
∠BCD = 60°
Therefore, the measure of ∠BCD = 60°.
Example 3: The lengths of the diagonals of a parallelogram are 6 cm and 8 cm respectively. One of the sides of a parallelogram is 5 cm. Check if the parallelogram is a rhombus.
Solution:
We know that the diagonals of a parallelogram are bisectors of each other, therefore,
\(BO~=~OD~=~\frac{1}{2}~BD\)
So, \(BO~=~\frac{1}{2}~\times~6~=~3\) cm [ As BD = 6cm is given.]
Similarly,
\(AO~=~OC~=~\frac{1}{2}~AC\)
So, \(AO~=~\frac{1}{2}~\times~8~=~4\) cm [ As AC = 8 cm is given.]
Let AB = 5 cm [ Given one side = 5 cm ]
To prove that ▱ ABCD is a rhombus, we need to check if AOB = 90°.
To check this we need to check △AOB. Is it the right triangle?
In △AOB
AO = 4 cm
BO = 3 cm
AB = 5 cm
The Pythagoras theorem is clearly satisfied here because,
\(AO^2~+~OB^2~=~AB^2\)
\(16~+~9~=~25\)
AO, BO and AB follow the Pythagoras theorem.
Hence the ▱ ABCD is a rhombus.
A square is a parallelogram that is equilateral and equiangular. The diagonals of a square are also equal and perpendicular bisectors of each other.
The key difference between a square and rectangle is that all the sides of a square are equal but in a rectangle only opposite sides are equal to each other.
All squares are rhombuses but all rhombuses are not squares. This is because a rhombus is not equiangular. When any of the interior angles of a rhombus become a right angle then we can say that the rhombus turns into a square.