Rhombus: Definition & Properties

Rhombus is a two-dimensional figure known as an equilateral quadrilateral. The basic property of a rhombus is that it has four equal and opposite parallel sides with equal opposite angles. The diagonals of the rhombus bisect each other at right angles. Different things around us, such as a kite, Kaju barfi (sweet), the structure of a building, mirrors, diamonds, jewellery, etc., are also of a Rhombus shape. Let us learn more about the rhombus properties, area formula, perimeter and examples.

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Quadrilateral: Definition

Let us discuss briefly what a quadrilateral is and how it looks like before we discuss the rhombus. A quadrilateral is a two-dimensional closed figure with four sides and the sum of its angles equal to \(360^\circ .\)
There are different types of quadrilaterals such as parallelograms, rectangles, squares, rhombuses, kites, etc. 

Rhombus: Definition

A two-dimensional shape whose four sides are equal, the opposite sides are parallel, and the opposite angles are equal is called a rhombus. Its diagonals bisect each other at right angles.

In the above rhombus \(ABCD,\,AC\) and \(BD\) are diagonals that intersect at point \(E\). Here, \(AB\) is parallel to \(CD\) and \(AD\) is parallel to \(BC,\) and the diagonal \(AC\) is perpendicular to the diagonal \(BD.\)

Rhombus: Properties

The properties of rhombus is listed as follows:

1. A rhombus has four equal sides.
2. The opposite sides of a rhombus are parallel.
3. A rhombus has equal opposite angles.
4. The diagonals of a rhombus bisect each other at right angles.
5. The diagonals bisect the angles of a rhombus.
6. The sum of two adjacent angles of a rhombus is equal to a straight angle.
7. The two diagonals of a rhombus form four right-angled congruent triangles.

Properties of Angles in a Rhombus

The rhombus is a special case of the parallelogram as its opposite sides are equal and parallel.
A parallelogram is a two-dimensional shape whose opposite sides are parallel and equal, the opposite angles are equal and the sum of the adjacent angles are \(180^\circ .\) So, for a rhombus opposite angles are equal and the sum of the adjacent angles of a rhombus is \(180^\circ .\)

Rhombus Shape and Diagonals

The rhombus is called a diamond geometric shape. It is a special type of parallelogram. We can compare this shape with a kite, earrings and even a section of the baseball field.

Diagonals

Diagonals bisect the angles of a rhombus since a rhombus is a parallelogram. The sum of two adjacent angles of a rhombus is equal to a straight angle.

Let us learn why the diagonals of a rhombus bisect each other at right angles. We know that a rhombus is a parallelogram and the diagonals of the parallelogram bisect each other.

Here, \(ABCD\) is a rhombus and \(AC\) and \(BD\) are the diagonals and they bisect each other at \(O.\)
Thus, \(BO = DO,\,AO = OC\)
Consider \(\Delta AOD\,\& \,COD.\)
\(AO = OC\) (diagonals bisect each other)
\(DA = CD\) (all sides are equal)
\(OD = OD\) (common side)
Hence, \(\Delta AOD \cong \Delta COD\) (\(SSS\) congruence)
The corresponding parts of the congruent triangles are equal.
Hence, \(\angle COD = \angle AOD\)
\(\angle CDO = \angle ADO\)
\(\angle DAO = \angle DCO\)
As \(\angle CDO = \angle ADO,\) the diagonal \(BD\) bisect the angle \(\angle ADC.\)
Hence, it is proved that the diagonals of a rhombus bisect the angles.
Consider the line \(AC.\)
So, \(\angle COD + \angle AOD = 180^\circ \) (linear pair angles)
\( \Rightarrow 2 \times \angle COD = 180^\circ \)
\( \Rightarrow \angle COD = \frac{{180^\circ }}{2} = 90^\circ .\)
\( \Rightarrow \angle COD = 90^\circ \)
\( \Rightarrow \angle AOD = \angle BOC = 90^\circ \) (vertically opposite angles)
\( \Rightarrow \angle AOB = \angle DOC = 90^\circ \) (vertically opposite angles)
Hence it is proved that the diagonals of a rhombus bisect each other at right angles.
From the above proof, we get to know that \( \Rightarrow \angle AOD = \angle BOC = 90^\circ ,\,\angle AOB = \angle DOC = 90^\circ \)
Hence, \(\Delta AOD,\,\Delta COD,\,\Delta AOB,\,\Delta BOC\) are four right-angled congruent triangles.

Relation between a Rhombus and a Parallelogram

We know that in a parallelogram, the opposite sides are parallel and equal. In a rhombus also, the opposite sides are parallel and equal. The rhombus has all sides equal. So, a rhombus is a special case of a parallelogram. Let us understand by a flow chart view:

So, we can conclude that all the rhombuses are parallelograms.

Are All Parallelograms Rhombus?

As we know there are lots of similarities between parallelograms and rhombuses but still, they are not completely the same. In a parallelogram opposite sides are equal and in a rhombus all sides are equal.

In the parallelogram, \(AC = BD\) and \(AB = CD.\) But in the rhombus \(ABCD,\,AC = BD = CD = AB.\)

Relation Between a Rhombus and a Square

We know all sides of a square are equal in length and all sides of a rhombus are equal.

Now, a question comes to my mind “ what is the difference between these two?”. The differences are:

The Length of the Diagonals is the Same

There are some formulae related to the rhombus for calculating the perimeter of a rhombus and the area of a rhombus.

Rhombus: Area & Perimeter

Perimeter means the boundary. The total distance covered along the boundary of a rhombus is known as the perimeter of a rhombus. We have discussed earlier that a rhombus has four equal sides.
Let us say, the length of a side of a rhombus is \(a\) units.
Thus, its perimeter will be \(a + a + a + a = 4a\) units.

A rhombus is a polygon. The area of any polygon is the amount of space it occupies or encloses.
There are some methods to calculate the area of the rhombus:
Method 1: \({\rm{Area}} = {\rm{base}} \times {\rm{height}}\)
Method 2: \({\rm{Area}} = \frac{1}{2} \times {\rm{product}}\,{\rm{of}}\,{\rm{the}}\,{\rm{diagonals}}.\)

Area of the Rhombus Using Base and Height

Step 1: The base of the rhombus is one of its edges, and the height is known as the altitude.
Step 2: Find the product of the base and the height.

Area of the Rhombus Using Diagonal Method

Think about a rhombus \(ABCD\) that has two diagonals, i.e. \(AC\,\& \,BD.\)

1. Step 1: Let us join vertices \(A\) and \(C.\,AC\) is one of the diagonals of the rhombus \(ABCD.\)
2. Step 2: Similarly, we will find the length of diagonal \(BD,\) that is the distance between \(B\) and \(D.\)
3. Step 3: Find the product of the diagonals \(AC\)and \(BD.\)
4. Step 4: Divide the product by \(2.\)
   Now, we can easily find the area of the rhombus \(ABCD.\)

Solved Problems

Q.1. If the perimeter of a rhombus is \(40\,{\rm{cm,}}\) then find the length of its edge?
Ans: Let the length of the edge of the rhombus be \(a\)
Given, the perimeter \( = 40\,{\rm{cm}}\)
\( \Rightarrow 4a = 40\)
\( \Rightarrow a = \frac{{40}}{4} = 10\)
Hence, the length of its edge is \( = 10\,{\rm{cm}}.\)

Q.2. Meera comes around the field which is rhombus in shape to fix the poles. She has taken \(44\) feet to come around and she measures the height of the field as \(10\) feet. Find the area of the field.
Ans: Given that she has taken \(44\) feet to come around.
Thus, the perimeter of the rhombus field is \(44\) feet.
If one side of the field is a unit, then, the perimeter \( = 4a\)
\( \Rightarrow 4a = 44\)
\( \Rightarrow a = \frac{{44}}{4} = 11\)
Base \( = 11\) feet
Now, area of the rhombus \( = {\rm{base}} \times {\rm{height}}\)
Thus, the area of the field \( = 11 \times 10 = 110\,{\rm{sq}}{\rm{.}}\)feet

Q.3. Rahul was given the area of a rhombus and the length of one diagonal. The area, she was told, was \( = 100\,{\rm{sq}}{\rm{.}}\,{\rm{cm}}\) and the length of a diagonal was \(20\,{\rm{cm}}.\) Find the length of the other diagonal?
Ans: The formula of the area of a rhombus \( = \frac{1}{2} \times {\rm{product}}\,{\rm{of}}\,{\rm{the}}\,{\rm{diagonals}}\)
Length of one of the diagonals \( = 20\,{\rm{cm}}.\)
Let us assume the length of the other diagonal is \(x\,{\rm{cm}}.\)
Thus, \(x \times 20 = 100\)
\( \Rightarrow x = \frac{{100}}{{20}} = 5\)
Hence, the length of the other diagonal is \( 5\,{\rm{cm}}.\)

Q.4. In the figure, \(ABCD\) is a rhombus with \(\angle ABC = 56^\circ ,\) determine \(\angle ACD.\)

Ans: Given, \(\angle ABC = 56^\circ \)
\(\angle ADC = 56^\circ \) (opposite angles),
\(\angle BCD = 180^\circ – 56^\circ = 124^\circ \)
We know that
\(\angle BCD = \angle ACB + \angle ACD\,\left( {AC\;bisects\;\angle BCD} \right)\)
\(\angle ACB = \angle ADC\)
\(\angle BCD = 2\angle ACD\)
Hence, \(\angle ACD = \frac{{124^\circ }}{2} = 62^\circ .\)

Q.5. If the diagonals of a rhombus are \(12\,{\rm{cm}}\) and \(16\,{\rm{cm,}}\) then find the length of each side.

Ans: Imagine \(ABCD\) is a rhombus and \(AC = 16\,{\rm{cm}}\,\& \,BD = {\rm{12}}\,{\rm{cm}}\) are diagonals that bisect each other at \(O.\)
In \(\Delta AOD,\)
\(AO = 8\,{\rm{cm}},\,OD = 6\,{\rm{cm}}\)
Now, applying the Pythagoras theorem we get,
\(AD = \sqrt {O{D^2} + A{O^2}} = \sqrt {{6^2} + {8^2}} = 10\,{\rm{cm}}\)
Hence, the length of the side is \( = 10\,{\rm{cm}}.\)

Q.6. The area of a rhombus is \(119\,{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{.}}\) and its perimeter is \(56\,{\rm{cm}}.\) Find its height.
Ans: One of the sides of the field \( = a\) unit
Now, the perimeter \( = 4a\)
\( \Rightarrow 4a = 56\)
Therefore, \(a = \frac{{56}}{4} = 14\,{\rm{cm}}\)
Base \( = 14\,{\rm{cm}}\)
area of the rhombus \( = {\rm{base}} \times {\rm{height}}\)
area of the rhombus \( = 14 \times {\rm{height = 119}}\)
\( \Rightarrow {\rm{height}} = \frac{{119}}{{14}}\,{\rm{cm}} = 8.5\,{\rm{cm}}\)
Hence, the height of the rhombus is \(8.5\,{\rm{cm}}{\rm{.}}\)

Q.7. The area of the rhombus is \(144\,{\rm{c}}{{\rm{m}}^{\rm{2}}}\) and one of its diagonals is double of the other. Find the length of its diagonals.
Ans: Let us say one diagonal \(\left( {{d_1}} \right)\) is \(x\,{\rm{cm}}.\)
The other diagonal \(\left( {{d_2}} \right)\) is \(2x\,{\rm{cm}}\)
We know, \(\frac{1}{2} \times {d_1} \times {d_2} = \frac{1}{2} \times x \times 2x = 144 \Rightarrow {x^2} = 144\)
Therefore, \(x = 12\,{\rm{cm}}\) and \(2x = 24\,{\rm{cm}}\)
Hence, the length of the diagonals are \(12\,{\rm{cm}}\) and \(24\,{\rm{cm}}{\rm{.}}\)

Q.8. If the length of the side of a square field is \(4\,{\rm{m}}.\) What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its side is \(2\,{\rm{m}}.\)
Ans: Length of the square is \(4\,{\rm{cm}}.\)
Area of the square is \(4\,{\rm{cm}} \times {\rm{4}}\,{\rm{cm = 16}}\,{\rm{c}}{{\rm{m}}^2}.\)
Area of the square \( = \) Area of rhombus (given)
We know, area of the rhombus \( = {\rm{base}} \times {\rm{height}} \Rightarrow 2 \times {\rm{height}} = 16\)
Therefore, height \( = \frac{{16}}{2}\,{\rm{cm}} = 8\,{\rm{cm}}.\)

Q.9. If the area of the rhombus be \(24\;{\rm{c}}{{\rm{m}}^2}\) and one of the diagonals be \(4\,{\rm{m}}.\) Find the perimeter of Rhombus.
Ans: We know that the area of the rhombus \( = \frac{1}{2} \times {d_1} \times {d_2}\)
Let us say \({d_1} = 4\,{\rm{cm}}\)

Thus, \(\frac{1}{2} \times 4 \times {d_1} = 24\;cm^2\)
or, \({d_2} = \frac{{24}}{2}\,{\rm{cm}} = 12\,{\rm{cm}}\)
Now, \(AO = 2\,{\rm{cm}}\) and \(BO = 6\,{\rm{cm}}\) (as diagonals bisect each other at \(O\))
Therefore, \(AB = \sqrt {{6^2} + {2^2}} = 2\sqrt {10} \,{\rm{cm}}\)
Now, perimeter of the rhombus is \(4 \times 2\sqrt {10} = 8\sqrt {10} \,{\rm{cm}}.\)

Q.10. Find the area of a rhombus having each side \(13\,{\rm{cm}}\) and one of whose diagonals is \(24\,{\rm{cm}}.\)
Ans:

Diagonals of a rhombus bisect each other at right angles.
So, \(AO = OC = \frac{{24}}{2}\,{\rm{cm}} = 12\,{\rm{cm}},\,AD = 13\,{\rm{cm}}\)
Now, we will apply the Pythagoras theorem in \(\Delta AOD.\)
Therefore, \(DO = \sqrt {A{D^2} – A{O^2}} = \sqrt {{{13}^2} – {{12}^2}} = 5\,{\rm{cm}}\)
The other diagonal is \(5 \times 2 = 10\,{\rm{cm}}.\)
Hence, \(\frac{1}{2} \times {d_1} \times {d_2} = \frac{1}{2} \times 24 \times 10 = 120\,{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{.}}\)

Related Topics to Study

Summary

A rhombus is a two-dimensional figure which has four equal sides and opposite angles are equal. It is a special case of a parallelogram. All rhombuses are parallelograms, but all parallelograms are not rhombuses. All squares are rhombuses, but all rhombuses are not squares. Also, we learnt the important properties of a rhombus here and we have discussed how to find the area of the rhombus using different formulas.

FAQs

Frequently asked questions related to rhombus is listed as follows:

Q.1. Does a rhombus have four right angles?
Ans: If a rhombus is a square then only it has four right angles otherwise not. The opposite angles of a rhombus are equal, and the sum of its adjacent angles are \(180^\circ .\)

Q.2. How do we identify a rhombus?
Ans: If the diagonals of a quadrilateral bisect themselves at \(90^\circ \) and opposite interior angles are equal, then only we can say the quadrilateral is a rhombus.

Q.3. What are the properties of the rhombus?
Ans: We know that the rhombus has four equal sides. The opposite sides of a rhombus are parallel. A rhombus has equal opposite angles. The Diagonals of a rhombus bisect each other at right angles. The Diagonals bisect the angles of a rhombus. The sum of two adjacent angles of a rhombus is equal to a straight angle. The two diagonals of a rhombus form four right-angled congruent triangles.

Q.4. Why is the rhombus not a square?
Ans: In a square, all interior angles are always right angles but, in a rhombus, all angles need not be right angles or equal. Hence, a rhombus cannot be a square.

Q.5. Are the diagonals of a rhombus equal?
Ans: No, the diagonals of a rhombus are not equal. If a rhombus is a square, then its diagonals are equal but if the rhombus is not a square then the length of the diagonals will be different.

Q.6. How to find the area of a rhombus if the length of the diagonals is given?
Ans: Let us say \({d_1}\) and \({d_2}\) are two diagonals. The area of the rhombus \( = \frac{1}{2} \times {d_1} \times {d_2}\)

Q.7. How to find the area of a rhombus if the base and height are given?
Ans: If the measure of one side and the height or altitude is given then we will apply, \({\rm{Area}} = {\rm{base}} \times {\rm{height}}\) The base can be considered as the side length.

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