A rhombus is a type of quadrilateral. It is a special case of a parallelogram, with its diagonals intersecting each other at 90 degrees. The area of the rhombus is calculated by taking the product of its diagonals and dividing it by 2.
![Rhombus Shape](https://protonstalk.com/wp-content/uploads/2022/06/Rhombus-Shape-1024x542.png)
The rhombus is an equilateral quadrilateral; all of its sides are equal in length, hence the term ‘rhombus’, which has been derived from the ancient Greek word ‘rhombos’, which means something that spins.
Index
Area of a Rhombus
The area of the rhombus is calculated by taking the product of its diagonals and dividing it by 2,
\(A = \frac{1}{2} d_1 \cdot d_2\),
Here, \(A\) is the area of the rhombus & \(d_1\), and \(d_2\) are its diagonals.
![Area of Rhombus](https://protonstalk.com/wp-content/uploads/2022/06/Rhombus-1024x703.png)
Let ABCD be a Rhombus
E being the center and AC and BD as diagonals.
The \(A\) can be written as:
\(A = 4 \cdot\) Area of \(\triangle AEB\)
\(A = 4 \cdot \frac{1}{2} (AE)(BE)\)
(Since, diagonals perpendicularly bisect each other in a rhombus.)
\(\begin{align}
A & = 4 \cdot \frac{1}{2} (\frac{1}{2} AC) (\frac{1}{2} BD)\\
& = 4 \cdot \frac{1}{8} AC \cdot BD\\
& = \frac{1}{2} AC \cdot BD\\
\end{align}
\)
Let \(AC = d_1\) & \(BD = d_2\)
Then, we can rewrite it as,
\(A = \frac{1}{2} d_1 \cdot d_2\)
Solved Examples
Question 1. Find the area of rhombus with the diagonals being 5cm and 10cm.
Solution. Area of a rhombus can be given as,
\(\begin{align}
A & = \frac{1}{2} d_1 \cdot d_2\\
& = \frac{1}{2} 5 \cdot 10\\
& = 5 \cdot 5 = 25cm^2\\
\end{align}
\)
Question 2. Find the rhombus ABCD, with center O. If the area of its triangle AOB, formed by the diagonals, is 7 cm2.
Solution.
Area of rhombus ABCD = Area of \(\triangle AOB\) + Area of \(\triangle BOC\) + Area of \(\triangle COD\) + Area of \(\triangle AOD\)
\(\begin{align}
\text{Area of rhombus ABCD} & = 4 \cdot \text{Area of } \triangle AOB\\
& = 4 \cdot 7 = 28cm^2\\
\end{align}
\)
FAQs
The formula of area of Rhombus is
\(A = \frac{1}{2} d_1 \cdot d_2\)
Here, \(d_1\) and \(d_2\) are the two diagonals to the rhombus.
The formula gives the perimeter of the rhombus
\(P = 4a\), where \(a\) is the length of a side.
Diagonals of the Rhombus bisect each other at \(90^\circ\).
Diagonals of a rhombus may or may not be equal.