Area of Rectangle - Definition, Formula, Derivation and Examples

A rectangle is a type of quadrilateral, a 2-D figure that has four sides and four vertices. In this article, we will learn how to calculate area of rectangle and its various applications in daily life, architecture, designing, etc. Rectangles are one of the most common geometric shapes in mathematics and found almost everywhere around us in the form of Books, windows, tables, laptops, tablets, etc.

What is Area of Rectangle?

The area of rectangle can be defined as the region covered within the boundary of the rectangle. Or the other way round, the amount of surface covered by a rectangle is called the area of rectangle. When we talk about the other dimensions of figure, like angle, all the four angles of the rectangle are measured as 90 degrees with opposite sides being equal and parallel.

The area of a rectangle is dependent on its sides. The formula for area of rectangle can be stated as follows:

Area of rectangle is equal to the product of the length and breadth of the rectangle. Whereas when we speak about the perimeter, it is equal to the sum of all its four sides. Hence, we can conclude that the region within the perimeter of the rectangle is the area of rectangle.

Formula of Area of Rectangle

Area of rectangle = Length $\times$ Breadth

$
{A}={lb}
$

In the case of a square, the area will become ${side}^2$. The main difference between these two figures is that the length and breadth are equal for square.

Derivation of Area of Rectangle Formula

Consider the rectangle PQRS with a diagonal PR. Diagonal PR divides the rectangle PQRS into two congruent triangles. The area of rectangle PQRS is the sum of the area of these two triangles.

Area of Rectangle PQRS = Area of Triangle PQR + Area of Triangle PRS
Since both the triangles (PQR and PRS) are congruent, we can express it as follows.
Area of Rectangle PQRS $=2 \times$ Area of Triangle PQR
Area of Rectangle PQRS $=2 \times(\frac{1}{2} \times$ Base $\times$ Height $)$
Area of Rectangle PQRS $=2 \times \frac{1}{2} \times P Q \times Q R$
Area of Rectangle $P Q R S=P Q \times Q R$
Hence, Area of Rectangle $=$ Length $\times$ Breadth.

How to find Area of Rectangle?

Now, let's see how to find area of rectangle.

Step 1: We write down the dimensions of length and breadth from the data that is given to us in the question.

Step 2: Next, we multiply the length and breadth to calculate the area of rectangle.

Step 3: Finally we write the answer in square units like sq m, sq cm along with its magnitude.

Surface Area of Rectangle

Surface area of rectangle is defined as the area under the rectangle in an xy plane. We can use the following steps to find the surface area of the rectangle:

Step 1: First we multiply the length ($l$) by the width ($b$) of rectangle.

Step 2: Hence we come across the expression $l \times b$, where $l$ is the length and $b$ is the width of rectangle.

We can say that surface area of rectangle is same as that of area of rectangle.

Area of Rectangle Using Different Conditions

Area of rectangle using different conditions like area of rectangle using diagonal and area of rectangle using perimeter.

Area of Rectangle Using Diagonal

The area of a rectangle can be calculated if we know the diagonal and any one side of the rectangle. We initially define the diagonal of a rectangle as the straight line inside the rectangle connecting its opposite vertices. There are two diagonals in a rectangle and both of them are equal.

There are 2 ways we can approach this problem:

Method 1: Pythagoras theorem helps us to find the unknown side and then we find the area.

^{Example: Find area of rectangle whose length is 2 cm and whose diagonal is 3 cm .
Solution:
The breadth of the rectangle is missing and it can be calculated using the Pythagoras theorem because the diagonals of a rectangle form 2 right-angled triangles.}

^{In this case, the breadth can be calculated using the formula, breadth $=\left[\sqrt{(\text { Diagonal })^2-(\text { length })}^2\right]$}

^{After substituting the given values, we get}

^{$
\begin{aligned}
& \text { breadth }=\sqrt{(3)^2-(2)^2} \\
& \text {breadth }=\sqrt{9-4 } \\
& \text { breadth }=\sqrt{5 } =2.23 \mathrm{~cm}
\end{aligned}
$}

^{Now, we know that the length $=2 \mathrm{~cm}$, breadth $=2.23 \mathrm{~cm}$. So, the area of the rectangle $=l \times b$. In this case,}

^{$
A=l \times b
$}

^{Area $=2 \times 2.23=6.44 \mathrm{~cm}^2$}

Method 2:

Here we apply a direct method:

For example, missing part is the length of the rectangle where diagonal and breadth are known, then we write the length in terms of the breadth and diagonal and then use the formula for area.

We know that $(\text { Diagonal })^2=(\text { Length })^2+(\text { breadth })^2$. Or $(\text { Length })^2=(\text { Diagonal })^2-$ $(\text { breadth })^2$. Further, this can be expressed as, Length $=\sqrt{ }\left[(\text { Diagonal })^2-(\text { breadth })^2\right]$. Here, the length comes out in terms of the diagonal and breadth. This formula can be used in place of the length.

We know that the formula to calculate the area of rectangle is,

Area of rectangle $=$ Length $\times$ Width.

Hence, Area of rectangle $=\sqrt{ }\left[(\text { Diagonal })^2-(\text { breadth })^2\right]$ $\times$ Width.

So, Area of a Rectangle $=$ breadth $\sqrt{ }\left[(\text { Diagonal })^2-(\text { Width })^2\right]$.

Let us use an example to understand this.

Example: Find the area of rectangle in which the width is 5 units, and the diagonal is 9 units.

Solution:

We know the diagonal and the breadth of the rectangle. So, let us apply the direct formula, Area of Rectangle $=$ Width $\sqrt{ }\left[(\text { Diagonal })^2-(\text { Width })^2\right]$

After substituting the values of diagonal $=9$ units, and width $=5$ units, we get, Area of Rectangle $=$ Breadth $\sqrt{ }\left[(\text { Diagonal })^2-(\text { Width })^2\right]$

Area $=5 \sqrt{ }\left[(9)^2-(5)^2\right]$
Area $=5 \sqrt{ }(81-25)$
Area $=5 \times \sqrt{ } 56$
Area of rectangle $=37.4$ square units

Area of Rectangle Using Perimeter

With the help of perimeter, we can find the unknown side and then find area using the same formula, Area of rectangle formula $=$ Length $\times$ breadth.

Example: Find the area of rectangle if the perimeter is 20 units and the length is 6 units.

Solution:

Perimeter of the rectangle $=20$ units, length $=6$ units. We can find the breadth using the formula,

Perimeter of rectangle $=2(l+b)$
$20=2(6+b)$
$10=6+b$
$b=4$ units
breadth $=4$ units, length $=6$ units,
Area of rectangle $=1 \times b$
Area $=4 \times 6=24$ square units

Solved Examples Based on Area of Rectangle

Example 1: Find the area of rectangle whose length is 12 cm and the width is 2 cm.
Solution:
Given,
Length $=12 \mathrm{~cm}$
Width $=2 \mathrm{~cm}$
Area of rectangle formula $=$ Length $\times$ Width

$
12 \times 2=24
$

So the area of rectangle $=24 \mathrm{~cm}^2$

Example 2: What is the area of rectangular blackboard whose length and breadth are 200 cm and 100 cm, respectively.

Solution:
Length of the blackboard $=200 \mathrm{~cm}=2 \mathrm{~m}$
Breadth of the blackboard $=100 \mathrm{~cm}=1 \mathrm{~m}$
Area of the blackboard $=$ area of rectangle $=$ length x breadth $=2 \mathrm{~m} \times 1 \mathrm{~m}=2$ square-metres

Example 3: The length of a rectangular screen is 10 cm. Its area is $80 \mathrm{sq}. \mathrm{cm}$. Find its width.
Solution:
Area of the screen $=80 \mathrm{sq} . \mathrm{cm}$.
Length of the screen $=10 \mathrm{~cm}$
Area of rectangle formula $=$ length x width
So, width = $\frac{\text{area}}{\text{length}}$
Thus, width of the screen $=\frac{80}{10}=8 \mathrm{~cm}$

Example 4: The length and breadth of a rectangular wall are 70 m and 30 m, respectively. Find the cost of painting the wall if the rate of painting is Rs 3 per sq. m.

Solution:

Length of the wall $=70 \mathrm{~m}$
The breadth of the wall $=30 \mathrm{~m}$

Area of rectangle $=$ length $x$ breadth $=70 \mathrm{~m} \times 30 \mathrm{~m}=2100$ sq. m
For $1 \mathrm{sq}. \mathrm{m}$ of painting costs Rs 3
Thus, for 2400 sq. m, the cost of painting the wall will be $=3 \times 2100=$ Rs 6300

Example 5: A floor whose length and width is 60 m and 20 m respectively needs to be covered by rectangular tiles. The dimension of each tile is $1 \mathrm{~m} \times 2 \mathrm{~m}$. Find the total number of tiles that would be required to fully cover the floor.

Solution:
Length of the floor $=60 \mathrm{~m}$
The breadth of the floor $=20 \mathrm{~m}$
Area of the floor=area of rectangle $=$ length $\times$ breadth $=60 \mathrm{~m} \times 20 \mathrm{~m}=1200 \mathrm{sq} . \mathrm{m}$
Length of one tile $=2 \mathrm{~m}$
The breadth of one tile $=1 \mathrm{~m}$
Area of one tile $=$ length $x$ breadth $=2 \mathrm{~m} \times 1 \mathrm{~m}=2 \mathrm{sq} . \mathrm{m}$
No. of tiles required $= \frac{\text{area of floor}}{\text{area of a tile}}=\frac{1200}{2}=600$ tiles

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