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October 11, 2024**Area of Rectangle:** The area of a rectangle is one of the basic concepts associated with Mathematics. Area is referred to as the total space that is enclosed by a 2-dimensional figure. Whereas, the area of a rectangle is the area that is enclosed by the rectangle. In other words, the space occupied within the perimeter (boundary) of a rectangle is called the area of the rectangle. The basic property of a rectangle is that it has four sides and each angle is 90 degrees.

There are many interesting applications of the area of a rectangle, such as determining the area of a rectangular floor, calculating the height of a building, etc. Also, when it comes to solving mensuration or geometry problems, the formula of area of rectangle formula plays an important role. In this article, let us understand the concept of the rectangle area in detail. Continue reading to learn more.

A rectangle’s area is determined by its sides. In general, the area of a rectangle is equal to the product of its length and width. The perimeter of a rectangle, on the other hand, is equal to the total of all of its four sides. As a result, the area of the rectangle is defined as the region enclosed by its perimeter.

A rectangle is a quadrilateral whose opposite sides are equal and parallel to each other. Since a rectangle has four sides, it has four angles. All angles of a rectangle are \(90\) degrees; thus, all the angles of the rectangle are right angles.

**Definition: **Area occupied by a rectangle within its boundary is called the area of the rectangle.

We have provided some examples of the rectangle that we encounter in our day-to-day lives such as doors of a room, computer screen, books, scale, etc.

The area of a rectangle is majorly dependent on its sides. For example, the area of a rectangle of a blackboard is dependent on the measure of length and breadth (or width).

Basically, the formula for the area of a rectangle is equal to the product of the length and breadth (or width) of the rectangle. Whereas, when we speak about the perimeter of a rectangle, it is equal to the sum of all its four sides.

Hence, the region enclosed by the perimeter of the rectangle is its area.

**Formula of Rectangle Area:**

Area of Rectangle \(= {\text{Length}} \times {\text{Breadth}}\)

The area of any polygon is the amount of space it occupies or encloses in the plane. The area is usually measured in square units like square meters, square feet, square inches, etc. Area of larger shapes such as fields or cities is measured in square kilometres, hectares, or acres.

The perimeter of the rectangle is the length of the boundary of the rectangle, which is given as the sum of all the sides of the rectangle.

The area of a rectangle is calculated in units by multiplying the breadth (or width) by the length of a rectangle.

Thus, the perimeter and the rectangle area is given by:

The Perimeter of Rectangle Formula | \(P = 2\left( {l + b} \right)\) |

Area of Rectangle Formula | \(A = l \times b\) |

The area of any rectangle is calculated when its length and breadth (or width) are known. By multiplying length and breadth, the rectangle’s area will be obtained in a square-unit dimension. In the case of a square, the area will become \(sid{e^2}.\)

The main difference between square and rectangle is that the length and breadth are equal for square, whereas the length and breadth are different in a rectangle.

The simplest way to find the rectangle area is to count the number of square units in it.

**Area of Rectangle = Number of Square Units Forming the Rectangle **

Example 1: Let us first calculate the area of the rectangle using the square method. Consider a rectangle \(ABCD\) of length \({\text{5 cm}}\) and width \({\text{4 cm}}\).

To find the area of this rectangle, we divide it into equal squares of \(1\) unit square each.

Now, count the number of squares in the above figure. There is a total of \(20\) squares.

Hence, the area of the rectangle \(ABCD\) is \({\text{20}}\,{\text{c}}{{\text{m}}^{\text{2}}}\).

Follow the steps below to find the area of a rectangle:

- Note the dimensions of length and breadth (or width) from the given data.
- Multiply length and breadth (or width) values.
- Write the answer in square units.

Firstly, the diagonals of the rectangle divide it into two equal right-angled triangles. Therefore, the area of the rectangle will be equal to twice the area of the right-angled triangle.

Suppose \(ABCD\) is a rectangle.

Now, let diagonal \(AC\) divide the rectangle into two congruent right-angled triangles, i.e., \(\Delta ABC\) and \(\Delta ADC.\) So, the area of both right-angled triangles will be equal.

Hence, area of rectangle \(ABCD\; = \) \(2 \times \) Area of \(\Delta ABC\).

**We know that area of the triangle** \(= \frac{1}{2} \times {\text{base}} \times {\text{height}}\)**Area of the rectangle** \( = 2 \times \frac{1}{2} \times AB \times BC\)

\(= 2 \times \left( {\frac{1}{2} \times l \times b} \right) = l \times b\)

Thus, the area of the rectangle \(= {\text{Length}} \times {\text{Breadth}}\)

We know that diagonal divides the rectangle into two congruent right-angled triangles. Thus, the diagonal becomes the hypotenuse of the corresponding right-angled triangle.

We know that the diagonal of a rectangle is calculated by using the Pythagoras theorem as follows:

\({\left( {{\text{Diagonal}}} \right)^{\text{2}}}{\text{=}}{\left( {{\text{Length}}}\right)^{\text{2}}}{\text{ + }}{\left( {{\text{Breadth}}} \right)^{\text{2}}}\)

From the above,

\({\text{Length=}}\sqrt {{{{\text{(Diagonal)}}}^{\text{2}}}{\text{-(Breadth}}{{\text{)}}^{\text{2}}}} \) and \({\text{Breadth=}}\sqrt {{{{\text{(Diagonal)}}}^{\text{2}}}{\text{-(Length}}{{\text{)}}^{\text{2}}}} \)

The area of the rectangle is given by

1. \({\text{Area}} = {\text{Length}} \times {\text{Breadth}} = {\text{Breadth}} \times \sqrt {{{\left( {{\text{Diagonal}}} \right)}^2} – {{\left( {{\text{Breadth}}} \right)}^2}} \)

2. \({\text{Area}} = {\text{Length}} \times {\text{Breadth}} = {\text{Length}} \times \sqrt {{{({\text{Diagonal}})}^2} – {{({\text{Length}})}^2}} \)

Let the length of the rectangle be \(“l”,\) breadth of the rectangle be \(“b”,\) the perimeter of rectangle be \(“P”\) and area of rectangle be \(“A”,\) then

1. Area of the rectangle \(= l \times b\)

2. The perimeter of the rectangle \(= 2\left( {l + b} \right)\)

3. Length of the rectangle \( = \frac{A}{b}\)

4. Length of the rectangle \( = \frac{P}{2} – b\)

5. The breadth of the rectangle \( = \frac{A}{l}\)

6. The breadth of the rectangle \( = \frac{P}{2} – l\)

7. Diagonal of the rectangle \( = \sqrt {{l^2} + {b^2}} \)

**1) Find the area, in a hectare, of the field whose length is \({\text{240}}\,{\text{m}}\) and width \({\text{110}}\,{\text{m}}\)**.

**Solution:**

Length \(\left( l \right) = 240\,{\text{m}}\) and width \((b) = 110~{\text{m}}\)

Area of the field (Rectangle) \( = l \times b\)

\( = 240 \times 110\)

\( = 26,400\,{{\text{m}}^2}\)

\( = \frac{{26,400}}{{10,000}} = 2.64\) hectare.

**2) The length of a rectangular screen is** \({\text{15}}\,{\text{cm}}\) **Its area is** \({\text{180}}\,{\text{sq}}{\text{.}}\,{\text{cm}}\). **Find its width.**

**Solution:** Area of the screen \(= 180\,{\text{sq}}.{\text{cm}}\)

Length of the screen \({\text{15}}\,{\text{cm}}\)

Area of the rectangle = \(l \times b\)

So, \({\text{Width=}}\frac{{{\text{Area}}}}{{{\text{Length}}}}\)

\( = \frac{{180}}{{15}} = 12~{\text{cm}}\)

**3) Find the perimeter and area of the rectangle of length** **\({\text{17}}\,{\text{cm}}\)** **and breadth** **\({\text{13}}\,{\text{cm}}\)**.

**Solution:**

Given: length \((l) = 17~{\text{cm}}\) and breadth \((b) = 13~{\text{cm}}\)

Perimeter of rectangle \( = 2\left( {l + b} \right)\)

\( = 2\left( {17 + 13} \right) = 2\left( {30} \right)\)

\(=60 \mathrm{~cm}\)

We know that area of rectangle \( = l \times b\)

\( = 17 \times 13\)

\({\text{=221}}\,{\text{sq}}{\text{.cm}}\)

**4) Find the area of a rectangle of length** **\({\text{43}}\,{\text{m}}\)** **and width** \({\text{13}}\,{\text{cm}}\).

**Solution:**

Given: length of the rectangle \((l) = 43\,{\text{m}}\).

Width of the rectangle \((b) = 13~{\text{m}}\)

We know that area of the rectangle \( = l \times b.\)

\( = 43 \times 13\)

\(= 559\,{{\text{m}}^2}\)

Area of the given rectangle is \(= 559\,{{\text{m}}^2}\).

**5) The length and width of the rectangular farm are** \(80\) **yards and** \(60\) **yards. Find the area of the farm.**

**Solution:**

Given: Length of the rectangle \((l) = 80\,{\text{yd}}\)

Width of the farm \((b) = 60\,{\text{yd}}\)

We know that area of the rectangle is length \( \times \) breadth.

So, area of the farm is \(= 80 \times 60 = 4800\,{\text{sq}}{\text{.yd}}\).

**6) The width of the rectangle is** \({\text{8}}\,{\text{cm}}\) **and its diagonal is** \({\text{17}}\,{\text{cm}}\). **Find the area of the rectangle and its perimeter.**

**Solution:**

Using Pythagoras theorem,

\(B{D^2} = C{D^2} + B{C^2}\)

⇒ \({17^2} = C{D^2} + {8^2}\)

⇒ \(C{D^2} = 289 – 64\)

⇒ \(C{D^2} = 225\)

⇒ \(CD = 15\)

Therefore, length of rectangle \({\text{=15}}\,{\text{cm}}\)

So, area of rectangle \( = l \times b\)

\(= 15 \times 8\,{\text{c}}{{\text{m}}^2}\)

\(= 120\,{\text{c}}{{\text{m}}^2}\)

Also, perimeter of rectangle \(= 2(15 + 8)\,{\text{cm}}\)

\(= 2 \times 23\,~{\text{cm}}\)

\(= 46\,~{\text{cm}}\).

**7) How many envelopes can be made from a sheet of paper** \({\text{100}}\,{\text{cm}}\) by \(75\,{\text{cm}}\), **supposing** \(1\) **envelope requires** \({\text{20}}\,{\text{cm}}\) **by** \({\text{5}}\,{\text{cm}}\) **piece of paper?**

**Solution:**

Area of the sheet \( = 100 \times 75\,{\text{c}}{{\text{m}}^2} = 7500\,{\text{c}}{{\text{m}}^2}\)

Area of envelope \( = 20 \times 5\,{\text{cm}} = 100\,{\text{c}}{{\text{m}}^2}\)

Number of envelopes that can be made \({\text{=}}\frac{{{\text{ Area of the sheet }}}}{{{\text{ area of the envelope }}}}\)

\( = \frac{{7500}}{{100}}\)

\( = 75\) envelopes

**8) The length and width of a rectangular wall are** \({\text{75}}\,{\text{m}}\) **and** \({\text{32}}\,{\text{m}}\), **respectively. Find the cost of painting the wall if the rate of painting is** \({\text{Rs}}{\text{.3}}\,{\text{per}}\,{\text{sq}}{\text{.}}\,{\text{m}}\)

**Solution:**

Given: Length of the wall \({\text{=75}}\,{\text{m}}\)

The breadth of the wall \({\text{=32}}\,{\text{m}}\)

Area of the wall \(= {\text{length}} \times {\text{breadth}}\)

\(= 75{\text{~m}} \times 32{\text{~m}}\)

\({\text{=2400}}\,{\text{sq}}{\text{.m}}\)

For \({\text{=1}}\,{\text{sq}}{\text{.m}}\) of painting costs \({\text{Rs}}.\,3\)

Thus, for \({\text{2400}}\,{\text{sq}}{\text{.m}}\) the cost of painting the wall will be \( = 3 \times 2400\)

\( = Rs.7200\)

**9) The length and breadth of a rectangular wall are**\( {\text{70}}\,{\text{m}}\) **and** \({\text{30}}\ {\text{m}}\), **respectively. Find the cost of painting the wall if the rate of painting is** \({\text{Rs 3 per sq}}{\text{. m}}\)

**Answer:**

Length of the wall \({\text{=70}}\,{\text{m}}\)

The breadth of the wall \({\text{=30}}\,{\text{m}}\)

Area of the wall \( = {\text{length}} \times {\text{breadth}} = 70~{\text{m}} \times 30~{\text{m}} = 2100\,{\text{sq}}.{\text{m}}\)

For \({\text{1}}\,{\text{sq}}{\text{.}}\,{\text{m}}\) of painting costs \({\text{Rs}}{\text{. 3}}\)

Thus, for \({\text{2100}}\,{\text{sq}}{\text{.}}\,{\text{m}}\) the cost of painting the wall will be \( = 3 \times 2100 = {\text{Rs}}\,6300\).

**10) A floor whose length and width is** \({\text{50}}\,{\text{m}}\) **and** \({\text{40}}\,{\text{m}}\), **respectively, needs to be covered by rectangular tiles. The dimension of each tile is** \({\text{1}}\,{\text{m}} \times 2\,{\text{m}}\). **Find the total number of tiles that would be required to fully cover the floor.**

**Answer:**

Length of the floor \(= 50\,{\text{m}}\)

Breadth of the floor \(= 40\,{\text{m}}\)

Area of the floor \(={\text{length}} \times {\text{breadth}} = 50\,{\text{m}} \times 40\,{\text{m}} = 2000\,{\text{sq}}{\text{.}}\,{\text{m}}\)

Length of one tile \(= 2\,{\text{m}}\)

Breadth of one tile \(= 1\,{\text{m}}\)

Area of one tile \(={\text{length}} \times {\text{breadth}} = 2\,{\text{m}} \times 1\,{\text{m}} = 2\,{\text{sq}}{\text{.}}\,{\text{m}}\)

No. of tiles required \( = \) area of floor/area of a tile \( = \frac{{2000}}{2}\;= \;1000\) tiles

**Check Properties of Rectangle Here**

Frequently asked questions related to the area of a rectangle are given below:

**Q.1.** **What is the formula for the area of the rectangle?****Ans:** The formula for the area of rectangle is:

\({\text{Area}} = {\text{Length}} \times {\text{Breadth}}\)

**Q.2.** **What is the perimeter of the rectangle?****Ans:** The perimeter of the rectangle is the sum of all its four sides.,

Hence, \({\text{Perimeter}}\left({{\text{rectangle}}} \right) = 2\left({{\text{Length}} + {\text{Breadth}}} \right)\)

**Q.3.** **What is the area of the rectangle?****Ans:** The area of the rectangle is the region occupied by the sides of the rectangle.

**Q.4.** **What is the unit of the area of the rectangle?****Ans:** The unit of area of any shape is a square unit. Hence, the unit of the area of rectangle is \({\text{c}}{{\text{m}}^{\text{2}}}{\text{,}}\,{{\text{m}}^{\text{2}}}{\text{,}}\,{\text{k}}{{\text{m}}^{\text{2}}}\) etc.

**Q.5.** **How to find length when the area and width of the rectangle are known?****Ans:** When the area and width of the rectangle is known, then length can be calculated by using:

\({\text{Length=}}\frac{{{\text{Area}}}}{{{\text{Width}}}}\)

**Q.6.** **How to find width when the area and length are known?****Answer:** When the area of the rectangle and length of the rectangle is known, then width can be calculated by using:

\({\text{Width=}}\frac{{{\text{Area}}}}{{{\text{Length}}}}\)

**Q.7.** **Why do we calculate the area of the rectangle?****Ans:** We calculate the area of a rectangle to find the area occupied by the rectangle within its perimeter.

**Q.8.** **Is the area of the rectangle the same as the area of the square?****Ans:** No, the area of the square is not necessarily the same as the area of the rectangle because every square is a rectangle with length and breadth equal, but all rectangles are not square. The formula to calculate the area of a rectangle is length×breadth and that of the square is \({\text{sid}}{{\text{e}}^2}\).

At Embibe, we provide Maths practice questions for Class \(8,9,10,11,\) and \(12\) along with detailed solutions:

Class 8 Maths Practice Questions |

Class 9 Maths Practice Questions |

Class 10 Maths Practice Questions |

Class 11-12 Maths Practice Questions |

So, make the best use of these resources and master the subject.

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