Area is one of the most fundamental concepts in geometry that measures the amount of space occupied by a two-dimensional shape or surface. Whether it is finding the floor space of a room, the size of a playground, or the surface covered by a piece of land, area calculations are used extensively in daily life. Understanding area helps students solve problems related to squares, rectangles, triangles, circles, polygons, and other geometric figures. This topic is frequently covered in school mathematics, mensuration, SSC, Banking, CUET, CAT, Railways, Defence, and other competitive examinations. In this article, we will explore the definition of area, area formulas for different shapes, properties, solved examples, and practical applications.
This Story also Contains
Area is one of the most important concepts in geometry and mensuration. It measures the amount of surface occupied by a two-dimensional shape or figure. Whether calculating the size of a room, a piece of land, a playground, or a geometric figure, area helps determine the space enclosed within its boundaries. Understanding area formulas is essential for school mathematics, competitive exams, engineering, architecture, and everyday measurements.
In simple words, area is the amount of space covered by a flat shape or surface.
For example:
The larger the surface covered by a shape, the greater its area.
Area is defined as the measure of the region enclosed by the boundary of a two-dimensional figure.
It is expressed in square units because both length and width are involved in the measurement.
For example:
Area is a fundamental concept that connects geometry, mensuration, algebra, and real-world measurements.
Area calculations are used in many practical situations.
| Situation | Use of Area |
|---|---|
| Flooring a room | Calculating floor space |
| Painting a wall | Finding paint coverage |
| Buying land | Measuring plot size |
| Designing gardens | Planning layouts |
| Construction projects | Material estimation |
Since area measures a two-dimensional region, its units are expressed as square units.
The standard unit of area in the International System of Units (SI) is the square metre.
Common units include:
Area involves multiplying two dimensions.
For example:
If a square has side length 4 cm,
Area $=4\times4=16 cm^2$
The notation $cm^2$ means 16 squares, each measuring 1 cm by 1 cm.
| Unit | Symbol |
|---|---|
| Square Millimetre | $mm^2$ |
| Square Centimetre | $cm^2$ |
| Square Metre | $m^2$ |
| Square Kilometre | $km^2$ |
| Hectare | ha |
| Acre | acre |
| Conversion | Value |
|---|---|
| $1 m^2$ | $10,000 cm^2$ |
| $1 cm^2$ | $100 mm^2$ |
| $1 km^2$ | $1,000,000 m^2$ |
| 1 hectare | $10,000 m^2$ |
| 1 acre | Approximately $4047 m^2$ |
Different geometric shapes have different area formulas based on their dimensions.
A square has four equal sides.
Formula:
$A=s^2$
where $s$ is the side length.
If side = 8 cm,
Area $=8^2=64 cm^2$
The area of a triangle depends on its base and height.
Formula:
$A=\frac{1}{2}bh$
If base = 12 cm and height = 8 cm,
Area $=\frac{1}{2}\times12\times8=48 cm^2$
From our previous knowledge we know that it is a triangle in which all the sides are equal. The perpendicular line that is drawn from the vertex of the triangle to the base divides the base into two equal parts as usual. Formula for area of equilateral triangle :
$A= \frac{(\sqrt{3})}{4} \times$ side $^2$

This type of triangle has two sides equal and the angles opposite to the equal sides are also equal. Formula for area of isosceles triangle used:
$A=\frac{1}{2} \times$ base $\times$ height
A rectangle is a four sided shape where opposite sides are equal and the diagonals are equal. The vertices of the rectangle makes $90^\circ$ with each other.
A rectangle has opposite sides equal.
Formula:
$A=l\times b$
where:
If length = 10 cm and breadth = 6 cm,
Area $=10\times6=60 cm^2$

Area $=$ Length $\times$ Breadth
$\mathrm{A}=\mathrm{lb}$
We follow the below steps to find the area of rectangle:
Step 1: We write dimensions of length, width as given in the question to us.
Step 2: Next, we multiply length, width values.
Step 3: At last, we write answer in square units along with the calculated numeric value of area.
A circle is a round shape with no corners or edges.

It is the space covered by the circle in an x-y plane. Or the other way around, the space occupied within the boundary/circumference of a circle.
We can find the area of circle once we know the diameter from which we find radius.
We use the following formulas:
- Area of circle $=\pi \times \mathrm{r}^2$, (r=radius)
- Area of circle in terms of diameter $=(\frac{\pi}{ 4}) \times \mathrm{d}^2, (\mathrm{~d}=$ diameter $)$
- Area of circle in terms of circumference $=\frac{C^2 }{4} \pi$, (C=circumference $)$
A square is a four sided shape where all the sides are equal and the diagonals are equal. The vertices of the square makes $90^\circ$ with each other.
Generally, area of square is stated as the number of square units required to fill the shape. In other words, the area of a square is the region within its boundary. It can also be calculated with the help of other dimensions, for example the diagonal and the perimeter of the square.

For finding area of square, we multiply the length of its two sides, which are always equal in measurement and so the unit of the area is given in square units.
Area of square $=$ Side $\times$ Side $=\mathrm{S}^2$
Area of square using diagonals $= \frac{Diagonal ^2}{ 2}$.
A trapezium is a four sided shape with two parallel sides and two non parallel sides.
We know from previous knowledge that a trapezium is a quadrilateral, which is defined as a shape with four sides and one set of parallel sides. Area of a trapezium depends upon the length of parallel sides and height of the trapezium. It is measured in square units.

Formula used to find the area of trapezium:
Area $=(\frac{1}{2}) h(a+b)$
where,
We take help from below mentioned steps to find area of trapezium:
A parallelogram is a four sided shape with parallel and equal opposite sides.
Area of parallelogram is the region covered by the parallelogram in a x-y plane. If we try to recall, a parallelogram is a special type of quadrilateral that has the pair of opposite sides as parallel. The opposite sides are always of equal length and opposite angles are of equal measures.

To find the area of parallelogram, we simply multiply the base of the perpendicular by its height. These both quantities are always perpendicular to each other, whereas the lateral side of the parallelogram is not perpendicular to the base.
Hence,
Area $=\mathrm{b} \times \mathrm{h}$ Square units(b=base, h=height)
We calculate area of parallelogram by using its base and height. It can also be calculated if its two diagonals are known along with any of their intersecting angles or if the length of the parallel sides is known, along with any of the angles between the sides.
Let us suppose p and q are the set of parallel sides of a parallelogram and h is the height, then based on the length of sides and height of it, formula for the area is :
Area $=$ Base $\times$ Height
$\mathrm{A}=\mathrm{p} \times \mathrm{h} \quad$ [sq.unit]
If the height of the parallelogram is unknown to us, then we use the concepts of trigonometry to achieve our aim.
Formula used :
Area $=a b \sin (x)$
Where a , b are the length of adjacent sides of the parallelogram and x is the angle between them.
It can also be calculated using its diagonal lengths. There are two diagonals for a parallelogram, which always intersect each other. Suppose the diagonals intersect each other at an angle p, then the area of the parallelogram is given by:
Area $=\frac{1}{2} \times d_1 \times d_2 \sin (p)$
A rhombus is a four sided shape with parallel and equal sides.
Area of Rhombus is defined as the amount of space covered by a rhombus in a two-dimensional space.It has four sides that are equal in length and are always congruent. It is also a type of a quadrilateral.

There exist three methods to find the area of a rhombus:
We consider a rhombus ABCD, that has two diagonals, AC & BD.
Step 1: First we find the length of diagonal $1, d_1$. It is the distance between $A$ and $C$. The diagonals of a rhombus are perpendicular to each other.
Step 2: Next , we find the length of diagonal 2, $d_2$, the distance between $B$ and $D$.
Step 3: Then, we multiply both the diagonals, $d_1$, and $d_2$.
Step 4: Finally we divide the result by 2 and get answer.
Step 1: We start by finding the base and height of the rhombus. The base of the rhombus is one of its sides, and the height is the altitude which is perpendicular distance from the chosen base to the opposite side.
Step 2: Then we multiply the base and the calculated height to get result.
We use this way when the side and one of its internal angles are given.
A cylinder is simply a three-dimensional structure having circular bases which are parallel to each other. It does not possess any vertices. It has 2 main values involved which are radius and its height.

Area of cylinder is defined as the sum of the curved surface and the area of two circular bases.
The Surface Area of Cylinder = Curved Surface + Area of Circular bases
S.A. (in terms of $\pi)=2 \pi r(h+r)$ sq.unit
$\pi(\mathrm{Pi})=3.142$ or $=\frac{22}{7}$
$r=$ Radius
h = Height
Surface Area of Cylinder: Surface area of cylinder is defined as the area of the curved surface of any cylinder having a base radius ‘r’, and height ‘h’, generally known as Lateral Surface Area (LSA). Formula for surface area of cylinder:
CSA or LSA $=2 \pi \times r \times h$ Square units
Base Area of Cylinder: It is a circular shape. Area of the circular bases of cylinder = $=2\left(\pi r^2\right)$
Total Surface Area of Cylinder: It is equal to the sum of the areas of all its faces. The total surface area with radius ‘r’, and height ‘h’ is equal to the sum of the curved area and circular areas of the cylinder.
TSA $=2 \pi \times r \times h+2 \pi r^2=2 \pi r(h+r)$ Square units
It is a solid curved surface in such a way that every point on the surface is the same distance from the centre. We can say that it is a 3D representation of a circle and cover areas in all x,y,z planes in geometry. We can find it in various forms around us like celestial bodies, football, basketball, etc.
Area of Sphere is the region covered by a surface of a spherical object in a three-dimensional space. As we have discussed earlier, if we try to spin a circle around a fixed axis and visualize this in our mind, we will find that we obtain the figure of a sphere.
Now to find the area of the sphere, we follow the steps:
Surface area of sphere is the area occupied by the curved surface of the sphere. Circular shapes take the shape of a sphere when we observe them as three-dimensional structures. For example, a globe, cricket ball or a soccer ball.

The formula for surface area of sphere just depends on the radius of the sphere.
Surface Area of Sphere $=4 \pi r^2 ; r=$ radius
In terms of diameter, $\mathrm{S}=4 \pi(\frac{\mathrm{d}}{2})^{2 }$, d=diameter
A cube is a 3 dimensional shape with equal sides. A cube is made up of 6 squares as their faces.
The definition of surface area of a cube states that if the total surface area is equal to the sum of all the areas of the faces of the cube. Since the cube has six faces, therefore, the total surface area of a cube will be equal to sum of all six faces of cube.

Total Surface Area of Cube (TSA) Formula
TSA of the cube is obtained by multiplying the square of its side length by 6 . Thus, the formula becomes " $6 \mathrm{a}^2$ ".
Total Surface Area of a Cube $=\left(6 \times\right.$ side $\left.^2\right)$ square units
Lateral Surface Area of Cube (LSA) Formula
LSA of the cube is obtained by multiplying the square of its side length by 4 . Thus, the formula becomes " $4 \mathrm{a}^2$ ".
Lateral Surface Area of a Cube $=\left(4 \times\right.$ side $\left.^2\right)$ square units.
Area formulas vary depending on the nature and structure of the geometric figure.
Two-dimensional shapes have length and width but no thickness.
Examples include:
Regular shapes have equal sides and equal angles.
Examples:
Their area formulas are fixed and straightforward.
Irregular shapes do not have equal sides or angles.
Their area can be found by:
| Shape | Area Formula |
|---|---|
| Square | $s^2$ |
| Rectangle | $lb$ |
| Triangle | $\frac{1}{2}bh$ |
| Circle | $\pi r^2$ |
| Parallelogram | $bh$ |
| Rhombus | $\frac{1}{2}d_1d_2$ |
| Trapezium | $\frac{1}{2}(a+b)h$ |
| Kite | $\frac{1}{2}d_1d_2$ |
Finding area involves identifying the shape and applying the appropriate formula.
Identify the geometric shape.
Measure the required dimensions.
Select the appropriate area formula.
Substitute the values into the formula.
Simplify and write the answer in square units.
Always identify the shape before selecting a formula.
For example:
Find the area of a square of side 9 cm.
Area $=9^2=81 cm^2$
Find the area of a rectangle with length 15 cm and breadth 4 cm.
Area $=15\times4=60 cm^2$
Area possesses several important mathematical properties.
If a figure is divided into non-overlapping parts, the total area equals the sum of the areas of the individual parts.
Area of whole figure = Sum of areas of all parts.
Area depends on two dimensions.
If a length doubles while the width remains unchanged, the area doubles.
| Area | Perimeter |
|---|---|
| Measures surface covered | Measures boundary length |
| Measured in square units | Measured in linear units |
| Two-dimensional concept | One-dimensional concept |
When a shape is cut and rearranged without stretching or overlapping, its area remains unchanged.
This principle is widely used in geometry proofs and mensuration.
Area calculations are widely used across mathematics, science, engineering, and everyday life.
Area helps compare shapes, solve geometric problems, and derive mathematical formulas.
Architects use area calculations while designing:
Engineers calculate areas for:
Common applications include:
The following area chart provides a quick revision guide for commonly used geometry formulas.
| Shape | Formula |
|---|---|
| Square | $s^2$ |
| Rectangle | $lb$ |
| Triangle | $\frac{1}{2}bh$ |
| Circle | $\pi r^2$ |
| Parallelogram | $bh$ |
| Rhombus | $\frac{1}{2}d_1d_2$ |
| Trapezium | $\frac{1}{2}(a+b)h$ |
| Kite | $\frac{1}{2}d_1d_2$ |
| Shape | Dimensions Required |
|---|---|
| Square | Side |
| Rectangle | Length, Breadth |
| Triangle | Base, Height |
| Circle | Radius |
| Rhombus | Diagonals |
| Trapezium | Parallel Sides, Height |
| From | To |
|---|---|
| $1 m^2$ | $10,000 cm^2$ |
| $1 cm^2$ | $100 mm^2$ |
| $1 km^2$ | $1,000,000 m^2$ |
| 1 hectare | $10,000 m^2$ |
| 1 acre | Approximately $4047 m^2$ |
A strong understanding of area formulas and mensuration concepts is essential for solving geometry and quantitative aptitude questions. The following books provide comprehensive theory, formulas, and practice problems.
| Book Name | Best For | Why It Helps |
|---|---|---|
| Quantitative Aptitude for Competitive Examinations R.S. Aggarwal | SSC, Banking, Railways | Covers mensuration and area formulas extensively |
| NCERT Mathematics Textbooks | School Students | Strong conceptual foundation |
| Mathematics for Class 9 & 10 R.D. Sharma | Board Exams | Detailed explanations with examples |
| Objective Mathematics Arihant Publications | Competitive Exams | Topic-wise geometry and mensuration questions |
| Fast Track Objective Arithmetic Rajesh Verma | Aptitude Preparation | Shortcut methods for mensuration problems |
Learning a few important shortcuts can help solve area-based questions faster and reduce calculation errors in examinations.
| Trick | Explanation |
|---|---|
| Square Area Shortcut | Area = side × side |
| Rectangle Area Shortcut | Area = length × breadth |
| Triangle Area Shortcut | Area = $\frac{1}{2} \times$ base $\times$ height |
| Circle Area Shortcut | Area = $\pi r^2$ |
| Parallelogram Shortcut | Area = base × height |
| Rhombus Shortcut | Area = $\frac{1}{2}d_1d_2$ |
| Trapezium Shortcut | Area = $\frac{1}{2}(a+b)h$ |
| Always use square units | Write answers in $cm^2$, $m^2$, etc. |
This formula table provides a quick revision guide for the most commonly used area formulas in geometry and mensuration.
| Shape | Area Formula |
|---|---|
| Square | $s^2$ |
| Rectangle | $lb$ |
| Triangle | $\frac{1}{2}bh$ |
| Circle | $\pi r^2$ |
| Parallelogram | $bh$ |
| Rhombus | $\frac{1}{2}d_1d_2$ |
| Trapezium | $\frac{1}{2}(a+b)h$ |
| Kite | $\frac{1}{2}d_1d_2$ |
| Regular Polygon | $\frac{1}{2}\times\text{Perimeter}\times\text{Apothem}$ |
Example 1: Find the area of a square with a side of 2 cm.
Solution:
Area of a square $=$ side $\times$ side
Given, side $= 2\ \mathrm{cm}$
Substituting the value in the formula:
Area $= 2 \times 2$
Area $= 4\ \mathrm{cm}^2$
Hence, the area of the square is $4\ \mathrm{cm}^2$.
Example 2: The dimensions of a rectangle are 10 cm and 8 cm. Find the area of the rectangle.
Solution:
The area of a rectangle is given by:
Area $=$ length $\times$ width
Given,
Length $= 10\ \mathrm{cm}$
Width $= 8\ \mathrm{cm}$
Substituting the values:
Area $= 10 \times 8$
Area $= 80\ \mathrm{cm}^2$
Hence, the area of the rectangle is $80\ \mathrm{cm}^2$.
Example 3: Can you find the area of a circle with a radius of 6 cm?
Solution:
Given,
Radius $= 6\ \mathrm{cm}$
Area of a circle $= \pi r^2$
Substituting the values:
Area $= \frac{22}{7} \times 6 \times 6$
Area $= \frac{792}{7}$
Area $= 113.14\ \mathrm{cm}^2$ (approximately)
Hence, the area of the circle is $113.14\ \mathrm{cm}^2$.
Example 4: The length of a rectangular field is 11 m and its width is 5 m. Find the area of the rectangular field.
Solution:
Given,
Length $= 11\ \mathrm{m}$
Width $= 5\ \mathrm{m}$
Area of a rectangle $=$ length $\times$ width
Substituting the values:
Area $= 11 \times 5$
Area $= 55\ \mathrm{m}^2$
Hence, the area of the rectangular field is $55\ \mathrm{m}^2$.
Example 5: Find the area of a triangle with base $b = 2\ \mathrm{cm}$ and height $h = 4\ \mathrm{cm}$.
Solution:
Area of a triangle $= \frac{1}{2} \times b \times h$
Given,
$b = 2\ \mathrm{cm}$
$h = 4\ \mathrm{cm}$
Substituting the values:
Area $= \frac{1}{2} \times 2 \times 4$
Area $= 4\ \mathrm{cm}^2$
Hence, the area of the triangle is $4\ \mathrm{cm}^2$.
Understanding area becomes easier when you explore related geometry and mensuration concepts that build a strong mathematical foundation. The following topics cover important formulas, properties, and problem-solving techniques frequently used in school mathematics and competitive examinations.
Frequently Asked Questions (FAQs)
Area of a shape is a 2-D quantity that is measured in square units like square inches or $\left(\right.$ in $\left.^2\right)$, etc.
The area of irregular shapes can be found by dividing the shape into unit squares. We can also approximate and find its value.
If a circle is folded into a triangle, the radius becomes the height of the triangle and the perimeter becomes its base which is $2 \times \pi \times r$. We know that the area of the triangle is found by multiplying its base and height and then dividing by 2 , which is: $\frac{1}{2} \times 2 \times \pi \times r \times r$. Therefore, the area of the circle is $\pi r^2$.
The total length of the boundary of a closed shape is called its perimeter. The perimeter of the triangle is the sum of three sides of the triangle.
The formulas for the area and perimeter of a square and a rectangle are as follows. Area of a square $=$ side $\times$ side. The perimeter of a square $=4 \times$ side. Area of a rectangle $=$ length $\times$ breadth. Perimeter of a rectangle $=2 \times$ (length + width )
On Question asked by student community
Hello Dear Student,
You will not be eligible for full fee reimbursement. While your income fits the criteria, the Telangana government caps the reimbursement amount for Backward Class (BC) candidates to a maximum of 35,000 per year.
Hope it helps!
Hello Dear Student,
For 7th-grade students in Maharashtra (including Pune), the premier option is the MSCE Pune Upper Primary Scholarship Examination (UPSE). Managed by the Maharashtra State Council of Examination, it awards qualifying 7th standard students 7,500 annually for three years.
You can check, find and access more information here:
Hello Student,
Agar aap NIOS kay student hai, aap DIPS mein admission le sakte hain. Halaki, DIPS CBSE board hai, by-laws ke according NIOS kay students class 11 mein admission lay skate hain.
Hello Dear Student,
A score of 188 in NCET places you in a competitive range, but your admission chances will depend on factors such as your category, state, preferred institution, and the cutoff trends for the current admission cycle.
In Uttar Pradesh, some institutions you may consider include:
Aligarh Muslim
Hello,
With a BITSAT score of 294 , JEE Main rank of 21k , and JEE Advanced rank of 17k , your best opportunities are likely through BITS rather than JEE.
Since you're interested in Computer Science and Mathematics & Computing , I would strongly consider:
BITS Goa CS
BITS