Area is defined as the total surface on a plane occupied by a closed shape. We measure the area in square units, and the standard unit is square meters \((\text {sq.m})\). To calculate the area of some plane figures such as triangle, rectangle, square, circle etc., we have formulas that are true for all values and easy to handle. In this article, we will learn different shapes and their formulas for areas that make our life easier.

What is Area?

The magnitude of the surface of the plane covered by a closed figure is called the area. In other words, the area is the measure of the entire interior region.

Concept of Area

Given below are open figures, and hence they do not occupy the area.

The general practice to calculate the area of any closed figure is to use square units and then find the number of square units present in the figure.
For example, a square of side \(1\) unit has an area of \(1 \,\text {sq. unit}\). It is written as \(\text {sq.m}\) when the unit is a meter (standard unit). Other units are \(\text {sq. decimeter, sq. km}\) etc.

The area of any figure can be measured by counting the number of square units occupied by the figure. Steps to find the area is as follows:

1. The number of full squares and the number of squares more than half are considered full squares.
2. Half squares are counted as half.
3. Squares that are less than half are ignored.
4. Lastly, the number of square units obtained in steps \((1), (2)\) and \((3)\) are added to get the area of the figure.

The area of the figure \((i)\) and \((ii)\) in the above figure can be easily measured because square units fit them evenly.
Area of figure \((i) =8 \,\text {sq.units}\). (Since the number of full squares \(=8\))
Area of figure \((ii) =9 \,\text {sq.units}\) (Since number of full square \(=9\))
Area of figure \((iii) =6+6 \times \frac{1}{2}=9 \,\text {sq.units}\) (full square \(=6\), half squares \(=6\))
Area of figure \((iv) =5+4 \times \frac{1}{2}=7 \,\text {sq.units}\) (full square \(=5\), half square \(=4\))
Area of figure \((\mathrm{v})=5 \,\text {sq.units}\) (square more than half \(=5\), square less than half \(=4\))
Area of figure \((\mathrm{vi})=2+5+2 \times \frac{1}{2}=8 \,\text {sq.units}\) (full square \(=2\), more than half square \(=5\), half square \(=2\))

From the above experiment, we have understood the difficulty of measuring the area of any shape when the square unit does not fit into them completely. It is impractical to measure the area of a certain shape by counting the units of squares occupied. It cannot always give us an accurate area. It can only be helpful to have an estimate. Hence, a generalised formula is used to find the area of closed figures that is valid for all the values.

Types of Shapes and Formula of Area

From the above discussion, we understood the importance of the area formula. Hence, we shall now learn the formulas of different shapes accepted for any values. The following are some of the shapes whose formulas will be discussed:

1. Area of Rectangle
2. Area of Square 
3. Area of Triangle
   a) Area of a triangle when base and height are given
   b) Area of a triangle with any side – Heron’s formula
   c) Area of an equilateral triangle – a special case of Heron’s formula
   d) Area of a triangle involving coordinates
   e) Area of a triangle involving trigonometric angles
4. Area of Parallelogram
5. Area of Rhombus
6. Area of a Trapezium
7. Area of Circle

Area of Rectangle

Let us assume a rectangle \(P Q R S\) of length \(l\) and breadth \(b\)

The area of rectangle \(\left( A \right) = {\rm{length}} \times {\rm{breadth}} = PQ \times QR\)
\(\Rightarrow A=l \times b\), where \(l\) is length and \(b\) is the breadth of the rectangle.
Thus, the area of a rectangle is given by the product of its length and breadth.

Area of Square

Consider a square whose length of the side is \(a\).

The area of square, \(A= \text {side} \times \text {side} =a^{2}\), where \(a\) is the side of the square. Thus, the area of a square is given by the square of the length of its side.

Area of Triangle

a) Area of a Triangle when Base and Height of the Triangle are given:

From the figure, the area of the triangle \(=\frac{1}{2} \times \text {base} \times \text {height} =\frac{1}{2} \times B C \times A D=\frac{1}{2} \times b \times h\).

b) Area of a Triangle when the Lengths of the Sides are given – Heron’s formula:

Heron’s formula is used to calculate the area of a triangle when the length of all three sides is known or given.

The area of a triangle according to Heron’s formula is given by,
area \((A)=\sqrt{s(s-a)(s-b)(s-c)}\), where, \(a, b\) and \(c\) are the lengths of sides of the triangle and \(s\) is the semi-perimeter of the triangle, given by \(s=\frac{a+b+c}{2}\).

c) Area of an Equilateral Triangle: A special case of Heron’s Formula:

So, Heron’s formula or Hero’s formula can be used to derive a special formula applicable to calculate the area of an equilateral triangle only.

In an equilateral triangle, all sides are equal in length. So, in this case, \(a=b=c\).

So, \(s=\frac{a+b+c}{2}=\frac{a+a+a}{2}=\frac{3 a}{2}\)

Hence, the area of an equilateral triangle is given by, \(A=\frac{\sqrt{3}}{4} \times(\text {side})^{2}=\frac{\sqrt{3}}{4} \times a^{2}\).

d) Area of an Isosceles Triangle:

Assume an isosceles triangle shown in the figure below:

The area of the isosceles triangle \( = \frac{1}{2} \times {\text{base}} \times {\text{height}} = \frac{1}{2} \times BC \times AD\)
\(=\frac{1}{2} \times b \times \sqrt{a^{2}-\frac{b^{2}}{4}}\)

e) Area of a Triangle involving Trigonometric Ratio:

In \(\triangle A B C, a, b\) and \(c\) are the side lengths of the triangle. (\(a\) is considered as the side opposite to \(\angle A\)). Similarly, \(b\) and \(c\) are considered as the sides opposite to the angles \(\angle B\) and \(\angle C\), respectively.
The formula for the area of the triangle is given by,
Area \(=\frac{1}{2} a b \sin C\)
Area \(=\frac{1}{2} b c \sin A\)
Area \(=\frac{1}{2} c a \sin B\)
Hence, the area of a triangle involving trigonometric ratio is,
area \(=\frac{1}{2} \times \text {product of any two sides} \times \text {sine of the angle including those two sides.}\)

f) Area of a Triangle when Coordinates of its Vertices are given:

The area of a triangle can be computed when the coordinates of the three vertices of a triangle on a Cartesian plane are given.

In the triangle \(A B C\), shown above, \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\) and \(C\left(x_{3}, y_{3}\right)\) represent the coordinates of the vertices of the triangle.
The area of the triangle can be calculated by using the formula,
area \(=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]\)

Area of Parallelogram

Assume a parallelogram \(A B C D\) with base “\(b\)” and height “\(h\)”

Now, area of parallelogram \( = {\rm{base}} \times {\rm{height}} = DC \times AE = b \times h\)

Thus, the product of the base and height of a parallelogram gives its area.

Area of Rhombus

Assume a rhombus has diagonals \(a\) and \(b\)

The area of the rhombus with diagonal lengths \(a\) and \(b=\frac{a \times b}{2}\)
Thus, the area of a rhombus is computed as half the product of its diagonals.

Area of Trapezium

Consider a trapezium \(M N O P\) of parallel sides \(a\) and \(b\) and height \(h\)

The area of trapezium is given by, \(A=\frac{1}{2} \times(a+b) \times h\)
Thus, the area of a trapezium is calculated as half the product of the sum of parallel sides and its height.

Area of a Circle

Assume a circle of radius \(r\) as shown in the figure below:

The circle area is given by, \(A=\pi r^{2}\), where \(r\) is the radius of the circle.

Solved Examples – Area Formulas

Q.1. The length of the rectangle is \(16 \,\text {cm}\), and breadth is \(4 \,\text {cm}\). Find the length of the side of the square if the area of the rectangle and square are equal.
Ans: Given, rectangle length, \(l=16 \mathrm{~cm}\), breadth, \(b=4 \mathrm{~cm}\)
We know that the area of rectangle \(=l \times b\)
\(=16 \times 4=64 \mathrm{~cm}^{2}\)
As the area of rectangle and square are same, area of rectangle \(=\) area of square Let the side length of the square be \(a \mathrm{~cm}\)
\(\Rightarrow a^{2}=64\)
\(\Rightarrow a=\sqrt{64}\)
\(\Rightarrow a=8 \mathrm{~cm}\)
Thus, the length of the side of the square is \(8 \mathrm{~cm}\).

Q.2. \(CAT\) is a right-angled triangle with \(\angle C=90^{\circ}, C A=5 \mathrm{~cm}\), and \(A T=13 \mathrm{~cm}.\) \(CO\) is perpendicular to \(AT\). Find the area of the triangle.

Ans: Given, \(C A=5 \mathrm{~cm}, A T=13 \mathrm{~cm}\) and \(\mathrm{CO} \perp A T\)
Applying Pythagoras theorem to right-angled \(\Delta C A T\)
\(C A^{2}+C T^{2}=A T^{2}\)
\(\Rightarrow 5^{2}+C T^{2}=13^{2}\)
\(\Rightarrow 25+C T^{2}=196\)
\(\Rightarrow C T^{2}=196-25=144\)
\(\Rightarrow C T=\sqrt{144}=12 \mathrm{~cm}\)
Area of triangle \(C A T=\frac{1}{2} \times b \times h=\frac{1}{2} \times 5 \times 12\)
\(=30 \mathrm{~cm}^{2}\)
Thus, the area of the given triangle is \(30 \mathrm{~cm}^{2}\).

Q.3. \(A B C D\) is a parallelogram. \(BP\) is the perpendicular on \(C D\) and \(B Q\) is the perpendicular on \(A D\). If the area of the parallelogram is \(216 \mathrm{~cm}^{2}\), find the lengths of \(B P.\)

Ans: Given, \(A B=18 \mathrm{~cm}, B C=13 \mathrm{~cm}\), area of parallelogram \(=216 \mathrm{~cm}^{2}\)
We know that the area of a parallelogram \(=b \times h\), where \(b\) is the base and \(h\) is the height
Taking \(C D\) as the base, height is \(B P\)
Area of parallelogram \(=C D \times B P\)
\(\Longrightarrow 216=18 \times B P\)
\(\Rightarrow B P=\frac{216}{18}=12 \mathrm{~cm}\)
Thus, the length of height \(\mathrm{BP}\) is \(12 \mathrm{~cm}\).

Q.4. Find the diameter of a circle whose area is \(154 \mathrm{~cm}^{2}\). [Take \(\pi=\frac{22}{7}.\)]
Ans: Given, area of circle \(=154 \mathrm{~cm}^{2}\)
We know that the formula to compute the area of a circle is, \(A=\pi r^{2}\), where \(r\) is the radius of the circle
\(\Rightarrow \frac{22}{7} \times r^{2}=154\)
\(\Rightarrow r^{2}=154 \times \frac{7}{22}=49\)
\(r=\sqrt{49}=7 \mathrm{~cm}\)
Diameter of the given circle \(=2 r=2 \times 7=14 \mathrm{~cm}\)
Thus, the diameter of the circle is \(14 \mathrm{~cm}\).

Q.5. Find the area of a rhombus whose diagonals are \(8 \mathrm{~cm}\) and \(10 \mathrm{~cm}\).
Ans: \(d_{1}=8 \mathrm{~cm}, d_{2}=10 \mathrm{~cm}\)
We know that area of a rhombus \(= d_{1} \times d_{2}\), where \(d_{1}, d_{2}\) are the diagonals of the rhombus
The area of rhombus \(=8 \times 10=80 \mathrm{~cm}^{2}\)
Thus, the area of the rhombus is \(80 \mathrm{~cm}^{2}\).

FAQs

Q.1. What is meant by area in math?
Ans: Area is a measure of the space occupied on a flat surface.

Q.2. How to compute the area of irregular shapes?
Ans: First, the irregular shape is plotted on a square grid. Then, the area of the shape is given by,
area of plane figure \(=\) area of full square \(+\) number of squares more than half-square \(+\frac{1}{2}\) (number of half squares).

Q.3. What is the formula for the area of a square?
Ans: Area of a square is computed as the square of its side.
The formula of the area of a square, \(A=a^{2}\), where a is the side length.

Q.4. What are the real-life uses of the area?
Ans: Some of the real-life uses of the area are floor covering, like tiles and carpets, wallpaper and paints etc. All these need an estimate of area. 

Q.5. Who invented Heron’s formula?
Ans: Heron’s formula is named after Heron of Alexandria, who invented the formula for finding the area of a triangle in terms of the lengths of its sides.

Study About Area Of Triangle

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