In this article, we will discuss in detail about Area Formula for Quadrilaterals. The magnitude of the measurement of the region enclosed by a closed plane figure is called its area. Let us recall what a quadrilateral is. A quadrilateral is a closed figure that is bounded by four-line segments. A quadrilateral may also be regular or irregular. A regular quadrilateral is a quadrilateral where all the sides are of equal length. A quadrilateral that is not regular is called an irregular quadrilateral.

The units of area are square centimetres (written as \(\mathrm{cm}^{2}\)), square meter (written as \(\text {m}^{2}\)), etc. The area of a square of side \(1 \mathrm{~cm}\) is \(1 \mathrm{~cm}^{2}\). The area of a square of side \(1 \mathrm{~m}\) is \(1 \mathrm{~m}^{2}\). Continue reading to know more about Area Formula for Quadrilaterals.

Area Formulas for All Quadrilaterals

A four-sided polygon is a quadrilateral. It has \(4\) sides, \(4\) vertices and \(4\) angles.

The area of a quadrilateral is nothing but the region surrounded by the sides of the quadrilateral. It is being measured in square units such as \(\text{cm}^{2}, \text {m}^{2}\) etc. The procedure of finding the area of a quadrilateral will vary depending on its type and the information that is available regarding the quadrilateral.

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If the quadrilateral does not belong to one of the types that are referred above, then we can find its area both by splitting it into two triangles or by using the formula (which is called the Bretschneider’s formula) of finding the area of quadrilateral using four sides. Now we can see the formulas to find the area of a quadrilateral which does not belong to either of the standard types.

There are \(6\) types of quadrilaterals.

1. Square
2. Rectangle
3. Parallelogram
4. Trapezium
5. Rhombus
6. Kite

Area of Square

A quadrilateral having all sides equal and each angle measure equal to right-angle, is called a square.

Let there be a square each of whose sides measures \(x\) units.

Then, we have:

7. Area of square \(x \times x=x^{2}\,\text {units}\)
8. Side of the square \(=\sqrt{\text { area }}\text {units}\)
9. Diagonal of the square \(=\sqrt{2} x\, \text {units}\)
10. Area of the square \(=\left[\frac{1}{2} \times(\text {diagonal})^{2}\right]\text{sq units}\)

Area of Rectangle

A quadrilateral having opposite sides equal, and each angle measures \(90^{\circ}\).

Consider a rectangle with length \(=l\) units and breadth \(=b\) units.

Then, we have:
Area of rectangle \(=(l \times b)\,\text {sq units}\)
Length \(=\left(\frac{\text { area }}{\text { breadth }}\right)\text {units}\)
Breadth \(=\left(\frac{\text { area }}{\text { length }}\right)\text {units}\)
Diagonal \(=\sqrt{l^{2}+b^{2}}\,\text {units}\)

Area of Parallelogram

Area of a parallelogram is the space covered inside the boundary (sides) of a parallelogram.

Area of a parallelogram \(=\text {(base×height) sq units}\)

Let \(P Q R S\) be a parallelogram. Join \(Q M\).
Then, \(Q M\) divides the parallelogram into two equal triangles.
Draw \(Q M \perp P S\).
Area of parallelogram \(P Q R S=2(\)area of \(\Delta P Q S)\)
\(=2\left(\frac{1}{2} \times P S \times Q M\right)\, \text {sq units}\) \(=(P S \times Q M)\, \text {sq units}\)
\(=\text {(base×height) sq units}\)

Therefore, the area of a parallelogram \(=\text {(base×height) sq units}\).

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Area of Trapezium

A trapezium is a quadrilateral in which one pair of opposite sides are parallel. The trapezium area can be calculated using the lengths of two of its parallel sides and the distance (height) between them. The formula to calculate the area \((A)\) of a trapezium using base and height is given as, \(A=\frac{1}{2}(a+b) \times h\) where,

1. \(a\) and \(b\) are two parallel sides of trapezium
2. \(h=\) height (the perpendicular distance between \(a\) and \(b\))

Area of Rhombus

A quadrilateral having all sides equal and opposite angles are equal is called a rhombus.

Area of rhombus \(=\frac{1}{2} \times(\text {product of diagonals})\)

Let \(P Q R S\) be a rhombus whose diagonals \(P R\) and \(Q S\) intersect at a point \(O\).
Let \(P R=d_{1}\) and \(Q S=d_{2}\).
We see that the diagonals of a rhombus bisect each other at right angles.
Now, in right \(\triangle P O Q\), we have:
\(ar(\Delta P O Q)=\left(\frac{1}{2} \times \frac{d_{1}}{2} \times \frac{d_{2}}{2}\right)=\frac{1}{8} d_{1} d_{2}\)

Therefore, the area of rhombus \(P Q R S=4 \times ar(P O Q)\)
\(=\left(4 \times \frac{1}{8} d_{1} d_{2}\right)=\left(\frac{1}{2} \times d_{1} \times d_{2}\right)\text {sq units}\)

Therefore, the area of a rhombus \(=\frac{1}{2} \times(\text {product of diagonals})\)

Area of Kite

A kite is a quadrilateral whose adjacent sides are equal.

The area of a kite represents half the product of the lengths of its diagonals. The formula to determine the area of a kite is area \(=\frac{1}{2} \times d_{1} \times d_{2}\)

Here \(d_{1}\) and \(d_{2}\) are long and short diagonals of a kite.

The area of kite \(P Q R S\) given below is \(\frac{1}{2} \times P R \times Q S\).

Here, \(P R\) is short diagonal, and \(Q S\) is long diagonal.

Area of Quadrilateral Formula- Coordinate Geometry

The area of a quadrilateral may be calculated when the coordinates of its vertices are known. Let us look at a quadrilateral in the coordinate plane as shown below,

In the quadrilateral given above, \(P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right), R\left(x_{3}, y_{3}\right)\) and \(S\left(x_{4}, y_{4}\right)\) are the vertices.

So, the area of the quadrilateral \(P Q R S\) is given as
\({\text{Area}} = \frac{1}{2}\left\{ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_1}} \right) – \left( {{x_2}{y_1} + {x_3}{y_2} + {x_4}{y_3} + {x_1}{y_4}} \right)} \right\}\)

Example: Find the area of a quadrilateral \(P Q R S\) to which the vertices are \(P(-3,1), Q(-1,4), R(3,2), S(1,-2)\)

Answer: Given, vertices of quadrilateral \(P Q R S\) are \(P(-3,1), Q(-1,4), R(3,2)\) and \(S(1,-2)\).

We know that area of quadrilateral \(=\frac{1}{2}\left\{ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_1}} \right) – \left( {{x_2}{y_1} + {x_3}{y_2} + {x_4}{y_3} + {x_1}{y_4}} \right)} \right\}\)

Now, \( = \frac{1}{2}\left\{ {\left({\left({ – 3} \right) \times 4 + \left({ – 1} \right) \times 3 + 3 \times \left({ – 2} \right) + 1 \times 1} \right) – \left( {\left( { – 1} \right) \times 1 + 3 \times 4 + 1 \times 2 + \left( { – 3} \right) \times \left( { – 2} \right)} \right)} \right\}\)

\(=\frac{1}{2}|(-12-3-6+1+1-12-2-6)|=19.5 \, \text {sq.units}\)

Area of Quadrilateral Formula with Sides

When the sides of a quadrilateral and two of its opposite angles are given, we will be able to find its area utilizing the Bretschneider’s formula. Let us look at a quadrilateral whose sides are \(a, b, c\), and \(d\), and two of its opposite angles are \(\theta_{1}\) and \(\theta_{2}\).

Then the area of the quadrilateral \(=\sqrt{(s-a)(s-b)(s-c)(s-d)-a b c d \cos ^{2} \frac{\theta}{2}}\), where

1. \(s=\) semi \(-\) perimeter of the quadrilateral \(=\frac{(a+b+c+d)}{2}\)
2. \(\theta=\theta_{1}+\theta_{2}\)

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Area Formula for Any Quadrilateral

Let \(PQRS\) be irregular quadrilateral.

Then area of quadrilateral \(P Q R S\)
\(=\frac{1}{2}\text {(Length of a diagonal)} \times \text {(Sum of the lengths of perpendiculars from the remaining two vertices on it)}\)
\(=\frac{1}{2} \times P R \times\left(h_{1}+h_{2}\right)\)

Area Formula for Cyclic Quadrilateral

Brahmagupta’s formula provides the area of a cyclic quadrilateral (i.e., a simple quadrilateral that is inscribed in a circle) with sides of length \(a, b, c\), and \(d\) as

Area \(=\sqrt{(s-a)(s-b)(s-c)(s-d)}\)

where \(s\) is the semi perimeter \(s=\frac{a+b+c+d}{2}\)

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Solved Examples– Area Formula for Quadrilaterals

Q.1. Find the area of a square, the length of whose diagonal is \(8 \sqrt{2} \mathrm{~m}\).
Ans: We know that area of the square \(=\left[\frac{1}{2} \times(\text { diagonal })^{2}\right]\text {sq units}\)
\(=\left(\frac{1}{2} \times(8 \sqrt{2}\, \text {m})^{2}\right) \text {m}^{2}=\left(\frac{1}{2} \times 64 \times 2\right) \text {m}^{2}=64 \, \text {m}^{2}\)
Hence, the area of the square is \(64 \mathrm{~m}^{2}\).

Q.2. Find the rectangle area whose dimensions are length \(=25 \mathrm{~m}\) and breadth \(=12 \mathrm{~m}\)
Ans: Given, length \(=25 \mathrm{~m}\) and breadth \(=12 \mathrm{~m}\)
We know that area of rectangle \(=(l \times b)\, \text {sq units}\)
\(=25 \mathrm{~m} \times 12 \mathrm{~m}=300 \mathrm{~m}^{2}\)
Hence, the area of the rectangle is \(300 \mathrm{~m}^{2}\).

Q.3. The base of a parallelogram is twice its height. If its area is \(512 \mathrm{~cm}^{2}\), find the base and the height.
Ans: Let the height be \(x \mathrm{~cm}\). Then, the base \(=2 x \mathrm{~cm}\).
Therefore, the area of the parallelogram \(=\text {(base×height)}\)
\(=(2 x \times x) \, \mathrm{cm}^{2}=\left(2 x^{2}\right)\, \mathrm{cm}^{2}\)
But, area of the parallelogram \(=512 \mathrm{~cm}^{2}\)
\(\therefore 2 x^{2}=512 \Longrightarrow x^{2}=256\)
\(\Rightarrow x=\sqrt{256}=16\)
Therefore, height \(=16 \mathrm{~cm}\) and base \(=32 \mathrm{~cm}\).
Hence, the base of the parallelogram is \(32 \mathrm{~cm}\), and its height is \(16 \mathrm{~cm}\).

Q.4. Find the area of a rhombus, the lengths of whose diagonals are \(30 \mathrm{~cm}\) and \(16 \mathrm{~cm}\)
Ans: Area of a rhombus \(=\frac{1}{2} \times(\text {product of diagonals})\)
\(=\frac{1}{2} \times(30 \times 16) \, \mathrm{cm}^{2}=240 \mathrm{~cm}^{2}\)
Hence, the area of the rhombus is \(240 \mathrm{~cm}^{2}\).

Q.5. The area of a trapezium is \(180 \mathrm{~cm}^{2}\) and its height is \(9 \mathrm{~cm}\). If one of the parallel sides is longer than the other by \(6 \mathrm{~cm}\), find the two parallel sides.
Ans: Let one of the parallel sides be of length \(x \mathrm{~cm}\). Later, the length of the other parallel side is \((x+6) \,\mathrm{cm}\).
Therefore, area of the trapezium \( = \left\{ {\frac{1}{2} \times \left( {2x + 6} \right) \times 9} \right\}~{\text{c}}{{\text{m}}^2}\)
\( = \left\{ {\frac{1}{2} \times \left( {2x + 6} \right) \times 9} \right\}~{\text{c}}{{\text{m}}^2} = \left( {9x + 27} \right)~{\text{c}}{{\text{m}}^2}\)
But, the area of the trapezium is given as \(180 \mathrm{~cm}^{2}\).
Therefore, \(9 x+27=180 \Longrightarrow 9 x=180-27=153\)
\(\Rightarrow x=\frac{153}{9}=17\)
Thus, the two parallel sides are of lengths \(17 \mathrm{~cm}\) and \((17+6) \,\mathrm{cm}=23 \mathrm{~cm}\).

Summary

In this article, we learned about area formulas for all quadrilaterals, area of the quadrilateral formula using coordinate geometry, area of quadrilateral formula with sides, area formula for any quadrilateral, area formula for cyclic quadrilateral, solved examples on area formula for quadrilaterals and frequently asked questions for this topic.

The learning outcome of this article is how to find the area of different types of quadrilaterals.

Types of Quadrilaterals and Their Properties

Frequently Asked Questions

Q.1. How do you find the area of an irregular quadrilateral?
Ans: The area of quadrilateral
\( = \frac{1}{2}\times\) Length of a diagonal\(\times\) Sum of the lengths of perpendiculars from the remaining two vertices on it

Q.2. How do you find the area of a quadrilateral with \(4\) unequal sides?
Ans: When the sides of a quadrilateral and two of its opposite angles are provided, we will be able to find its area using Bretschneider’s formula. Let us look at a quadrilateral whose sides are \(a, b, c\), and \(d\), and two of its opposite angles are \(\theta_{1}\) and \(\theta_{2}\).
Then the area of the quadrilateral \(=\sqrt{(s-a)(s-b)(s-c)(s-d)-a b c d \cos ^{2} \frac{\theta}{2}}\), where
3. \(\quad s=\) semi \(-\) perimeter of the quadrilateral \(=\frac{(a+b+c+d)}{2}\)
4. \(\quad \theta=\theta_{1}+\theta_{2}\)

Q.3. What is the area formula?
Ans: The word area refers to space within the border or perimeter of a closed shape. The geometry of these kinds of shapes contains at least three sides connected together to form a border. Figurative representation of such space in mathematics refers to the area formula. 

Q.4. What is the area?
Ans: The magnitude of the measurement of the region enclosed by a closed plane figure is called its area.

Q.5. How do you find the area of a quadrilateral with diagonals?
Ans: If the diagonal and the length of the perpendiculars from the vertices are given, then the area of the quadrilateral is calculated as:
Area of quadrilateral \( = \frac{1}{2}\times\) Length of a diagonal \(\times\) Sum of the lengths of perpendiculars from the remaining two vertices on it

Now you are provided with all the necessary information regarding the area formula of quadrilaterals. Practice more questions and master this concept.

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