A parallelogram is a special quadrilateral in which opposite sides are parallel and equal in length. One of the most important concepts related to parallelograms is finding their area, which represents the region enclosed within the figure. The area of a parallelogram is widely used in geometry, mensuration, engineering, architecture, and real-world measurement problems. Understanding the area formula and its derivation helps students solve various academic mathematics and competitive examination questions efficiently. In this article, we will discuss the definition of the area of a parallelogram, important formulas, properties, derivations, solved examples, and practical applications.
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The area of a parallelogram is the measure of the region enclosed within its four sides. It represents the amount of two-dimensional space covered by the parallelogram and is expressed in square units such as $cm^2$, $m^2$, or $km^2$. Understanding the area of a parallelogram is an important part of geometry and mensuration and is frequently tested in school examinations, SSC, Banking, CUET, CAT, Railways, Defence, and other competitive exams.
In simple words, the area of a parallelogram tells us how much surface is covered inside the shape.
For example, if a parallelogram-shaped garden occupies a certain piece of land, its area tells us the size of the land covered by that garden.
Like rectangles and squares, the area of a parallelogram is measured in square units.
The area of a parallelogram is defined as the product of its base and the corresponding perpendicular height.
Mathematically,
$A=b\times h$
where:
This formula gives the exact amount of space enclosed within the parallelogram.

If we wish to find the area of parallelogram, we have to multiply the base of the perpendicular of parallelogram by its height. We must remember certain points that the base and the height of the parallelogram are always perpendicular to each other, which means they form an angle of 90 degrees wherever they meet or intersect, whereas the lateral side of the parallelogram is not perpendicular to the base, hence angle measuring 90 degrees is not formed at their meeting point. The formula for area of parallelogram is given as:
Area $=b \times h$ Square units
Where "b" = base and " $h$ " = height
Area of parallelogram can be calculated using its base and height. Apart from that, the area of parallelogram 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. Following are the ways to find the area of parallelogram depending upon which sides are unknown to us and which are known:
If the base and height of a parallelogram are known, its area can be calculated directly using the standard area formula. This is the most commonly used method in geometry and mensuration problems.
The area of a parallelogram is given by:
$A=b\times h$
where:
Find the area of a parallelogram whose base is 10 cm and height is 5 cm.
Solution:
Given,
Base $=10\ \text{cm}$
Height $=5\ \text{cm}$
Using the formula:
Area of parallelogram $= b\times h$
$=10\times5$
$=50\ \text{cm}^2$
Hence, the area of the parallelogram is $50\ \text{cm}^2$.
Sometimes the perpendicular height of a parallelogram is not given. In such cases, the area can be calculated using trigonometry if the lengths of two adjacent sides and the included angle are known.
The formula is:
$A=ab\sin x$
where:
This formula is particularly useful in trigonometry and coordinate geometry problems.
The area of a parallelogram can also be calculated using its diagonals and the angle between them.
Suppose:
Then,
$A=\frac{1}{2}d_1d_2\sin y$
This method is useful when the lengths of the diagonals are known instead of the base and height.
| Method | Formula |
|---|---|
| Using Base and Height | $A=bh$ |
| Using Trigonometry | $A=ab\sin x$ |
| Using Diagonals | $A=\frac{1}{2}d_1d_2\sin y$ |
| Symbol | Meaning |
|---|---|
| $b$ | Base of the parallelogram |
| $h$ | Perpendicular height |
| $a$ | Length of one side |
| $x$ | Angle between adjacent sides |
| $d_1$ | First diagonal |
| $d_2$ | Second diagonal |
| $y$ | Angle between diagonals |
When the sides of a parallelogram are represented as vectors, the area can be determined using vector algebra.
Let:
Then the area is equal to the magnitude of their cross product.
The area of the parallelogram is:
$A=|\vec a\times\vec b|$
where $|\vec a\times\vec b|$ denotes the magnitude of the cross product of vectors $\vec a$ and $\vec b$.
Let: $d_1=\vec a+\vec b$ and $d_2=\vec b-\vec a$
Then, $d_1\times d_2=(\vec a+\vec b)\times(\vec b-\vec a)$
Expanding,
$= \vec a\times\vec b-\vec a\times\vec a+\vec b\times\vec b-\vec b\times\vec a$
Since, $\vec a\times\vec a=0$ and $\vec b\times\vec b=0$
we get, $=\vec a\times\vec b-\vec b\times\vec a$
Using the property: $\vec a\times\vec b=-\vec b\times\vec a$
Therefore, $d_1\times d_2=2(\vec a\times\vec b)$
Thus, $\vec a\times\vec b=\frac{1}{2}(d_1\times d_2)$
Hence, the area of the parallelogram in terms of diagonals is:
$A=\frac{1}{2}|d_1\times d_2|$
The area and perimeter of a parallelogram are related through its side lengths.
The area of a parallelogram is:
$A=bh$
where:
The perimeter of a parallelogram is:
$P=2(a+b)$
where:
From the perimeter formula:
$P=2(a+b)$
Dividing both sides by 2:
$\frac{P}{2}=a+b$
Therefore,
$b=\frac{P}{2}-a$
Substituting this value into the area formula:
$A=bh$
$A=\left(\frac{P}{2}-a\right)h$
This gives the area of a parallelogram in terms of its perimeter, side length, and height.
The area formula of a parallelogram is important because it helps calculate the size of various geometric and real-world objects.
Parallelogram-shaped structures are commonly found in daily life and engineering designs.
| Application | Use |
|---|---|
| Architecture | Roof and wall designs |
| Engineering | Structural frameworks |
| Flooring | Tile arrangements |
| Land Measurement | Plot calculations |
| Graphic Design | Geometric modelling |
A parallelogram has several unique geometric properties that distinguish it from other quadrilaterals.
In a parallelogram:
If $ABCD$ is a parallelogram, then:
$AB=CD$ and $BC=AD$
The opposite angles of a parallelogram are equal.
For parallelogram $ABCD$:
$\angle A=\angle C$
$\angle B=\angle D$
Also, adjacent angles are supplementary.
$\angle A+\angle B=180^\circ$
The diagonals of a parallelogram bisect each other.
If diagonals $AC$ and $BD$ intersect at point $O$, then:
$AO=OC$
$BO=OD$
However, unlike a rectangle or square, the diagonals are generally not equal.
The area of a parallelogram depends on its base and perpendicular height.
The side chosen as the reference side is called the base.
The perpendicular distance between two parallel sides is called the height.
A common mistake is using the slant side instead of the perpendicular height.
Only the perpendicular height should be used in area calculations.
Several formulas can be used to calculate the area of a parallelogram depending on the available information.
The most commonly used formula is:
A=bh
where:
If the base and corresponding height are known, simply multiply them.
Base $=12\ cm$
Height $=5\ cm$
Area $=12\times5=60\ cm^2$
If two sides and the included angle are known, the area can be calculated using trigonometry.
$A=ab\sin\theta$
where:
$a=8\ cm$
$b=6\ cm$
$\theta=30^\circ$
Area $=8\times6\times\sin30^\circ$
Area $=48\times\frac{1}{2}$
Area $=24\ cm^2$
| Quantity | Meaning |
|---|---|
| Base ($b$) | Any side chosen as reference |
| Height ($h$) | Perpendicular distance to opposite side |
The area depends on the perpendicular height and not on the slant side length.
The area formula can be derived using basic geometric concepts.
Consider a parallelogram with base $b$ and height $h$.
If a triangular portion from one side is cut and moved to the opposite side, the figure transforms into a rectangle.
The resulting rectangle has:
Area of rectangle:
$=b\times h$
Therefore,
Area of parallelogram $=b\times h$
A parallelogram can be divided into triangles.
By rearranging these triangles, a rectangle with the same base and height is obtained.
Since area remains unchanged during rearrangement:
Area of parallelogram $=b\times h$
The area depends entirely on:
If the height increases while the base remains fixed, the area increases.
If the height decreases, the area decreases.
Finding the area is straightforward when the required dimensions are known.
Identify the base.
Find the corresponding perpendicular height.
Apply the formula:
$A=b\times h$
Substitute the values.
Simplify and write the answer in square units.
Base $=15\ cm$
Height $=8\ cm$
Area $=15\times8$
Area $=120\ cm^2$
When height is not given but two sides and the included angle are known:
Use:
$A=ab\sin\theta$
$a=10\ cm$
$b=8\ cm$
$\theta=60^\circ$
Area $=10\times8\times\sin60^\circ$
Area $=80\times\frac{\sqrt3}{2}$
Area $=40\sqrt3\ cm^2$
Find the area of a parallelogram with base 14 cm and height 9 cm.
Area $=14\times9$
Area $=126\ cm^2$
Find the area of a parallelogram with sides 12 cm and 7 cm and included angle $30^\circ$.
Area $=12\times7\times\sin30^\circ$
Area $=84\times\frac{1}{2}$
Area $=42\ cm^2$
Several common quadrilaterals are special forms of parallelograms.
A rectangle is a parallelogram with all angles equal to $90^\circ$.
Formula:
$A=l\times b$
A rhombus is a parallelogram with all sides equal.
Formula:
$A=\frac{1}{2}d_1d_2$
where $d_1$ and $d_2$ are diagonals.
A square is a parallelogram with equal sides and right angles.
Formula:
$A=s^2$
where $s$ is the side length.
All these shapes follow the same fundamental area concept.
| Shape | Area Formula |
|---|---|
| Parallelogram | $bh$ |
| Rectangle | $lb$ |
| Square | $s^2$ |
| Rhombus | $\frac{1}{2}d_1d_2$ |
The area of a parallelogram has numerous practical applications.
Used for:
Architects use parallelogram area formulas while designing:
Engineers use area calculations in:
Surveyors often divide irregular plots into parallelogram-shaped sections to simplify area calculations.
These concepts form the foundation of area of parallelogram formulas, mensuration, quadrilateral geometry, and competitive exam mathematics.
The following books help build a strong understanding of parallelograms, geometry concepts, and mensuration formulas frequently used in examinations.
| Book Name | Best For | Why It Helps |
|---|---|---|
| NCERT Mathematics Textbooks | School Students | Covers parallelogram concepts clearly |
| Mathematics for Class 9 & 10 - R.D. Sharma | Board Exams | Detailed derivations and examples |
| Plane Geometry - S.L. Loney | Advanced Geometry | Strong theoretical foundation |
| Quantitative Aptitude - R.S. Aggarwal | Competitive Exams | Geometry and mensuration practice |
| Objective Mathematics - Arihant Publications | Entrance Exams | Exam-oriented questions and shortcuts |
Understanding a few key relationships can help solve area of parallelogram questions quickly and accurately.
| Trick | Explanation |
|---|---|
| Main Formula | Area = base × height |
| Do not use slant side | Always use perpendicular height |
| Same Base and Height | Equal area for all parallelograms with same base and height |
| Rectangle Relation | Rectangle is a special parallelogram |
| Height is Important | Area changes with height, not side length |
| Trigonometric Formula | Area = $ab\sin\theta$ |
| Units Check | Final answer must be in square units |
This formula sheet summarizes the most important formulas related to the area of a parallelogram and related figures.
| Concept | Formula |
|---|---|
| Area Using Base and Height | $A=bh$ |
| Area Using Two Sides and Included Angle | $A=ab\sin\theta$ |
| Perimeter | $2(a+b)$ |
| Rectangle Area | $lb$ |
| Rhombus Area | $\frac{1}{2}d_1d_2$ |
| Square Area | $s^2$ |
Example 1: What is the area of a parallelogram with a base of 2 cm and height of 5 cm?
Solution:
Given,
Base, $b = 2\ \mathrm{cm}$
Height, $h = 5\ \mathrm{cm}$
We know that,
Area of a parallelogram $=$ base $\times$ height
$A = b \times h$
Substituting the given values:
$A = 2 \times 5$
$A = 10\ \mathrm{cm}^2$
Therefore, the area of the parallelogram is $10\ \mathrm{cm}^2$.
Example 2: Find the area of a parallelogram whose base is 4 cm and height is 10 cm.
Solution:
Given,
$b = 4\ \mathrm{cm}$
$h = 10\ \mathrm{cm}$
The area of a parallelogram is:
$A = b \times h$
Substituting the values:
$A = 4 \times 10$
$A = 40\ \mathrm{cm}^2$
Therefore, the area of the parallelogram is $40\ \mathrm{cm}^2$.
Example 3: The base of a parallelogram is thrice its height. If the area of the parallelogram is $190\ \mathrm{cm}^2$, find the base and height.
Solution:
Let the height of the parallelogram be $h$ cm.
Then, the base of the parallelogram is $3h$ cm.
Given,
Area of parallelogram $= 190\ \mathrm{cm}^2$
We know that,
Area of parallelogram $=$ base $\times$ height
$190 = 3h \times h$
$190 = 3h^2$
$h^2 = \frac{190}{3}$
$h^2 = 63.33$
$h \approx 7.96\ \mathrm{cm}$
Therefore, height $\approx 8\ \mathrm{cm}$
Now,
Base $= 3h$
$= 3 \times 7.96$
$= 23.88\ \mathrm{cm}$
Therefore, the height of the parallelogram is approximately $8\ \mathrm{cm}$ and the base is approximately $23.88\ \mathrm{cm}$.
Example 4: The area of a parallelogram is $500\ \mathrm{cm}^2$. Its height is twice its base. Find the height and base.
Solution:
Given,
Area $= 500\ \mathrm{cm}^2$
Height $= 2 \times$ Base
Let the base be $b$ cm.
Then,
$h = 2b$
Using the formula:
Area of parallelogram $=$ base $\times$ height
$500 = b \times 2b$
$500 = 2b^2$
$b^2 = 250$
$b = \sqrt{250}$
$b \approx 15.81\ \mathrm{cm}$
Now,
$h = 2b$
$= 2 \times 15.81$
$= 31.62\ \mathrm{cm}$
Therefore, the base of the parallelogram is approximately $15.81\ \mathrm{cm}$ and the height is approximately $31.62\ \mathrm{cm}$.
Example 5: Calculate the area of a solar sheet that is in the shape of a parallelogram, given that the base measures 10 in and the altitude measures 8 in.
Solution:
Given,
Base $= 10\ \mathrm{in}$
Height (Altitude) $= 8\ \mathrm{in}$
We know that,
Area of parallelogram $=$ base $\times$ height
$A = 10 \times 8$
$A = 80\ \mathrm{in}^2$
Therefore, the area of the solar sheet is $80\ \mathrm{in}^2$.
For more such examples refer Area of parallelogram worksheet.
Understanding the area of a parallelogram becomes easier when you study related geometry and mensuration topics. The following concepts help strengthen your understanding of area formulas, properties of quadrilaterals, and problem-solving techniques commonly asked in school and competitive examinations.
Frequently Asked Questions (FAQs)
The area of parallelogram is expressed as :
Area of $=$ base $\times$ height.
If we wish to find the perimeter of parallelogram, we add all the sides together. The following formula gives the perimeter of any parallelogram:
Perimeter of parallelogram $=2(a+b)$
The area of a perpendicular with height 2 cm and base 4 cm will be:
Area of parallelogram is $=\mathrm{b} \times \mathrm{h}$
Or, Area of parallelogram $=2 \times 5=10 \mathrm{~cm}^2$
The area of parallelogram can be calculated when the adjacent sides or diagonals are given in vector form. Formula of area of parallelogram is given by: $|\mathbf{a} \times \mathbf{b}|$, where $\mathbf{a}$ and $\mathbf{b}$ are adjacent side vectors. Also, the area of parallelogram formula using diagonals in vector form is, area of parallelogram $= \frac{1}{2} \left|\left(\mathbf{d}_{\mathbf{1}} \times \mathbf{d}_{\mathbf{2}}\right)\right|$, where $\mathbf{d}_{\mathbf{1}}$ and $\mathbf{d}_{\mathbf{2}}$ are diagonal vectors.
Yes. Two parallelograms with the same base and height have equal areas.
