| Classwise Concept with Examples | ||||||
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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
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| Perimeter and Area of Squares & Rectangles | Area of a Parallelogram & Triangle | Circumference and Area of a Circle |
Chapter 11 Perimeter and Area (Concepts)
This chapter shifts our geometric focus towards two fundamental measurements associated with plane figures (two-dimensional shapes): Perimeter and Area. These concepts are essential for quantifying the spatial characteristics of shapes and have immense practical relevance. We begin by defining Perimeter as the total distance around the boundary of a closed two-dimensional figure. Imagine walking exactly along the edge of a shape; the total distance covered is its perimeter. For polygons, such as triangles, squares, rectangles, and other figures composed of straight line segments, calculating the perimeter is straightforward: it's simply the sum of the lengths of all its sides.
We derive specific, efficient formulas for common regular polygons:
- For a Square with side length $s$, the Perimeter is $P = s + s + s + s = 4 \times s$.
- For a Rectangle with length $l$ and breadth $b$, the Perimeter is $P = l + b + l + b = 2l + 2b = 2 \times (l + b)$.
Next, we explore Area, defined as the measure of the surface or region enclosed within the boundary of a closed figure. It quantifies the 'space' the figure occupies in a 2D plane. We revisit familiar formulas for basic shapes:
- For a Square with side length $s$, the Area is $A = s \times s = s^2$.
- For a Rectangle with length $l$ and breadth $b$, the Area is $A = l \times b$.
Throughout these calculations, careful attention is paid to the units of measurement. Perimeter, being a length, is measured in linear units like centimeters (cm), meters (m), or kilometers (km). Area, representing a surface, is measured in square units, such as square centimeters ($cm^2$), square meters ($m^2$), or square kilometers ($km^2$). Understanding conversions between these units, especially for area (e.g., noting that $1 \, m^2 = 1 \, m \times 1 \, m = 100 \, cm \times 100 \, cm = 10,000 \, cm^2$), is crucial. The chapter frequently incorporates application-based problems to solidify understanding, such as calculating the area of pathways constructed around rectangular gardens or circular ponds, determining the cost of fencing a field (perimeter) or tiling a floor (area, potentially using $\textsf{₹}$ for cost), and finding the area of composite figures formed by combining simpler shapes like rectangles, squares, triangles, and circles (or parts like semicircles). This practical relevance makes the chapter vital for everyday problem-solving and foundational for fields like design, architecture, and engineering.
Perimeter and Area of Squares & Rectangles
In geometry, we study various shapes and their properties. Mensuration is the branch of mathematics that deals with the measurement of geometric figures. Two fundamental measurements for two-dimensional (plane) figures are Perimeter and Area. The perimeter is the distance around the boundary of a shape, while the area is the measure of the surface enclosed by the shape. In this section, we will focus on calculating the perimeter and area of two common quadrilaterals: squares and rectangles.
Perimeter and Area of a Square
A Square is a special type of quadrilateral where all four sides are equal in length, and all four interior angles are right angles ($90^\circ$).
Perimeter of a Square:
The perimeter of any polygon is the total length of its boundary, which is the sum of the lengths of all its sides. For a square, since all four sides have the same length, the perimeter is simply four times the length of one side.
Perimeter of a Square = Side Length + Side Length + Side Length + Side Length
Let '$s$' be the length of one side of a square. The perimeter $P$ is:
Perimeter of a Square $= 4 \times \text{Side Length} $
$P = 4s $
Area of a Square:
The area of a square is the measure of the two-dimensional space enclosed within its boundaries. The area is calculated by multiplying the length of one side by itself.
Area of a Square = Side Length $\times$ Side Length
Let '$s$' be the length of one side of a square. The area $A$ is:
$A = s \times s = s^2 $
The unit of area is always expressed in square units. If the side length is in centimetres (cm), the area is in square centimetres (cm$^2$). If the side length is in metres (m), the area is in square metres (m$^2$), and so on.
Perimeter and Area of a Rectangle
A Rectangle is a quadrilateral where opposite sides are equal in length and parallel, and all four interior angles are right angles ($90^\circ$). It is defined by its length (usually the longer side) and its breadth or width (usually the shorter side).
Perimeter of a Rectangle:
The perimeter of a rectangle is the sum of the lengths of its four sides. Since opposite sides are equal, a rectangle has two sides of length equal to its length ($l$) and two sides equal to its breadth ($b$).
Perimeter of a Rectangle = Length + Breadth + Length + Breadth
Let '$l$' be the length and '$b$' be the breadth of a rectangle. The perimeter $P$ is:
Perimeter of a Rectangle $= l + b + l + b = 2l + 2b $
We can factor out the common factor 2:
Perimeter of a Rectangle $= 2 \times (\text{Length} + \text{Breadth}) $
$P = 2(l+b) $
Area of a Rectangle:
The area of a rectangle is the measure of the two-dimensional space it covers. We find the area by multiplying its length by its breadth.
Area of a Rectangle = Length $\times$ Breadth
Let '$l$' be the length and '$b$' be the breadth of a rectangle. The area $A$ is:
$A = l \times b $
The unit of area is in square units, corresponding to the units of length and breadth (e.g., cm$^2$, m$^2$).
Example 1. A square park has a side length of 25 metres. Find its perimeter and area.
Answer:
Given: Side length of the square park, $s = 25$ m.
To Find: Perimeter and Area of the square park.
Using the formula for the perimeter of a square:
Perimeter $= 4 \times \text{Side Length}$
Substitute the side length $s=25$ m:
Perimeter $= 4 \times 25 \text{ m}$
Perimeter $= 100$ m
Using the formula for the area of a square:
Area $= \text{Side Length} \times \text{Side Length}$
Substitute the side length $s=25$ m:
Area $= 25 \text{ m} \times 25 \text{ m}$
Perform the multiplication: $25 \times 25 = 625$. The unit is $\text{m} \times \text{m} = \text{m}^2$.
Area $= 625 \text{ m}^2$
The perimeter of the square park is 100 m and its area is 625 m$^2$.
Example 2. The length and breadth of a rectangular field are 50 m and 30 m respectively. Find its perimeter and area.
Answer:
Given:
Length of the rectangular field, $l = 50$ m.
Breadth of the rectangular field, $b = 30$ m.
To Find: Perimeter and Area of the rectangular field.
Using the formula for the perimeter of a rectangle:
Perimeter $= 2 \times (\text{Length} + \text{Breadth})$
Substitute the given values $l=50$ m and $b=30$ m:
Perimeter $= 2(50 \text{ m} + 30 \text{ m})$
First, add inside the parenthesis: $50 + 30 = 80$.
Perimeter $= 2(80 \text{ m})$
Multiply by 2: $2 \times 80 = 160$. The unit is metres.
Perimeter $= 160$ m
Using the formula for the area of a rectangle:
Area $= \text{Length} \times \text{Breadth}$
Substitute the given values $l=50$ m and $b=30$ m:
Area $= 50 \text{ m} \times 30 \text{ m}$
Perform the multiplication: $50 \times 30 = 1500$. The unit is $\text{m} \times \text{m} = \text{m}^2$.
Area $= 1500 \text{ m}^2$
The perimeter of the rectangular field is 160 m and its area is 1500 m$^2$.
Example 3. The area of a square plot is 400 m$^2$. Find the side length and the perimeter of the plot.
Answer:
Given: Area of the square plot, $A = 400$ m$^2$.
To Find: Side length ($s$) and Perimeter ($P$) of the square plot.
Using the formula for the area of a square: Area $= s^2$.
Substitute the given area:
$400 \text{ m}^2 = s^2 $.
To find the side length 's', we need to find the square root of 400. Remember that $s$ must be a positive value since it's a length.
$s = \sqrt{400 \text{ m}^2} $.
We know that $20 \times 20 = 400$. So, the square root of 400 is 20.
$s = 20$ m.
The side length of the square plot is 20 m.
Now, using the formula for the perimeter of a square: Perimeter $= 4 \times \text{Side Length}$.
Substitute the side length $s=20$ m:
Perimeter $= 4 \times 20$ m
Perimeter $= 80$ m
The side length of the square plot is 20 m and its perimeter is 80 m.
Example 4. A rectangular sheet of paper has length 12 cm and breadth 10 cm. Find its area. If the perimeter of the sheet is 44 cm, verify this using the given dimensions.
Answer:
Given:
Length of rectangular sheet, $l = 12$ cm.
Breadth of rectangular sheet, $b = 10$ cm.
Perimeter of the sheet is given as 44 cm.
To Find: Area of the sheet.
To Verify: The given perimeter (44 cm) using the calculated perimeter from the dimensions.
Using the formula for the area of a rectangle: Area $= l \times b$.
Substitute the given dimensions:
Area $= 12 \text{ cm} \times 10 \text{ cm}$
Multiply: $12 \times 10 = 120$. Unit is $\text{cm} \times \text{cm} = \text{cm}^2$.
Area $= 120 \text{ cm}^2$
The area of the rectangular sheet is 120 cm$^2$.
Now, using the formula for the perimeter of a rectangle to verify the given perimeter (44 cm): Perimeter $= 2(l+b)$.
Substitute the given dimensions $l=12$ cm and $b=10$ cm:
Perimeter $= 2(12 \text{ cm} + 10 \text{ cm})$
Add inside the parenthesis: $12 + 10 = 22$.
Perimeter $= 2(22 \text{ cm})$
Multiply by 2: $2 \times 22 = 44$. The unit is centimetres.
Calculated Perimeter $= 44 \text{ cm}$
The calculated perimeter (44 cm) is equal to the given perimeter (44 cm).
The verification is successful.
Area of a Parallelogram & Triangle
In the previous section, we revisited the perimeter and area of squares and rectangles. Now, we will extend our understanding of area to two other important plane figures: the Parallelogram and the Triangle. Finding the area of these shapes is essential in various geometric and practical applications.
Area of a Parallelogram
A Parallelogram is a quadrilateral (a four-sided polygon) with two pairs of parallel sides. Unlike a rectangle, the angles of a parallelogram are not necessarily right angles ($90^\circ$), but opposite angles are equal. The sides also have specific relationships: opposite sides are equal in length.
To calculate the area of a parallelogram, we need two measurements: its base and its corresponding height.
- The base of a parallelogram is the length of any one of its sides.
- The height (or altitude) corresponding to that base is the perpendicular distance from the opposite side to the base (or the extension of the base). The height is a perpendicular line segment between the base and its opposite side.
Let the base be $b$ and the corresponding height be $h$.
Derivation of the Area Formula (from a Rectangle):
We can understand the formula for the area of a parallelogram by relating it to the area of a rectangle. Consider a parallelogram ABCD. Draw a perpendicular from vertex D to side AB (or its extension), meeting AB at point E. This forms a right-angled triangle ADE. Now, imagine cutting this triangle ADE off from the parallelogram and sliding it over to the other side, attaching it to side BC. The figure formed by joining triangle ADE to the right side of the remaining quadrilateral EBCD is a rectangle ECDF.
The area of the parallelogram ABCD is equal to the area of the rectangle ECDF because we just rearranged the parts without losing or adding any area. The length of the rectangle ECDF is equal to the base AB of the parallelogram, and the breadth of the rectangle ECDF is equal to the height DE of the parallelogram.
Area of rectangle = Length $\times$ Breadth = Base of parallelogram $\times$ Height of parallelogram.
Therefore, the formula for the area of a parallelogram is:
Area of a Parallelogram = Base $\times$ Height
Let '$b$' be the length of the base and '$h$' be the corresponding height of a parallelogram. Then the area $A$ is:
$A = b \times h $
The unit of area is in square units (e.g., cm$^2$, m$^2$), just like for rectangles and squares.
Area of a Triangle
A Triangle is the simplest polygon, having three sides, three vertices, and three angles. We can calculate the area of a triangle if we know the length of one of its sides (which we call the base) and the length of the corresponding height (or altitude) from the opposite vertex to that base.
- The base of a triangle can be any one of its three sides.
- The corresponding height (or altitude) is the perpendicular distance from the vertex opposite the chosen base to that base (or the extension of that base if the foot of the perpendicular lies outside the triangle, as in obtuse triangles).
Let the base be $b$ and the corresponding height be $h$.
Derivation of the Area Formula (from a Parallelogram):
We can understand the area of a triangle by relating it to the area of a parallelogram. Consider any triangle, say $\triangle ABC$. Make an identical copy of this triangle ($\triangle DCB$). Now, place the two congruent triangles together such that a pair of corresponding sides (say, BC) coincides, and the other two vertices (A and D) are on opposite sides of BC. If you arrange them correctly (for example, with AB parallel to CD and AC parallel to BD), they will form a parallelogram ABDC.
The base of the parallelogram ABDC is the same as the base of the triangle BC, and the height of the parallelogram (the perpendicular distance from A to BC) is the same as the height of the triangle corresponding to the base BC. The area of the parallelogram ABDC is Base $\times$ Height.
Since the parallelogram ABDC is made up of two identical copies of $\triangle ABC$, the area of one triangle is half the area of the parallelogram.
Area of a Triangle = $\frac{1}{2} \times$ Area of the formed Parallelogram
Area of a Triangle = $\frac{1}{2} \times ($Base of Parallelogram $\times$ Height of Parallelogram$)$
Since the base and height of the parallelogram are the same as the base and height of the triangle:
Area of a Triangle $= \frac{1}{2} \times$ Base of Triangle $\times$ Height of Triangle
Let '$b$' be the length of the base and '$h$' be the corresponding height of a triangle. Then the area $A$ is:
$A = \frac{1}{2} \times b \times h $
The unit of area is in square units, consistent with the units of the base and height.
Example 1. Find the area of a parallelogram with base 8 cm and corresponding height 4 cm.
Answer:
Given:
Base of the parallelogram, $b = 8$ cm.
Corresponding height of the parallelogram, $h = 4$ cm.
To Find: Area of the parallelogram.
Using the formula for the area of a parallelogram: Area $= b \times h$.
Substitute the given values:
Area $= 8 \text{ cm} \times 4 \text{ cm}$
Perform the multiplication: $8 \times 4 = 32$. The unit is $\text{cm} \times \text{cm} = \text{cm}^2$.
Area $= 32 \text{ cm}^2$
The area of the parallelogram is 32 cm$^2$.
Example 2. Find the area of a triangle with base 10 m and height 6 m.
Answer:
Given:
Base of the triangle, $b = 10$ m.
Corresponding height of the triangle, $h = 6$ m.
To Find: Area of the triangle.
Using the formula for the area of a triangle: Area $= \frac{1}{2} \times b \times h$.
Substitute the given values:
Area $= \frac{1}{2} \times 10 \text{ m} \times 6 \text{ m}$
Perform the multiplication. We can multiply $\frac{1}{2}$ by 10 first: $\frac{1}{2} \times 10 = 5$.
Area $= 5 \text{ m} \times 6 \text{ m}$
Now multiply $5 \times 6 = 30$. The unit is $\text{m} \times \text{m} = \text{m}^2$.
Area $= 30 \text{ m}^2$
The area of the triangle is 30 m$^2$.
Circumference and Area of a Circle
We have discussed the perimeter and area of polygons like squares, rectangles, parallelograms, and triangles, which are formed by straight line segments. Now, let's explore a perfectly round shape: the Circle. A circle is defined as the set of all points in a plane that are at a fixed distance from a fixed point. The fixed point is called the center of the circle, and the fixed distance is called the radius ($r$). The straight line segment that passes through the center and connects two points on the circle is called the diameter ($d$). The diameter is always twice the radius, i.e., $d = 2r$.
The measurements related to a circle's boundary and the space it covers are called its circumference and area, respectively.
Circumference of a Circle
For polygons, the perimeter is the sum of the side lengths. For a circle, the boundary is a curve. The perimeter of a circle is given a special name: the Circumference. The circumference is the total distance around the circle.
There is a constant mathematical relationship between the circumference of any circle and its diameter. If you divide the circumference of any circle by its diameter, you will always get approximately the same value. This constant value is represented by the Greek letter pi ($\pi$).
$\frac{\text{Circumference}}{\text{Diameter}} = \pi $.
The value of $\pi$ cannot be expressed exactly as a simple fraction or a terminating decimal; it is an irrational number. However, for calculations, we use approximate values for $\pi$, such as $\frac{22}{7}$ or $3.14$ (or sometimes $3.14159$). For most problems in Class 7, using $\pi \approx \frac{22}{7}$ is common.
From the definition $\frac{\text{Circumference}}{\text{Diameter}} = \pi$, we can rearrange to get the formula for the circumference:
Circumference = $\pi \times$ Diameter.
Since the diameter ($d$) is twice the radius ($r$, $d = 2r$), we can also express the circumference formula in terms of the radius:
Circumference = $\pi \times (2 \times \text{Radius})$.
Circumference = $2\pi \times$ Radius.
Let 'C' be the circumference and 'r' be the radius of a circle. Then the circumference is:
$C = 2\pi r $.
Or, if 'd' is the diameter:
$C = \pi d $.
The unit of circumference is a linear unit (e.g., cm, m), representing length.
Area of a Circle
The area of a circle is the measure of the two-dimensional space enclosed within its boundary (the circumference). We can visualise the formula for the area of a circle by imagining dividing the circle into many small sectors (like slices of a pizza) and rearranging them.
Imagine cutting a circle into a large number of very thin sectors. If we arrange these sectors side-by-side, alternating the direction of the pointed ends, they form a shape that is very close to a rectangle. The more sectors we divide the circle into, the closer the shape gets to a perfect rectangle.
In this approximate rectangle:
- The 'height' or 'breadth' is equal to the radius ($r$) of the circle.
- The 'length' is equal to half the circumference of the circle (because the curved edges of the sectors along the top form half the circumference, and the curved edges along the bottom form the other half). Half the circumference $= \frac{1}{2} \times 2\pi r = \pi r$.
The area of this approximate rectangle is Length $\times$ Breadth $= (\pi r) \times r = \pi r^2$. As the number of sectors approaches infinity, this approximation becomes exact.
So, the formula for the area of a circle is:
Area of a Circle $= \pi \times$ Radius $\times$ Radius
Let 'A' be the area and 'r' be the radius of a circle. Then the area is:
$A = \pi r^2 $.
The unit of area is in square units (e.g., cm$^2$, m$^2$), just like for other plane figures.
Example 1. Find the circumference and area of a circle with radius 14 cm. (Use $\pi = \frac{22}{7}$).
Answer:
Given:
Radius of the circle, $r = 14$ cm.
Value of $\pi$ to use = $\frac{22}{7}$.
To Find: Circumference (C) and Area (A) of the circle.
Calculate Circumference (C):
Use the formula C $= 2\pi r$.
Substitute the given values $r=14$ and $\pi=\frac{22}{7}$:
Circumference $= 2 \times \frac{22}{7} \times 14 \text{ cm} $.
Perform the multiplication. We can simplify by cancelling the 7 in the denominator with the 14 in the numerator ($14 \div 7 = 2$).
Circumference $= 2 \times 22 \times \frac{\cancel{14}^{2}}{\cancel{7}_1} \text{ cm} $.
Circumference $= 2 \times 22 \times 2 \text{ cm} $.
Multiply the numbers: $2 \times 22 = 44$, and $44 \times 2 = 88$. The unit is cm.
Circumference $= 88 \text{ cm} $.
Calculate Area (A):
Use the formula A $= \pi r^2$.
Substitute the given values $r=14$ and $\pi=\frac{22}{7}$. Remember $r^2 = r \times r = 14 \times 14$.
Area $= \frac{22}{7} \times (14 \text{ cm})^2 $.
Area $= \frac{22}{7} \times 14 \text{ cm} \times 14 \text{ cm} $.
Simplify by cancelling the 7 in the denominator with one of the 14s in the numerator ($14 \div 7 = 2$).
Area $= 22 \times \frac{\cancel{14}^{2}}{\cancel{7}_1} \times 14 \text{ cm}^2 $.
Area $= 22 \times 2 \times 14 \text{ cm}^2 $.
Multiply the numbers: $22 \times 2 = 44$. Then $44 \times 14$.
Area $= 616 \text{ cm}^2 $.
The circumference of the circle is 88 cm and its area is 616 cm$^2$.
Example 2. The circumference of a circular garden is 308 m. Find the radius and the area of the garden. (Use $\pi = \frac{22}{7}$).
Answer:
Given:
Circumference of the circular garden, $C = 308$ m.
Value of $\pi$ to use = $\frac{22}{7}$.
To Find: Radius ($r$) and Area ($A$) of the garden.
Calculate Radius (r):
Use the formula for the circumference of a circle: C $= 2\pi r$.
Substitute the given values $C=308$ and $\pi=\frac{22}{7}$:
$308 \text{ m} = 2 \times \frac{22}{7} \times r $.
We need to solve this equation for $r$. Rearrange the terms to isolate $r$. Multiply both sides by 7 and divide both sides by 2 and 22.
$r = \frac{308 \times 7}{2 \times 22} \text{ m} $.
Perform the calculation, simplifying by cancellation:
$r = \frac{\cancel{308}^{154} \times 7}{\cancel{2}_{1} \times 22} \text{ m} $ (Divide 308 by 2)
$r = \frac{154 \times 7}{22} \text{ m} $.
Divide 154 by 22. $22 \times 7 = 154$.
$r = \frac{\cancel{154}^{7} \times 7}{\cancel{22}_{1}} \text{ m} $.
$r = 7 \times 7 \text{ m} $.
$r = 49 \text{ m} $.
The radius of the circular garden is 49 m.
Calculate Area (A):
Use the formula A $= \pi r^2$.
Substitute the calculated radius $r=49$ m and $\pi=\frac{22}{7}$. Remember $r^2 = r \times r = 49 \times 49$.
Area $= \frac{22}{7} \times (49 \text{ m})^2 $.
Area $= \frac{22}{7} \times 49 \text{ m} \times 49 \text{ m} $.
Simplify by cancelling the 7 in the denominator with one of the 49s in the numerator ($49 \div 7 = 7$).
Area $= 22 \times \cancel{49}^{7} \times 49 \text{ m}^2 $.
Area $= 22 \times 7 \times 49 \text{ m}^2 $.
Multiply the numbers: $22 \times 7 = 154$. Then $154 \times 49$.
Area $= 7546 \text{ m}^2 $.
The radius of the circular garden is 49 m and its area is 7546 m$^2$.