Perimeter and Area Class 7 Notes CBSE Maths Chapter 9 (Free PDF Download)

• The area of a closed figure is the part of the plane occupied by the closed-form, whereas the perimeter is the distance surrounding it.

• The area of a plane or region enclosed by it is measured in square metres.

• In a previous lecture, we learned how to calculate the perimeter and area of a square and rectangle. They are as follows:

• Perimeter of a square \[\text{= 4  }\!\!\times\!\!\text{  side}\]

• Perimeter of a rectangle \[\text{= 2  }\!\!\times\!\!\text{  }\left( length+breadth \right)\]

• Area of a square \[\text{= side  }\!\!\times\!\!\text{  side}\]

• Area of a rectangle \[\text{= length  }\!\!\times\!\!\text{  breadth}\]

Example:

1. A wire is in the shape of a square of side \[\text{12 cm}\]. If the wire is rebent into a rectangle of length \[\text{16 cm}\], find its breadth. Which encloses more area, the square or the rectangle?

Ans:

Side of the Square \[\text{12 cm}\]

Here,

Length of the wire \[\text{=}\] Perimeter of the square

Therefore,

\[\text{Length of the wire = 4  }\!\!\times\!\!\text{  side}\]

\[\text{Length of the wire = 4 }\times \text{ 12 cm}\]

\[\text{Length of the wire = 48 cm}\]

Therefore, length of the wire is \[\text{48 cm}\].

We have given, the Length of the rectangle.

Let’s, consider \[\text{l}\] be the length of the rectangle and \[\text{b}\] be the breadth of the rectangle.

Now, 

Perimeter of rectangle \[\text{=}\] Length of wire

Hence,

\[\text{Perimeter of rectangle = 48 cm ---}\left( 1 \right)\]

Since,

Perimeter of a rectangle \[\text{= 2  }\!\!\times\!\!\text{  }\left( length+breadth \right)\]

Eq, \[\left( 1 \right)\] becomes,

\[\text{ 48 = 2  }\!\!\times\!\!\text{  }\left( \text{16 + b} \right)\]

\[\dfrac{\text{48}}{\text{2}}\text{ = 16 + b}\]

\[\text{ 24 = 16 + b}\]

\[\text{   b = 24 - 16}\]

Therefore,   

\[\text{ b = 8 cm }\]

Hence, breadth of the rectangle is \[\text{8 cm }\].

Area of a square \[\text{= }{{\left( \text{side} \right)}^{2}}\]

Therefore,

\[\text{Area of a square = }{{\left( \text{side} \right)}^{\text{2}}}\]

\[\text{                           = }{{\left( \text{12} \right)}^{\text{2}}}\]

\[\text{                           = 12 }\!\!\times\!\!\text{ 12}\]

\[\text{Area of a square = 144 c}{{\text{m}}^{\text{2}}}\]

\[\text{Area of a rectangle = }\left( \text{l  }\!\!\times\!\!\text{  b} \right)\]

\[\text{                               = }\left( \text{16  }\!\!\times\!\!\text{  8} \right)\]

\[\text{Area of a rectangle = 128 c}{{\text{m}}^{\text{2}}}\]

Therefore, the square encloses more area even though its perimeter is the same as that of the rectangle.

• Triangles as Parts of Rectangles:

• The area of the rectangle is equal to the sum of the two triangles' areas. 

• The area of both triangles is the same.

• For above rectangle,

\[\text{Area of each triangle = }\dfrac{\text{1}}{\text{2}}\left( \text{Area of the rectangle} \right)\]

\[\text{Area of each triangle = }\dfrac{\text{1}}{\text{2}}\times \text{ l }\times \text{ b}\]

• For below square,

\[\text{Area of each triangle = }\dfrac{\text{1}}{4}\left( \text{Area of the square} \right)\]

\[\text{Area of each triangle = }\dfrac{\text{1}}{\text{4}}{{\left( side \right)}^{2}}\]

Example:

Ans:

\[\text{Area of each triangle = }\dfrac{\text{1}}{\text{2}}\times \text{ l }\times \text{ b}\]

\[\text{Area of each triangle = }\dfrac{\text{1}}{\text{2}}\times \text{ 12 }\times \text{ 4}\]

\[\text{Area of each triangle = 24 c}{{\text{m}}^{2}}\]

• Area of a Parallelogram:

• Any side of a parallelogram can be considered as the base of the parallelogram.

• The perpendicular drawn on that side from the opposite vertex is known as height (altitude).

• \[\text{Area of a Parallelogram = Base  }\!\!\times\!\!\text{  Height}\]

Example:

1. One of the sides and the corresponding height of a Parallelogram are \[\text{17 cm}\] and \[\text{7 cm}\] respectively. Find the area of the parallelogram.

Ans:

Here, Base of Parallelogram is, \[\text{b = 17 cm}\]

and, Height of Parallelogram is \[\text{h = 7 cm}\]

Therefore,

\[\text{Area of a Parallelogram = Base  }\!\!\times\!\!\text{  Height}\]

\[\text{                                       = 17  }\!\!\times\!\!\text{  7}\]

\[\text{Area of a Parallelogram = 144 }{{\text{m}}^{\text{2}}}\]

Therefore, area of a Parallelogram is \[\text{144 }{{\text{m}}^{\text{2}}}\].

• Area of a Triangle:

• Area of a triangle  \[\text{= }\dfrac{1}{2}\left( \text{Area of the parallelogram generated from it} \right)\]

Therefore,

\[\text{Area of a triangle = }\dfrac{\text{1}}{\text{2}}\text{  }\!\!\times\!\!\text{  Base  }\!\!\times\!\!\text{  Height}\]

• \[\text{Area of an equilateral triangle = }\dfrac{\sqrt{3}}{4}\text{ }{{\left( side \right)}^{2}}\]

• All congruent triangles have the same area, however triangles with the same area do not have to be congruent.

Example:

Find the area of the given triangle.

Ans:

Here, Base of triangle is \[\text{b = 7 cm}\]

and, Height of triangle is \[\text{h = 3}\text{.5 cm}\]

Therefore,

\[\text{Area of a triangle = }\dfrac{\text{1}}{\text{2}}\text{  }\!\!\times\!\!\text{  Base  }\!\!\times\!\!\text{  Height}\]

\[\text{Area of a triangle = }\dfrac{\text{1}}{\text{2}}\text{  }\!\!\times\!\!\text{  B  }\!\!\times\!\!\text{  H}\]

\[\text{                             = }\dfrac{\text{1}}{\text{2}}\text{  }\!\!\times\!\!\text{  7  }\!\!\times\!\!\text{  3}\text{.5}\]

\[\text{Area of a triangle = 12}\text{.25 c}{{\text{m}}^{\text{2}}}\]

Therefore, area of a triangle is \[\text{12}\text{.25 c}{{\text{m}}^{\text{2}}}\].

• Circumference of a Circle:

• The circumference of a circular location is the distance around it.

• The ratio of a circle's circumference to diameter of a circle is constant and denoted by \[\pi \text{ }\left( pi \right)\].

• Circumference of a circle in terms of diameter:

\[\text{Circumference of a circle }\left( C \right)\text{ = }\pi \text{d}\]

Where, \[\text{d}\] is the diameter of a circle.

$\text{ }\!\!\pi\!\!\text{  = }\dfrac{\text{22}}{\text{7}}$

$\text{ }\!\!\pi\!\!\text{  = 3}\text{.14 (Approx}\text{.)}$

• Circumference of a circle in terms of radius:

\[\text{Circumference of a circle }\left( \text{C} \right)\text{ =  }\!\!\pi\!\!\text{ d}\]

\[\text{Circumference of a circle }\left( \text{C} \right)\text{ =  }\!\!\pi\!\!\text{   }\!\!\times\!\!\text{  2r}\]

\[\text{Circumference of a circle }\left( \text{C} \right)\text{ = 2 }\!\!\pi\!\!\text{ r}\]

Example:

1. What is the circumference of a circular disc of radius \[\text{21 cm}\]?

Ans:

Here, radius of a circular disc is \[\text{r = 21 cm}\]

Thus, circumference of a disc,

\[\text{C = 2 }\!\!\pi\!\!\text{ r}\]

\[\text{C = 2  }\!\!\times\!\!\text{  }\dfrac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  21}\]

\[\text{C = 132 cm}\]

Therefore, the circumference of a circular disc is \[\text{132 cm}\].

• Area of Circle:

\[\text{Area of a Circle =  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\]

Where, \[\text{r}\] is the radius of the circle.

Example:

1. Diameter of a circular garden is \[\text{10}\text{.6 m}\]. Find its area.

Ans:

Diameter of a given circular garden is 

\[\text{d = 10}\text{.6 m}\]

Therefore, the radius of a circle is

\[\text{r = }\dfrac{d}{2}\]

\[\text{r = }\dfrac{\text{10}\text{.6}}{2}\]

\[\text{r = 5}\text{.3 m}\]

Thus,

\[\text{Area of a Circle =  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\]

\[\text{Area of a Circle = }\dfrac{\text{22}}{\text{7}}{{\left( \text{5}\text{.3} \right)}^{\text{2}}}\]

\[\text{Area of a Circle = }\dfrac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  5}\text{.3  }\!\!\times\!\!\text{  5}\text{.3}\]

\[\text{Area of a Circle = 88}\text{.28 }{{\text{m}}^{\text{2}}}\]

Therefore, area of a circular garden is \[\text{88}\text{.28 }{{\text{m}}^{\text{2}}}\].

• Conversion of Units:

• The units of area can be translated using the same method as the units of lengths that we learned earlier:

• \[\text{    1 c}{{\text{m}}^{\text{2}}}\text{ = 100 m}{{\text{m}}^{\text{2}}}\]

\[\text{ 1 }{{\text{m}}^{\text{2}}}\text{ = 10000 c}{{\text{m}}^{\text{2}}}\]

\[\text{1 hectare = 10000 }{{\text{m}}^{\text{2}}}\]

• The number of units produced by converting a unit of area to a smaller unit will be greater.

Example:

\[\text{10000 c}{{\text{m}}^{\text{2}}}\text{ = 10000  }\!\!\times\!\!\text{  100 m}{{\text{m}}^{\text{2}}}\]

\[\text{10000 c}{{\text{m}}^{\text{2}}}\text{ = 1000000 m}{{\text{m}}^{\text{2}}}\]

• The number of larger units will be smaller when we convert a unit of area to a larger unit.

\[\text{10000 c}{{\text{m}}^{\text{2}}}\text{ = }\dfrac{\text{10000}}{10000}\text{ }{{\text{m}}^{\text{2}}}\]

\[\text{10000 c}{{\text{m}}^{\text{2}}}\text{ = 1 }{{\text{m}}^{\text{2}}}\]

Example:

1. A rectangular park is \[\text{55 m}\] long and \[\text{20 m}\] wide. A path \[\text{3}\text{.5 m}\] wide is constructed outside the park. Find the area of the path.

Consider \[\text{KLMN}\] rectangle which represent the rectangular park and the shaded region represent the path \[\text{3}\text{.5 m}\] wide.

To find the area of the path, first we need to find \[\left( \text{Area of rectangle ABCD - Area of rectangle KLMN} \right)\]

Now,

\[\text{AB = 3}\text{.5 + 55 + 3}\text{.5}\]

\[\text{AB = 62 m}\]

\[\text{AD = 3}\text{.5 + 20 + 3}\text{.5}\]

\[\text{AD = 27 m}\]

\[\text{Area of rectangle ABCD = l }\times \text{ b}\]

\[\text{Area of rectangle ABCD = AB }\times \text{ AD}\]

\[\text{Area of rectangle ABCD = 62 }\times \text{ 27}\]

Therefore,

\[\text{Area of rectangle ABCD = 1,674 }{{\text{m}}^{2}}\]

\[\text{Area of rectangle KLMN = l }\times \text{ b}\]

\[\text{Area of rectangle KLMN = KL }\times \text{ KN}\]

\[\text{Area of rectangle KLMN = 55 }\times \text{ 20}\]

Hence,

\[\text{Area of rectangle KLMN = 1,100 }{{\text{m}}^{2}}\]

Thus,

\[\text{Area of Path = }\left( \text{Area of rectangle ABCD - Area of rectangle KLMN} \right)\]

\[\text{Area of Path = 1674 - 1100}\]

Therefore,

\[\text{Area of Path = 574 }{{\text{m}}^{2}}\].

Chapter Summary - Perimeter and Area

"Perimeter and Area" in Class 7 introduces the fascinating world of shapes and measurements. Learn how to find the distance around shapes (perimeter) and the space they cover (area). Delve into exciting examples like calculating the boundary of a playground or the carpet needed for a room. Discover the importance of units in measurement, making your math skills practical for real-life situations. By mastering Perimeter and Area, Class 7 students gain essential skills for understanding space, measuring accurately, and solving everyday challenges with math. Explore the fun of geometry while becoming a confident problem solver!

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