## Area and Perimeter of Plain Figures

Subject: Compulsory Maths

#### Overview

A part of the plane enclosed by a simple closed figure is called a plane region and the measurement of it, is called its area. The area is measured in square units. The length of the boundary of a closed figure is called the perimeter. The units of the perimeter are same as that of length, i.e., m, cm, mm, etc.

##### Area and Perimeter of Plain Figures

Triangle, Rectangle, Square, Circle etc are the plane figures. The total length of the boundary lines of a plane figure is called its perimeter.

### Area and Perimeter of a Square

We can divide the square into small squares of 1 cm side length to find the area of a square by the method of counting squares.

Consider a square that has a side length of 4 cm using the method of counting squares, we find that the area of the square = 16 cm2

Clearly, the square contains 4 rows of 4 squares. Therefore, Area = 4cm x 4cm = 16cm2

This suggests that:

The area of a square is equal to its side-length multiplied by its side-length. That is

Area = length x length

= (length)2

Using A for area and l for length, we can write it as:

A = l2

$\therefore$ Area of square = l2

Also,Perimeter of a square = a + a + a + a

= 4a

### Area and Perimeter of a Rectangle

To find the area of a rectangle by the method of counting squares, we divide the rectangle into small squares of 1 cm side length.

Consider a rectangle of length 5 cm and width 3 cm

Using the method of counting squares, we find that the area of the rectangle is 15cm2.

Clearly, the rectangle contains 3 rows of 5 squares.

Therefore, area = 5cm x 3cm = 15 cm2

This suggests that:

The area of a rectangle is equal to its length multiplied by its width. That is,

Area = Length x Width

Using A for area, l for length and b for width, we can write it simply as:

A = l x b

$\therefore$ Area of rectangle = l x b

Also, perimeter of a rectangle = 2(l+b)

### Area of a Parallelogram

Let, ABCD be the parallelogram. Let DE? AB and BN? DC

Here, AB// DC, DE = BN

Area of Parallelogram = area of $\triangle$ DAB + area of$\triangle$BCD

= $\frac{1}{2}$ AB X DE + $\frac{1}{2}$ DC x BN

=$\frac{1}{2}$ AB X DE + $\frac{1}{2}$ AB X DE ( AB = DC, DE =BN )

Thus, area of a parallelogram = base x height

### Area of Trapezium

We know that a trapezium is a quadrilateral whose one pair of opposite sides is parallel. If two non-parallel sides of a trapezium, it is called an isosceles trapezium.

Let, ABCD be a trapezium having parallel sides AB and DC. Draw DF? AB and CE? AB. Let DF = CE = h. Then area of the trapezium ABCD = area of $\triangle$AFD + area of rectangle FECD + area of $\triangle$EBC

=$\frac{1}{2}$AF x DF + FE x DF + $\frac{1}{2}$ EB xCE

=$\frac{1}{2}$AF x h + FE x h+ $\frac{1}{2}$ EB xh

=$\frac{1}{2}$ h(AF + 2FE + EB)

= $\frac{1}{2}$h(AF + FE + EB + FE) [AF+FE+EB = AB and FE = DC]

= $\frac{1}{2}$h (AB+DC)

Thus, area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x distance between them.

### Area of Rhombus

We know that the rhombus is a parallelogram having sides equal. We also know that the diagonals of a rhombus bisect each other at right angles.

Consider a rhombus ABCD whose diagonals AC and BD bisect each other at right angles at a point O.

Let, AC =d1 and BD =d2

Then, AO =$\frac{1}{2}$d1 and BO =$\frac{1}{2}$d2

Area of $\triangle$AOB = $\frac{1}{2}$ AO x BO = $\frac{1}{2}$ x $\frac{1}{2}$d1x $\frac{1}{2}$d2 = $\frac{1}{8}$d1d2

Since diagonals of a rhombus divide it in to four congruent right angled triangles,

Area of rhombus = 4 x area of $\triangle$ AOB

=4 x $\frac{1}{8}$d1d2

=$\frac{1}{2}$d1d2

Thus, area of rhombus ABCD = $\frac{1}{2}$ xproduct of diagonals.

Note: Since square is also a rhombus having equal diagonals, area of a square = $\frac{1}{2}$d2

### Area and Perimeter of a Triangle

A triangle is a polygon with three vertices, and three sides or edges that are line segments. To find the area of a triangle by the method of counting squares, firstly we divide a rectangle into small squares of 1 cm side length. Secondly, we draw the largest triangle to divide the rectangle into three parts as shown below:

Finally, we estimate the area of a triangle by counting the squares. Area of rectangle = 7x 5 = 35cm2

Area of Triangle = 17.5cm2

This shows that area of triangle =$\frac{1}{2}$ base x height =$\frac{1}{2}$bh

Also, If a, b and c denote three sides of a $\triangle$ABC, Perimeter of $\triangle$ABC = AB + BC + CA

= c + a + b

= a + b + c

##### Things to remember
• Area of a square= length x length = (length)2 = l2
• Perimeter of a square = 4a
• Area of rectangle = l x b
• Perimeter of a rectangle = 2(l+b).
• Area of a parallelogram = base x height
• Area of a rhombus = $\frac{1}{2}$ x d1 x d2.
• Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x h
• Area of the circle = $\frac{1}{2}$ circumference
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.

Solution:

Given, side(a) = 20m

Perimeter of square flower-bed = 4a

=4 x 20

=80

$\therefore$ perimeter of square flower-bed = 80 m

Solution:

Given, side(a) = 35m

Now,

Area of square= a2

= (35m)2

= 1225m2

$\therefore$ area of square = 1225

Also,

Primeter of square = 4a

= 4 x 35m

= 140m

$\therefore$ Perimeter of square = 140m

Solution:

Given, length (l) =12 m and wide (b)=20 m

Area of rectangle field = l x b

=12m x 20m

= 240m2

Perimeter of rectangle field = 2 (l+b)

= 2 (12m + 20m)

= 2 x 32m

= 64m

Solution:

Given, length (l) = 25m and wide (b) = 30m

Area of a rectanguler room = lxb

= 25m x 30m

= 750m2

Perimeter of a rectanguler room = 2(l+b)

= 2(25m+30m)

= 2 x 55m

= 110m

Solution:

Here, base = 14 cm and height = 16.5 cm

Area of the parallelogram = base x height

= 14 cm x 16.5cm

= 231cm2

Solution:

Here, base = 10cm and height = 12.5cm

Area of a parallelogram = base x height

= 10cm x 12.5cm

=120cm2

Solution:

Area of the trapezium = $\frac{1}{2}$ x (sum of parallel sides)xh

= $\frac{1}{2}$x(14+26)x10

= $\frac{1}{2}$x40x10 = 200cm2

Solution:

Area of the trapezium = $\frac{1} {2}$ x (sum of parallel sides) x h

= $\frac{1}{2}$ x (20+30) x 10

=$\frac{1}{2}$ x 50 x 10

= 250cm2

Solution:

Area of the rhombus = $\frac{1}{2}$ xd1d2

= $\frac{1}{2}$ x 24 x 16.5

=198cm2

Solution:

Area of the rhombus = $\frac{1}{2}$ x d1d2

= $\frac{1}{2}$ x 12cm x 14.5cm

= 87 cm2

Soution:

The circumference of the circle is given by

c =2$\pi$r

=2$\times$$\frac{22}{7}$$\times$10.5

=66 cm

Thus, circumference=66 cm

Solution:

We know that, c=2$\pi$r

or, 88=2$\times$$\frac{22}{7}$$\times$r

or, r=$\frac{88\times7}{2\times22}$$\times$r

or, r=$\frac{88 x 7 }{2 x 22}$

$\therefore$ r=14 cm

Solutions:

Note that, in 1 revolution the car covers a distance equal to the circumference of the wheel.

Now, the diameter of the wheel=63 cm

Therefore, radius(r)=$\frac{63}{2}$cm

Circumference of the wheel= 2$\pi$r

=2$\times$$\frac{22}{7}$$\times$$\frac{63}{2}$

=198 cm

=1.98 m

Here, the distance covered in 1 revolution=1.98 m

Distance covered in 1000 revolutions=1.98$\times$1000

=1980 m

Solution:

Circumference= 44 cm

So, 2$\pi$r=44

or, r=$\frac{44}{2\pi}$

or, r=$\frac{44\times7}{2\times22}$

$\therefore$ r=7 cm

Area of the circle=$\pi$r2

=$\frac{22}{7}$$\times$7 $\times$7

=154cm2

Solution:

Area of a circle=154 cm2

So, $\pi$r2=154

or, r2=$\frac{154}{\pi}$

or, r2=$\frac{154\times7}{22}$

or, r2=49

or, r=$\sqrt{49}$

$\therefore$ r =7cm

Solution:

Here, base=14 cm and height= 16.5 cm

Area of the parellogram= base$\times$height

=14 cm$\times$16.5 cm

=231 cm2

Solution:

Base(b)=10 cm, height(h)=6 cm

Area of $\triangle$PQR=$\frac{1}{2}$b$\times$h

=$\frac{1}{2}$QR$\times$PQ

=$\frac{1}{2}$10cm$\times$6cm

=30cm2

Therefore, area of $\triangle$PQR=30cm2

Solution:

Given, AB=6cm, BC=5cm and CA=7cm

Perimeter of $\triangle$ABC = AB + BC + CA

= 6cm + 5cm + 7cm

= 18cm

Solution:

Length of the rectangle(l)=5cm

Area of the rectangle=?

By using formula,

A= l$\times$b

=5cm$\times$3.6cm

=18.0cm2

Solution:

Given: length = 17 cm, breadth = 13 cm

Perimeter of rectangle = 2 (length + breadth)

= 2 (17 + 13)cm

= 2 × 30cm

= 60cm

Also,

Area of rectangle = length × breadth

= 17cm × 13cm

= 221cm2

Solution:

Given, Area of Rectangular plot = 660m2 and Length = 33m

Now, We know,

Area of rectangle = Length $\times$ breadth

or, 660m2 = 33m $\times$ breadth

or, breadth = $\frac{660m^2}{33m}$

$\therefore$ breadth = 20m

Again,

Perimeter of the rectangular plot = 2 (length + breadth)

= 2 (33m + 20m)

= 2 × 53m

= 106m

Solution:

Given, Perimeter of rectangle = 48cm and Breadth = 6cm

We know, Perimeter of rectangle = 2(length + breadth)

So, 48cm = 2(length + 6cm)

or, $\frac{48cm}{2} = length + 6cm or, 24cm = length + 6cm or, length = 24cm - 6cm \(\therefore$ length = 18cm

Now,

Area of rectangle = length × breadth

= 18cm × 6cm

= 108cm2

Solution:

Perimeter of rectangle = 2(length + breadth)

= 2(25cm + 17cm)

= 2 × 42cm

= 84cm

Let the side of square be x.

Then,

Perimeter of square = 4x

We know,

Perimeter of rectangle = Perimeter of Square

So, 84cm = 4x

or, x = $\frac{84cm}{4}$

$\therefore$ x = 21cm

Therefore, sides of square = 21cm.

Solution:

Circumference of circle = 2πr
= 2 × 22/7 × 7
= 44 cm

Area of circle = πr2
= 22/7 × 7 × 7 cm2
= 154 cm2