Area and Perimeter of Plain Figures

Subject: Compulsory Maths

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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.

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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

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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

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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

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\(\therefore\) 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.

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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.

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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.
Questions and Answers

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

Thus, radius=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

Breadth of the rectangle(b)=3.6cm

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

Quiz

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