Note on Area and Perimeter of Plain Figures

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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

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

.

\(\therefore\) 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.

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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

  • 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
.

Very Short Questions

Solution:

Given, side(a) = 20m

Perimeter of square flower-bed = 4a

=4 x 20

=80

\(\therefore\) perimeter of square flower-bed = 80 m

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

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:

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

Perimeter of \(\triangle\)ABC = AB + BC + CA

= 6cm + 5cm + 7cm

= 18cm

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

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