 ## Area of Plane Figures

Subject: Compulsory Maths

#### Overview

The plane surface enclosed by the boundary line of a plane closed figure is known as its area . The area is always measured in the square unit. For example,cm2 , mm2,m2etc . This note gives information about different solid figure and also its formula.
##### Area of Plane Figures

The plane surface enclosed by the boundary line of a plane closed figure is known as its area . The area is always measured in a square unit. For example,cm2 , mm2,m2etc.

Area of rectangles :

In the adjoining graph , the area of each square room is 1 cm2. So , the surface enclosed by the rectangle is 10 cm2.

$\therefore$ Area of rectangle = 10 cm2

i.e . 5 rooms along length x 2 rooms breadth = 10cm2

$\therefore$ Area of rectangle = length x breadth = l x b

Area of triangles

Here , area of the triangle EFG= length x breadth  =FG xGH

=base x height

=b x$\frac{1}{2}$h

$\frac{1}{2}$bh

So the area of triangle =$\frac{1}{2}$base x height = $\frac{1}{2}$b xh

Area of trianglesgle

Here , area of triangle = area of rectangle FGHI

=FG XGH

=base x hieight

=b x$\frac{1}{2}$

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

So , the area of tringle =$\frac{1}{2}$base x height = $\frac{1}{2}$ b xh

Area of Parallelogrms

Here , area of the parallelogram EFGH =Area of the rectangle of EFGH

=EF x FI

= base x height

= b x h

So, the area of parallelogram = base x height

= b x h

Area of circles :

The length of rectangle EFGH (l) =$\frac{1}{2}$ x circumference = $\frac{1}{2}$ x2 πr

The breadth of rectangle EFGH (b) = r

area of the circle = Area of rectangle EFGH

=πr x r

=πr2

So , the area of circle =πr2

##### Things to remember
• Area of rectangle = length x breadth = l x b.
• The area of parallelogram = base x height.
• The area of circle =πr2
• 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.
##### Videos for Area of Plane Figures ##### Perimeter and Area- Plane figures - Maths - Class 7

Solution:

Here,
area of two rectangles = l × b
= 20 cm × 15 cm
= 300 cm2
Also,
area of two triangles = 2 ($\frac{1}{2}$ ×b × h)
= 2 × $\frac{1}{2}$ × 15 × 10 cm2
= 150 cm2
Again,
∴ Area of the figure = 300 cm2 + 150 cm2
= 450 cm2

Solution:

Area of the rectangle = l × b
= 20 cm × 15 cm
= 300 cm2
Radius of each semi-circle = $\frac{14}{2}$
= 7 cm
Area of two semi-circles = 2($\frac{1}{2}$πr2)
= $\frac{22}{7}$ × 7 × 7 cm2

= 154 cm2
∴ Area of the figure = 350 cm2 + 154 cm
= 504 cm2

Solution:

Here,
Area of parallelogram = b × h
= 16 cm × 10cm
= 160 cm2
Area of triangle = $\frac{1}{2}$ b × h
= $\frac{1}{2}$ × 16 cm × 10 cm
= 80 cm2
∴ Area of the shaded region = Area of parallelogram -Area of triangle
= 160 cm2 - 80 cm2
= 80 cm2

Solution:

Area of rectangle = l × b
= 12 cm × 9 cm
= 108 cm2
Area of parallelogram = b × h
= 6 cm × 4 cm
= 24 cm2
∴ Area of the shaded region = Area of rectangle - Area of parallelogram
= 108 cm2 - 24 cm2
= 84 cm2

Solution:

Here,
Area of bigger rectangle = l × b
= 20 cm × 16 cm
= 320 cm2
Area of smaller rectangle = l × b
= 14 cm × 10 cm
= 140cm2
∴ Area of the shaded region = Area of bigger recatngle - Area of smaller rectangle
= 320 cm2 - 140 cm2
= 180 cm2

Solution:

The perimeter of the square field = 96 m
or, 4 l = 96 m
or, l = $\frac{96}{4}$
or, l = 24 m
∴ The length of the field (l) = 24m

Solution:

Let, the breadth (b) pf the room be x m.
So, the length (l) of the room will be 2x m.
Now,
The perimeter of the rectangular room = 48 m
or, 2 (l + b) = 48 m
or, 2 (2x + x) = 48 m
or, 6x  = 48 m
or, x = $\frac{48}{6}$ m
or, x = 8 m
∴ The breadth of the room (b) = x = 8 m
The length of the room (l) = 2x
= 2 × 8m
= 16 m
Again.
Area of the room = l  × b
= 16 m × 8 m
=  128 m2
∴ Thw area of a room is 128 m2

Solution:

Here,
The perimeter of circular ground = 220 m
or, 2πr = 220 m
or, 2 × $\frac{22}{7}$ × r = 220 m
or, r = $\frac{220 × 7}{2 × 22}$ m
or, r = 35 m
Now,
Area of the circular ground = πr2
= $\frac{22}{7}$ × 35 m × 35 m
= 3850 m2