Area of Plane Figures

Subject: Compulsory Maths

Find Your Query

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

Scholarships after +2 Abroad Studies Opportunities

=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

= length x breadth

=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

= length x breadth

=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

= length x breadth

=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
Areas of Plane Figures
Perimeter and Area- Plane figures - Maths - Class 7
Questions and Answers

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 

Quiz

© 2019-20 Kullabs. All Rights Reserved.