PERIMETER AND AREA OF SIMPLE GEOMETRICAL FIGURE
The sum of length of all the sides of a geometrical figure is called its perimeter.
Example: Planning the construction of a house. Since you have to pour a concrete foundation, within the housing constraints you want to maximize the area within the constraints which are related to the perimeter (like you can only get so close to a neighbor’s house, etc.)
Perimeter of simple geometrical figures
Some simple geometrical figures are
Find the perimeter of the Geometrical figures.
|GIVEN EF=3CM, FG=5CM ,GH=7CM ,HI=5CM ,IE=4CM |
Perimeter = EF+ FG+ GH+ HI+ IE
|GIVEN JK8C, JN=2CM,KL=6CM, MN-5CM ML 6CM |
Perimeter= JK+ JN +KL+MN +ML
8CM + 2CM+ 6CM+ 5CM+ 6CM
|Given PQ=4CM,QR=2CM,RS=2CM,ST=4CM,TU=3CM,PU=3CM |
Perimeter = 18CM
|GIVEN= 7CM, XC=8CM, NC=4CM, RN=6CM, MR=3C MW=2CM |
II. In list ‘A’ plane figures and in list ‘B’ their perimeters are given. Match list ‘A’ with list ‘B’
Perimeter( sum of 4 sides) = 24cm
Sum of 3 sides =22cm
Length of the 4th side =2cm
Sum of given 3 sides = 2cm+ 10cm+ 10cm
Length and four side = perimeter – sum of 3 sides
Length of 4th side =24cm -22cm = 22cm.
Perimeter (sum of 5 sides)=24cm
=sum of 4 sides =19cm
Length of the 5 sides =5cm
Sum of given 4 sides=3cm+ 4cm+ 4cm+ 8cm
Length and fifth side = perimeter-sum of 3 sides
Length of 5 sides =24cm -19cm =5cm.
In general area is expressed in square unit. ∴Units of Area : Sq cms, Sq mtrs, Sq kms… etc.
Count the number of squares inside the shape. Totally there are 7 squares. Area of given shape = 7 sq cm.
On the graph sheet given, the area of each square is 1 sq cm. Find the area of the given shapes.
There are totally 5 square.
Area of given shape =5 sq cm
There are totally 21 square
Area of given shape=21 sq cm.
There are totally 16 Square
Area of given shape=16 sq cm.
There are totally 20 Square
Area of given shape=20 sq cm.
There are totally 12 Square
Area of given shape=12 sq cm.