Class:-6th
MENSURATION
Prepared by:
Harishankar Meena
TGT,Maths
JNV,Nawanagar, Buxar, Bihar
Chapter No. :- 10
MENSURATION
Introduction �
Mensuration :-
mensuration is the branch of mathematics that deals with
the geometrical shapes, their perimeter , area and volumes.
PERIMETER
Look at the following figures ,
If you start from the point A , in each case and move along the
line segments then you again reach the Point A, So distance
Covered in a complete round of the shape in each case , is
known as the perimeter of the given shapes.
A A A
Perimeter :- perimeter is the total length or total distance covered along the boundary of a closed figure.
For Example
Find the perimeter of the given figure
Perimeter = Total distance
covered in one
round
= AB+BQ+QC+CD+DS+SE+
EF+FR+RG+GH+HP+PA
= 1+3+3+1+3+3+1+3+3+1+3+3
= 28cm
Use of perimeter in our daily life
(i) A farmer who wants to fence his field.
(ii) A person preparing a track to conduct sports.
(iii) Planning the construction of a house.
(iv) Building a swimming pool.
(v) Framing a photograph
PERIMETER OF TRIANGLE
A
B
C
PERIMETER OF RECTANGLE
Perimeter of rectangle
= Sum of the lengths of all sides
= L + B + L+ B
= 2L + 2B
= 2( L+B)
= 2 ( Length + Breadth)
Note:- In a rectangle opposite sides are equal and each angle
is a right angle
L
B
PERIMETER OF SQUARE
Perimeter of square
= sum of the lengths of all sides
let length one side of the square is ‘a’
then ,
Perimeter of square is
= a +a + a + a
= 4 a
= 4 X side
a
Note :- In a square all sides and angles are equal.
Some Applications
Q. 3 Find the cost of fencing a rectangular park of length
175m and breadth 125m , at the rate of Rs.12per meter.
Solution:- perimeter = 2 (l+ b)
= 2(175+125)
= 2 (300)
= 600m
given , 1 m cost = Rs. 12
so, 600m cost = 12x 600
= Rs. 7200
Rectangular Park
Question 4.
Pinkey runs around a square field of side 75m, Bobby runs around a rectangular field, with length 160m and breadth 105m .Who covers more distance by how much.
Solution:-
75m
160m
105m
Distance covered by Pinky in one round
= 4x(side)
= 4x 75
= 300m
Distance covered by Boby in one round.
= 2x(l+b)
= 2x (160+105)
= 2x(265)
= 530 m
Differences in distances = 530m – 300m
= 230m
Hence, Bobby covers more distance and by 230m
PERIMETER OF REGULAR POLYGONS
Some Applications
AREA
AREA
Calculating the area of a closed plane figure by using a square or graph paper
Example 1.
By counting squares, estimate the area of the given fig.
Sol.
Covered area | Number | Area (sq.unit) |
Fully -filled squares | 11 | 11 |
Half- filled squares | 3 | |
More than half- filled squares | 7 | 7 |
Less than half –filled | 5 | 0 |
Example 2.
By counting squares, estimate the area of the given fig.
Covered area | Number | Area (sq.unit) |
Fully -filled squares | 6 | 6 |
Half- filled squares | 7 | |
More than half- filled squares | 0 | 0 |
Less than half –filled | 0 | 0 |
Area of rectangle
| | | | | |
| | | | | |
| | | | | |
| | | | | |
| | | | | |
Find the area of a rectangle whose length is 6cm and breadth is 5cm.
Sol. Draw a rectangle on graph paper having
1 cm x 1 cm squares .
The rectangle covers 30 squares completely.
so, the area is = 30 sq. cm
which can be written as 6 x 5 = 30 sq .cm
Area of rectangle = ( length X Breadth) sq. unit
6cm
5cm
1cm
1cm
Area of The Square
| | | |
| | | |
| | | |
| | | |
4cm
4cm
1cm
1cm
Note:- When we find the perimeter or area , the length
* Various units of measurement are connected by the
following relations:
1 kilometre = 1000 metre(m)
1 metre = 100 centimetre(cm)
1 decimetre = 10centimetre (cm)
1metre = 10 decimetre (dm)
1 centimetre = 10 milimetre(mm)
1 foot = 12 inches
1 yard = 3 feet
Some Applications
��Q.3 One side of a square plot is 300m. Find its area .Also,find � the cost of levelling it at the rate of Rs. 0.50 per sq. metre.��
4M
3M
TILE
Wall
Q.6 Split the following shapes into rectangle and find the area
of each.
Total Area = (Area of rectangle with
sides 12 and 2 units) +
( Area of rectangle with
sides 8 and 2 units)
= (12 x 2) + ( 8 x 2)
= (24 + 16)
= 40 sq. units
Total area =( Area of rectangle with
sides 6 and 4 units) +
( Area of rectangle with
sides 4 and 3 units)
= ( 6 x4) + ( 4x 3)
= 24 + 12
= 36 sq. cm.
6-2= 4cm
THANK
YOU
Self Assessment Test
Multiple choice questions:-
Q.1 Perimeter of a rectangle is ?
A. 2 x length + breadth B. length x breadth C. 2x( length+ breadth)
Q.2 Perimeter of a square is ?
A. 2 X length of a side B. 4 X length of a side C. 3 X length of a side
Q.3 The perimeter of an equilateral triangle of side 7 cm is?
A. 7cm B. 14 cm C. 21 cm
Q.4 Perimeter of a regular pentagon with side measuring 4 cm is ?
A. 15 cm B. 20cm C. 25cm
Q.5 The side of a square is 6 cm, if its side is doubled , then its new
perimeter is ?
A. 24cm B . 60 cm C. 48 cm
Short answer type questions:-
Q.8
Q.9 The length and breadth of a rectangle are 10cm and 8cm . Respectively
, if its length is doubled , then find its new area.
Q.10 Find the breadth of the rectangle whose area is 120sq. Cm
and length is 15 cm ?
Q.11 There is a rectangular lawn 10 m long and 4m wide in
front of Raju’s house. It is fenced along the two smaller
sides and one longer side leaving a gap of 1 m for the
entrance . Find the length of fencing.
Long answer type questions:-
Q.12 The cost of fencing a square park at the rate of Rs. 30
park.
Q.13 What will happen to the area of a square if its side is
(i) doubled (ii) halved
Q.14 The total cost of flooring a room at Rs. 8.50 per square
metre is Rs. 510. If the length of the room is 8 metre ,
find its breadth.
Q.15 What is the length of outer boundary of the park shown
figure ? What will be the total cost of fencing it at the
rate of Rs.20 per metre? There is a rectangular flower
bed in the centre of the park. Find the cost of menuring
the flower bed at the rate of Rs. 50 per square metre.
THANK YOU