Class – 6 CH-10 MENSURATION
MATHS NCERT SOLUTIONS
Exercise 10.1
Question
1:
Find the perimeter of each of the following figures:
Solution 1:
(a)
Perimeter
= Sum of all the sides
= 4 cm + 2 cm + 1 cm + 5 cm =
12 cm
(b)
Perimeter
= Sum of all the sides
=
23 cm + 35 cm + 40 cm + 35 cm = 133 cm
(a) Perimeter = Sum of all the sides
= 15 cm + 15 cm + 15 cm + 15 cm = 60 cm
(b) Perimeter = Sum of all the sides
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
(c) Perimeter = Sum of all the sides
1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm =
15 cm
(d) Perimeter = Sum of all the sides
= 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2
cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm
= 52 cm
2
The lid of a rectangular box of sides 40 cm by 10 cm is
sealed all round with tape. What is the length of the tape required?
Solution 2:
Total length of tape
required = Perimeter of rectangle
= 2 (length + breadth)
= 2 (40 + 10)
= 2 x 50
= 100 cm
= 1 m
Thus, the total length of tape required is 100 cm or 1 m.
Question
3: A table-top measures 2 m 25 cm by 1 m
50 cm. What is the perimeter of the table-top?
Solution 3:
Length
of table top = 2 m 25 cm = 2.25 m
Breadth
of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 x (length + breadth)
=
2 x (2.25 + 1.50)
=
2 x 3.75
= 7.50 m
Thus, the perimeter of table top is 7.5 m.
Question
4:
What is the length of the wooden
strip required to frame a photograph of length and breadth 32 cm and 21 cm
respectively?
Solution 4:
|
Length of wooden strip |
= Perimeter of photograph |
|
Perimeter of photograph |
= 2 x (length + breadth) |
|
|
= 2 (32 + 21) |
|
|
= 2
x 53 cm |
= 106 cm
Thus, the length of the wooden strip required is equal to
106 cm.
5
A rectangular piece of land
measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What
is the length of the wire needed?
Solution 5:
Since the 4 rows of wires are
needed.
Therefore the total length of
wires is equal to 4 times the perimeter of rectangle.
Perimeter
of field = 2 x (length + breadth)
= 2 x (0.7 + 0.5)
= 2 x 1.2
= 2.4 km
=
2.4 x 1000 m
= 2400 m
Thus, the length of wire = 4 x 2400 = 9600 m = 9.6 km
Question
6:
Find the perimeter of each of
the following shapes:
(a) A triangle
of sides 3 cm, 4 cm and 5 cm.
(b) An
equilateral triangle of side 9 cm.
(c) 
An
isosceles triangle with equal sides 8 cm each and third side 6 cm Solution 6:
(a) Perimeter
of ABC = AB + BC + CA
= 3 cm + 5 cm + 4 cm
= 12 cm
(b) Perimeter
of equilateral ABC = 3 x side
= 3 x 9 cm
= 27 cm
(c) Perimeter
of ABC = AB + BC + CA
= 8 cm + 6 cm + 8 cm
= 22 cm
7
Find the perimeter of a triangle
with sides measuring 10 cm, 14 cm and 15 cm.
Solution 7:
Perimeter of triangle = Sum of all three sides
= 10 cm + 14 cm + 15 cm
= 39 cm
Thus, the perimeter of triangle is 39 cm.
Question
8: Find the perimeter of a regular hexagon
with each side measuring 8 cm.
Solution 8:
Perimeter of Hexagon = 6 x
length of one side
= 6 x 8 m
= 48 m
Thus, the perimeter of hexagon is 48 m.
Question
9: Find the side of the square whose
perimeter is 20 m.
Solution 9:
Perimeter of square = 4 x
side
20
= 4 x side
Side = 
= 5 cm
Thus, the side of square is 5 cm.
Question
10: The perimeter of a regular pentagon is
100 cm. How long is its each side?
Solution 10:
Perimeter of regular pentagon =
100 cm
5 x side = 100 cm
Side = 
= 20 cm
Thus, the side of regular
pentagon is 20 cm.
11
A piece of string is 30 cm long. What will be the length of
each side if the string is used to form: (a) a square (b) an equilateral triangle (c) a regular hexagon?
Solution 11:
Length of string = Perimeter of
each figure
(a) Perimeter
of square = 30 cm
4 x side = 30 cm
Side = 
= 7.5 cm
Thus, the length of each side of
square is 7.5 cm.
(b) Perimeter
of equilateral triangle = 30 cm
3 x side = 30 cm
Side = 
= 10 cm
Thus, the length of each side of
equilateral triangle is 10 cm.
(c) Perimeter
of hexagon = 30 cm
6 x side = 30 cm
Side = 
= 5 cm
Thus, the side of each side of hexagon is 5 cm.
Question
12:
Two sides of a triangle are 12
cm and 14 cm. The perimeter of the triangle is 36 cm. What is the third side?
Solution 12:
Let the length of third side be x cm.
Length of other two side are 12
cm and 14 cm.
Now, Perimeter of triangle = 36 cm
12 14 x 36
26 x 36
x 36 26
x10 cm
Thus, the length of third side
is 10 cm.
13
Find the cost of fencing a
square park of side 250 m at the rate of ₹20 per
meter.
Solution 13:
|
Side of square |
= 250 m |
|
Perimeter of square |
= 4 x side |
|
|
= 4 x 250 |
= 1000 m
Since,
cost of fencing of per meter =
₹ 20
Therefore, the cost of fencing
of 1000 meters = 20 x 1000 = ₹20,000
Question
14:
Find the cost of fencing a
rectangular park of length 175 m and breadth 125 m at the rate of ₹12
per meter.
Solution 14:
Length
of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter
of park =
2 x (length + breadth)
=
2 x (175 + 125)
=
2 x 300 = 600 m
Since,
the cost of fencing park per meter = ₹ 12
Therefore, the cost of fencing park of 600 m = 12 x 600 = ₹
7,200
Question
15:
Sweety runs around a square park
of side 75 m. Bulbul runs around a rectangular park with length of 60 m and
breadth 45 m. Who covers less distance? Solution 15:
Distance covered by Sweety = Perimeter of square park
Perimeter
of square = 4 x side
=
4 x 75 = 300 m Thus, distance covered by Sweety is 300 m.
Now,
distance covered by Bulbul = Perimeter of rectangular park
Perimeter
of rectangular park = 2 x (length + breadth)
=
2 x (60 + 45)
=
2 x 105 = 210 m
Thus, Bulbul covers the distance
of 210 m and Bulbul covers less distance.
16
What is the perimeter of each of the following figures? What
do you infer from the Solution?
|
|
= 4 x 25 = 100 cm |
|
(b) Perimeter of rectangle |
= 2 x (length + breadth) |
|
|
= 2 x (40 + 10) |
|
|
= 2 x 50 = 100 cm |
|
(c) Perimeter of rectangle |
= 2 x (length + breadth) |
|
|
= 2 x (30 + 20) |
|
|
= 2 x 50 = 100 cm |
|
(d) Perimeter of triangle |
= Sum of all sides |
|
|
= 30 cm + 30
cm + 40 cm = 100 cm |
Thus, all the figures have same perimeter.
17
Avneet buys 9 square paving slabs, each with a side 
m. He lays them in the form of a square
(a) What
is the perimeter of his arrangement?
(b) Shari does
not like his arrangement. She gets him to lay them out like a cross. What is
the perimeter of her arrangement?
(c) Which
has greater perimeter?
(d) Avneet
wonders, if there is a way of getting an even greater perimeter. Can you find a
way of doing this? (The paving slabs must meet along complete edges, i.e., they
cannot be broken.) 
Solution 17:
(a)
6 m
(b)
10 m
(c)
Second arrangement has greater perimeter.
(d) Yes, if all the squares are arranged in row, the perimeter be 10 cm
Exercise 10.2
Question
1:
Find the areas of the following figures by counting squares:
(a)
Number of filled square = 9
Area
covered by squares = 9 x 1 = 9 sq. units
(b)
Number of filled squares = 5
Area
covered by filled squares = 5 x 1 = 5 sq. units
(c)
Number of full filled squares = 2
Number of half-filled squares = 4
Area covered by full filled
squares =
2 x 1 = 2 sq. units
And Area covered
by half-filled squares = 4 x 
= 2 sq. units
Total
area = 2 + 2 = 4 sq. units
(d)
Number of filled squares =
8
Area covered by filled
squares =
8 x 1 = 8 sq. units
(a)
Number of filled squares =
10
Area
covered by filled squares = 10 x 1 = 10
sq. units
(b) Number
of full filled squares = 2
Number
of half-filled squares = 4
Area covered by full filled
squares =
2 x 1 = 2 sq. units
And Area covered
by half-filled squares = 4 x 
= 2 sq. units
Total
area = 2 + 2 = 4 sq. units
(c) Number
of full filled squares = 4
Number
of half-filled squares = 4
Area covered by full filled
squares =
4 x 1 = 4 sq. units
And Area covered
by half-filled squares = 4 x 
= 2 sq. units
Total
area = 4 + 2 = 6 sq. units
(d)
Number of filled squares =
5
Area
covered by filled squares = 5 x 1 = 5
sq. units
(e)
Number of filled squares =
9
Area
covered by filled squares = 9 x 1 = 9
sq. units
(f)
Number of full filled squares = 2
Number of half-filled squares = 4
Area
covered by full filled squares = 2 x 1 = 2 sq. units
And Area covered
by half-filled squares = 4 x 
= 2 sq. units
Total
area = 2 + 2 = 4 sq. units
(g)
Number of full filled squares = 4
Number of half-filled squares = 2
Area
covered by full filled squares = 4 x 1 = 4 sq. units
And Area covered
by half-filled squares = 2 x 
= 1 sq. units
Total
area = 4 + 1 = 5 sq. units
(h)
Number of full filled squares = 3
Number of half-filled squares =
10
Area
covered by full filled squares = 3 x 1 = 3 sq. units
And Area covered
by half-filled squares = 10 x 
= 5 sq. units
Total
area = 3 + 5 = 8 sq. units
(i)
Number of full filled squares = 7
Number of half-filled squares =
14
Area
covered by full filled squares = 7 x 1 = 7 sq. units
And Area covered
by half-filled squares = 14 x 
= 7 sq. units
Total
area = 7 + 7 = 14 sq. units
(j)
Number of full filled squares = 10
Number of half-filled squares =
16
Area
covered by full filled squares = 10 x 1 = 10 sq. units
And Area covered
by half-filled squares = 16 x = 8 sq. units
Total
area = 10 + 8 = 18 sq. units
Exercise 10.3
Question
1:
Find the areas of the rectangles whose sides are:
(a)
3 cm and 4 cm (b)
12 m and 21 m
(c)
2 km and 3 km (d)
2 m and 70 cm
Solution 1:
(a)
Area of rectangle = length x breadth
= 3 cm x 4 cm = 12 cm2
(b)
Area of rectangle = length x breadth
= 12 m x 21 m = 252 m2
(c)
Area of rectangle =
length x breadth
= 2 km x 3 km = 6 km2
(d)
Area of rectangle = length x breadth
= 2 m x 70 cm = 2 m x 0.7 m = 1.4 m2
Question
2:
Find the areas of the squares whose sides are:
(a)
10 cm (b)
14 cm (c)
5 cm
Solution 2:
(a)
Area of square = side x side = 10 cm x 10 cm =
100 cm2
(b)
Area of square = side x side = 14 cm x 14 cm =
196 cm2
(c) Area
of square = side x side = 5 m x 5 m = 25 m2
Question
3:
The length and the breadth of three rectangles are as given
below:
(a)
9 m and 6 m (b)
17 m and 3 m (c)
4 m and 14 m
Which one has the largest area and
which one has the smallest? Solution 3:
(a)
Area of rectangle = length x breadth = 9 m x 6 m = 54 m2
(b)
Area of rectangle = length x breadth= 3 m x 17 m = 51 m2
(c)
Area of rectangle = length x breadth= 4 m x 14 m = 56 m2
Thus, the rectangle (c) has
largest area, and rectangle (b) has smallest area.
Question
4: The area of a rectangle garden 50 m
long is 300 m2, find the width of the garden.
Solution 4:
Length of rectangle = 50 m and
Area of rectangle = 300 m2
Since, Area of rectangle = length x breadth
Area
of rectangle 300
Therefore,
Breadth
= = = 6 m Length 50
Thus, the breadth of the garden is 6 m.
Question
5:
What is the cost of tilling a
rectangular plot of land 500 m long and 200 m wide at the rate of ₹8
per hundred sq. m?
Solution 5:
Length of land = 500 m and
Breadth of land = 200 m
Area of land = length x breadth = 500 m x 200 m =
1,00,000 m2 
Cost
of tilling 100 sq. m of land = ₹ 8
Cost
of tilling 1,00,000 sq. m of land = 
= ₹ 8000
Question 6: A table-top measures 2 m by 1 m 50 cm.
What is its area in square meters?
Solution 6:
Length of table = 2 m
Breadth of table = 1 m 50 cm = 1.50 m Area of
table = length x breadth
= 2 m x 1.50 m = 3 m2
Question
7:
A room us 4 m long and 3 m 50 cm
wide. How many square meters of carpet is needed to cover the floor of the
room? Solution 7:
Length of room = 4 m
Breadth of room = 3 m 50 cm =
3.50 m
Area of carpet = length x breadth
=
4 x 3.50 = 14m2
Question
8:
A floor is 5 m long and 4 m
wide. A square carpet of sides 3 m is laid on the floor. Find the area of the
floor that is not carpeted.
Solution 8:
Length of floor = 5 m and
breadth of floor = 4 m
Area
of floor = length x breadth
=
5 m x 4 m = 20 m2
Now, Side of square carpet = 3 m
Area of square carpet = side x
side = 3 x 3 = 9 m2
Area of floor that is not carpeted = 20 m2 – 9
m2 = 11 m2
Question
9:
Five square flower beds each of
sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of
the remaining part of the land?
Solution 9:
Side of square bed = 1 m
Area of square bed = side x side
= 1 m x 1 m = 1 m2
Area of 5 square beds = 1 x 5 = 5 m2
Now, Length of land
= 5 m Breadth of land = 4 m
Area of land =
length x breadth
= 5 m x 4 m = 20 m2
Area of remaining part = Area of
land – Area of 5 flower beds
= 20 m2 – 5 m2 = 15 m2
Question
10:
By splitting the following figures into rectangles, find
their areas. (The measures are given in centimetres)
Solution 10:
(a) Area
of HKLM = 3 x 3 = 9 cm2
Area of IJGH = 1 x 2 = 2 cm2
Area of FEDG = 3 x 3 = 9 cm2
Area of ABCD = 2 x 4 = 8 cm2
Total area of the figure = 9 + 2 + 9 + 8 = 28 cm2
(b) 
Area of ABCD = 3 x
1 = 3 cm2 Area of BDEF = 3 x 1 = 3 cm2
Area of FGHI = 3 x 1 = 3 cm2
Total area of the figure = 3 + 3 + 3 = 9 cm2
Question
11:
Split the following shapes
into rectangles and find their areas. (The measures are given in centimetres)
Solution 11:
(b) There
are 5 squares each of side 7 cm. Area of one square = 7 x 7 = 49 cm2
Area of 5 squares = 49 x 5 = 245 cm2
(c) Area
of rectangle ABCD = 5 x 1 = 5 cm2
Area of rectangle EFGH = 4 x 1 = 4 cm2
Total area of the figure = 5 + 4 cm2
Question
12:
How many tiles whose length and
breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular
region whose length and breadth are respectively?
(a) 100 cm and 144 cm (b) 70
cm and 36 cm Solution 12:
(a)
Area of region = 100 cm x 144 cm = 14400 cm2
Area of one tile = 5 cm x 12 cm =
60 cm2
Area of region
Number
of tiles =
Area
of one tile
=
= 240
Thus, 240 tiles are required.
(b)
Area of region = 70 cm x 36 cm = 2520 cm2
Area of one tile = 5 cm x 12 cm =
60 cm2
Area of region
Number
of tiles =
Area
of one tile
=
= 42
Thus, 42 tiles are required.
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