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Ncert exemplar Solutions Class 6 Mathematics Mensuration
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of NCERT Class 6 Mathematics Book, Unit 6, Mensuration. Here students can easily find step by step solutions of all the problems for Mensuration, Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get Unit 6 solutions.
Mensuration Unit 6 Solution :
Multiple Choice Questions :
Question no – (1)
Following figures are formed by joining six unit squares. Which figure has the smallest perimeter in Fig. 6.4?
Solution :
Perimeter of fig.1 is 10 unit.
Perimeter of fig.2 is 12 unit.
Perimeter of fig.3 is 12 unit.
Perimeter of fig.4 is 12 unit.
∴ The correct answer is option – (D) – (i)
∴ Fig.1 has smallest perimeter.
Question no – (2)
A square shaped park ABCD of side 100 m has two equal rectangular flower beds each of size 10m × 5m Fig. 6.5. Length of the boundary of the remaining park is
Solution :
A square shaped park ABCD of side 100 m.
Perimeter of square park = 4 x length
Perimeter of square park = 4 x 100 = 400 m.
Thus, the correct answer is option – (B) 400 m
Question no – (3)
The side of a square is 10cm. How many times will the new perimeter become if the side of the square is doubled?
Solution :
The side of a square is 10 cm
Perimeter of square = 4 x length
Perimeter of square = 4 x 10 = 40 m
Now the side of the square is doubled i.e. 20 cm
Perimeter of square = 4 x length
Perimeter of square = 4 x 20 = 80 m
Perimeter is 2 times of 1 square.
Hence, correct answer is option – (A) 2 times
Question no – (4)
Length and breadth of a rectangular sheet of paper are 20 cm and 10 cm, respectively. A rectangular piece is cut from the sheet as shown in Fig. 6.6. which of the following statements is correct for the remaining sheet
Solution :
The correct option is – (A) Perimeter remains same but area changes.
Length and breadth of a rectangular sheet of paper are 20 cm and 10 cm.
Perimeter of a rectangular sheet of paper = 2(length + breadth)
Perimeter of rectangular sheet of paper = 2 x (20 cm + 10 cm)
Perimeter of rectangular sheet of paper = 2 x (30 cm) = 60 cm.
Area of rectangular sheet of paper = length x breadth
20 cm x 10 cm = 200 cm
Area of rectangular piece is cut from the sheet = length x breadth
Length of piece = 5 cm.
Breadth of piece = 2 cm.
Area of rectangular piece is cut from the sheet = 5 x 2 = 10 cm
Area of rectangular sheet of paper when rectangular piece is removed = 200 cm – 10
cm = 190 cm.
Perimeter remains same but area changes is true.
(B) Area remains the same but perimeter changes.
Area changes but perimeter is same.
(C) Both area and perimeter are changing.
Only area changes, perimeter is not changing.
(D) Both area and perimeter remain the same.
Only perimeter is same, Area changes.
Question no – (5)
Two regular Hexagons of perimeter 30cm each are joined as shown in Fig. 6.7. The perimeter of the new figure is
Solution :
Each side of regular hexagon = 30 / 6 = 5 cm.
Two regular Hexagons of perimeter 30 cm each.
Sum of perimeter of Two regular Hexagons = 30 + 30 = 60 cm
Now, when we joining two hexagon there is 1 side is common between them.
We subtract 2 side from total perimeter.
Because the common side is inside the two hexagon which we never count.
Perimeter is sum of all exterior sides.
60 cm – 10 cm = 50 cm
So, the correct answer is option – (D) 50 cm.
Question no – (6)
In Fig. 6.8 which of the following is a regular polygon. all have equal side except (i)
Solution :
Regular polygon is the fig. in which all sides of figure is same and also each angle is same.
In fig. 1, 3 and 4
Sides are equal but angles are not equal.
Fig. 2 is a regular polygon.
Thus, the correct answer is option (B) (ii).
Fill in the blanks :
Question no – (9)
Solution :
Perimeter of the shaded portion in Fig. 6.9 is given by
AB + BM + MD + DE + EN + NG + GH + HA
Question no – (10)
Solution :
The amount of region enclosed by a plane closed figure is called its Area.
Question no – (11)
Solution :
Area of a rectangle with length 5 cm and breadth 3 cm is 15 sq. cm
Area of rectangle = length x breadth
= 5 cm x 3 cm
= 15 sq. cm
Question no – (12)
Solution :
(a) The area of the rectangle is 12 sq. unit.
Area of rectangle = length x breadth
Length of rectangle = 6 cm
Breadth of rectangle = 2 cm.
Area of rectangle = 6 x 2 = 12 sq.cm
(b) The area of the square is 16 sq. unit
A rectangle and a square have the same perimeter.
Length of rectangle = 6 cm.
Breadth of rectangle = 2 cm.
Perimeter of a rectangle = 2(length + breadth)
Perimeter of rectangle = 2 x (6 cm + 2 cm)
Perimeter of rectangle = 2 x (8 cm) = 16 cm.
A rectangle and a square have the same perimeter.
Perimeter of square = 16 cm
Side of square = 16 / 4 = 4 cm.
Area of square = length x length
Length of square = 4 cm
= 4 cm x 4 cm
= 16 sq. cm
Area of square = 16 sq. cm
Question no – (13)
Solution :
(a) 1 m = 100 cm
(b) 1 sq. cm = 1 cm × 1cm.
(c) 1 sqm = 1m × 1 m = 100cm × 100 cm
(d) 1 sqm = 10000 sq.cm
True or False :
Question no – (14)
Solution :
If length of a rectangle is halved and breadth is doubled then the area of the rectangle obtained remains same.
Given statement is True.
Explanation :
Let, length of the rectangle = x
and breadth of the rectangle = y
Area the rectangle,
= x × y
= x y
If, length of the rectangle halved so x/2
breadth of the rectangle doubled so it became 2y
Area of the rectangle,
= x/2 × 2 y
= x y
∴ It is same.
Question no – (15)
Solution :
Area of a square is doubled if the side of the square is doubled.
Given statement is False.
Explanation :
Let, the side of square = a
If it is doubled square = 2a
area of the square = a2
after doubling side area of the square
= (2a)2
= 4a2
∴ Area of square is 4 times of before area it the side of the square doubled.
Question no – (16)
Solution :
Perimeter of a regular octagon of side 6 cm is 36 cm
This statement is False.
Because, regular octagon has 8 sides.
Each side of regular octagon is 6 cm.
Perimeter of a regular octagon = 8 x side of regular octagon.
Perimeter of a regular octagon = 8 x 6 = 48 cm.
Alternative Solution,
Given statement is False.
Explanation :
Side of the octagon = 6 cm
Perimeter = 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6
= 48 cm
Question no – (17)
Solution :
A farmer who wants to fence his field, must find the perimeter of the field –
This statement is True.
Explanation : A farmer must have to know the perimeter to fence the field because he have to fence all the sides of the field.
Question no – (18)
Solution :
An engineer who plans to build a compound wall on all sides of a house must find the area of the compound.
This statement False.
Because, An engineer who plans to build a compound wall on all sides of a house must find the Perimeter of the compound.
Question no – (19)
Solution :
To find the cost of painting a wall we need to find the perimeter of the wall
This statement is False.
Because, To find the cost of painting a wall we need to find the Area of the wall.
Question no – (20)
Solution :
To find the cost of a frame of a picture, we need to find the perimeter of the picture.
Given statement is True.
Explanation : To field the cost of the frame of a picture we need to find the perimeter of the picture.
Question no – (21)
Solution :
Four regular hexagons are drawn so as to form the design as shown in Fig. 6.11.
Perimeter of the design is 28 cm.
Perimeter is the sum of all exterior sides of regular hexagon.
There are total 14 exterior sides of regular hexagon.
Each side of regular hexagon = Perimeter of the design / Total sides of regular hexagon.
Each side of regular hexagon = 28 / 14
Each side of regular hexagon = 2 cm.
Question no – (22)
Solution :
Isosceles triangle is the triangle having two sides are equal in length.
One of the two equal sides is 18 cm.
Other side is also 18 cm.
Perimeter of an isosceles triangle is 50 cm.
Perimeter of an isosceles triangle = sum of all three sides.
Perimeter of an isosceles triangle = 18 + 18 + Third side.
50 cm = 36 cm + Third side.
Third side = 50 cm – 36 cm
Third side = 14 cm.
Question no – (23)
Solution :
Perimeter of the rectangle is 40 cm.
We know, Perimeter of a rectangle = 2 (length + breadth)
Let ‘X’ be the breadth of rectangle.
Then Length of a rectangle is three times its breadth.= 3 x X
Perimeter of a rectangle = 2(length + breadth)
Perimeter of a rectangle = 2 x ( 3 x X + X )
Perimeter of a rectangle = 2 x ( 4 x X)
Perimeter of a rectangle = 8 x X
But Perimeter of the rectangle is 40 cm.
40 = 8 x X
X = 40 / 8
X = 5 cm.
X is the breadth of rectangle.
Breadth of rectangle is 5 cm.
Then Length of a rectangle is three times its breadth
= 3 x 5
= 15 cm
Then Length of a rectangle is 15 cm.
Question no – (24)
Solution :
There is a rectangular lawn 10 m long and 4m wide in front of Meena’s house.
We know, Length of rectangular lawn = 10 m.
Breadth of rectangular lawn = 4 m.
Perimeter of a rectangular lawn = 2 (length + breadth)
Perimeter of rectangular lawn = 2 x (10 m + 4 m)
Perimeter of rectangular lawn
= 2 x (14 m)
= 28 m.
It is fenced along the two smaller sides and one longer side leaving a gap of 1 m for the entrance.
We have to subtract 1 larger side from perimeter and also gap of 1 m.
Perimeter of rectangular lawn = 28 cm.
Length of rectangular lawn = 10 m.
The length of fencing = 28 – (10 + 1)
The length of fencing = 28 – 11
The length of fencing = 17 m.
Alternative Solution,
The length of fencing,
= (4 + 4 + 9)
= 17 m
∴ The length of fencing is 17 cm.
Question no – (25)
Solution :
1 box is taken as 1 unit.
There are 13 units.
Area of the region = Sum of all units given in fig.
Area of the region = 13 sq. units.
Alternative Solution,
The region given in fig 6.13 is measured by taking 13 squint as a unit.
Question no – (26)
Solution :
Tahir measured the distance around a square field as 200 rods (lathi).
The length of this rod was 140 cm.
Distance around a square field = we are multiplying 200 rods with 140 cm length.
Distance around a square field = 200 x 140
Distance around a square field = 28,000 cm
This distance is nothing but the perimeter of square field.
Perimeter of square field = Sum of measure of four sides.
Perimeter of square field = 4 x length
Length of square field = 28,000 cm / 4
Length of square field = 7,000 cm
We know,
1 m = 100 cm
7000 cm = 70 m
Length of square field is 70 m
Alternative Solution :
Distance,
= 200 × 140
= 28000 cm
∴ Length of side of square,
= 28000/4 cm
= 7000 cm
= 70 m.
Question no – (27)
Solution :
The length of a rectangular field is twice its breadth.
We know, Perimeter of a rectangular field = 2(length + breadth)
Let ‘X’ be the breadth of rectangular field
Then Length of a rectangular field is three times its breadth.= 2 x X
Perimeter of a rectangular field = 2(length + breadth)
Perimeter of a rectangular field = 2 x (2 x X + X)
Perimeter of a rectangular field = 2 x (3 x X)
Perimeter of a rectangular field = 6 x X
Jamal jogged around it four times and covered a distance of 6 km.
4 times perimeter of rectangular field
= 4 x (6 x X)
= 24 x X
24 x X = 6 km
X = 6 km / 24 = 1 /4 km
X = 250 m
X is the breadth of rectangle.
Breadth of rectangular field is 250 m.
Then Length of a rectangular field is two times its breadth
= 2 x 250
= 500 m
Then, Length of a rectangular field is 500 m.
Alternative Solution,
Let, breadth is = x
Length is = 2x
area = 6000/4 m
= 1500 m
2 (x + 2x) = 1500
2 × 3x = 1500
= x 1500/3 × 3 = 250
∴ breadth = 250 m
∴ Length = 250 × 2 = 500 m
Question no – (28)
Solution :
Three squares are joined together as shown in Fig. 6.14.
Their sides are 4 cm, 10 cm and 3 cm.
The perimeter of the figure,
The square 1 has side 4 cm.
We are see only 3 sides of square.
Perimeter of square 1 is 3 x 4 = 12 cm.
The square 2 has side 10 cm.
Perimeter of square 2 = 4 x length of side
Perimeter of square 2 = 4 x 10 = 40 cm
From this we have to subtract 2 sides of square 1.
Perimeter of square 2 = 40 – 8 = 32 cm.
The square 3 has side 3 cm.
We are see only 3 sides of square.
Perimeter of square 3 is 3 x 3 = 9 cm.
Perimeter of figure = Sum of perimeter (square 1 + square 2 + square 3)
Perimeter of figure = 12 + 32 + 9
Perimeter of figure = 53 cm
Alternative Solution :
The perimeter of the figure,
= (4 + 4 + 4 + 6 + 10 + 7 + 3 + 3 + 3 + 10)
= 54 cm
∴ The perimeter of the figure is 54 cm.
Question no – (29)
Solution :
In Fig. 6.15 all triangles are equilateral and AB = 8 units.
Other triangles have been formed by taking the mid points of the sides.
The perimeter of the figure = Sum of all exterior sides of figure.
The perimeter of the figure,
Suppose, Triangle 1 is small triangle having length 1 unit.
Perimeter of Triangle 1 = we have seen only 6 sides.
Perimeter of Triangle 1
= 1 x 6
= 6 units
Triangle 2 having length 2 units.
Perimeter of Triangle 2 = we have seen only 3 sides. 3 half sides.
Perimeter of Triangle 2
= 2 x 3 + 3
= 9 units
Triangle 3 having length 4 units.
Perimeter of Triangle 3 = we have seen only 3 sides. 3 half sides.
Perimeter of Triangle 3
= 4 x 3 + 6
= 18 units
Triangle 4 having length 8 units.
Perimeter of Triangle 4 = we have seen only 3 half sides.
Perimeter of Triangle 4
= 4 x 3
= 12 units
The perimeter of the figure = sum of perimeter (Triangle 1 + Triangle 2 + Triangle
3 + Triangle 4)
The perimeter of the figure
= 6 + 9 + 18 + 12
= 45 units.
The perimeter of the figure is 45 units.
Question no – (30)
Solution :
Length of a rectangular field is 250 m and width is 150 m.
We know, Length of rectangular field = 250 m.
Breadth of rectangular field = 150 m.
Perimeter of a rectangular field = 2(length + breadth)
Perimeter of rectangular field
= 2 x (250 m + 150 m)
Perimeter of rectangular field
= 2 x (400 m)
= 800 m.
Anuradha runs around this field 3 times.
Total distance she run
= 800 x 3
= 2400 m
Total distance she run is 2.4 km.
To run 4 km she should run around the field,
We are dividing 4 km by 800 m
4 km = 4000 m
She should run around the field
= 4000 /800
= 5 rounds.
Therefore, She should run around the field 5 times.
Alternative Solution,
Perimeter,
= 2 (150 + 250)
= 2 × 400 m
= 800 m
Anuradha runs,
= 800 × 3 m
= 2400 m
∴ She should run,
= 4 km/800 m
= 4000 m/800 m
= 5
∴ She run 2400 m and she should run 5 times around the field.
Question no – (31)
Solution :
Bajinder runs ten times around a square track and covers 4 km.
Distance he travel in 1 round of square track =
We are dividing 4 km by 10
Distance he travel in 1 round of square track
= 4000 m / 10
Distance he travel in 1 round of square track = 400 m.
Distance he travel in 1 round of square track is nothing but the perimeter of square track.
Perimeter of square track = 400 m.
Perimeter of square track = Sum of measure of four sides.
Perimeter of square track = 4 x length
Length of square track = 400/4
Length of square track = 100 m.
Alternative Solution,
The length of the track,
= 4 km/10
= 400 m/10
= 400 m
∴ The length of the track 400 m.
Question no – (32)
Solution :
The lawn in front of Molly’s house is 12m × 8m.
Perimeter of rectangular lawn in front of Molly’s house
= 2(length + breadth)
Perimeter of rectangular lawn in front of Molly’s house
= 2 x (12 m + 8 m)
= 2 x 20
Perimeter of rectangular lawn in front of Molly’s house = 40 m.
The lawn in front of Dolly’s house is 15m × 5m.
Perimeter of rectangular lawn in front of Dolly’s house = 2(length + breadth)
Perimeter of rectangular lawn in front of Dolly’s house
= 2 x (15 m + 5 m)
= 2 x 20
Perimeter of rectangular lawn in front of Dolly’s house = 40 m.
A bamboo fencing is built around both the lawns. Fencing is required for both =
Fencing is required for both = Sum of perimeter of rectangular lawn (Molly’s house
+ Dolly’s house)
Fencing is required for both
= 40 + 40
= 80 m
Therefore, Fencing is required for both is 80 m.
Alternative Solution,
Fence in front of Dolly’s house,
= 2 (12 + 8)
= 2 × 20
= 40 m
Fence in front Dolly’s house,
= 2 × (15 + 5)
= 2 × 20 m
= 40 m
∴ (40 + 40) = 80 m fencing is required for both.
Question no – (33)
Solution :
Regular pentagon has 5 sides.
The perimeter of a regular pentagon = 5 x Length of side
The perimeter of a regular pentagon is 1540 cm.
1540 cm = 5 x Length of side
Length of side regular pentagon
= 1540 cm/5
= 308 cm.
Length of side regular pentagon is 308 cm.
Alternative Solution,
The side of pentagon,
= 1540 cm/5
= 308 cm
Thus, its each side is 308 cm long.
Question no – (34)
Solution :
The perimeter of a triangle is 28 cm.
Isosceles triangles is the triangle having two sides are equal in length.
The perimeter of Isosceles triangle = Sum of all its 3 sides.
One of its sides is 8 cm.
Other side is also 8 cm.
The perimeter of Isosceles triangle = 8 + 8 + Third side
28 cm = 16 cm + third side
Third side,
= 28 – 16
= 12 cm
Sides of Isosceles triangles is 8 cm, 8 cm and 10 cm.
Also, One of its sides is 8 cm.
We subtract this side from the perimeter of Isosceles triangle.
= 28 – 8 = 20 cm
Isosceles triangles is the triangle having two sides are equal in length.
Length of equal sides
= 20 / 2
= 10 cm.
Sides of Isosceles triangles is 8 cm, 10 cm and 10 cm.
Alternative Solution,
The possible sides of isosceles triangles are,
10 cm × 10 cm × 8 cm;
8 cm × 8 cm × 12 cm
Question no – (35)
Solution :
The length of an Aluminium strip is 40 cm.
We have given the perimeter of rectangular frames.
Perimeter of rectangular frames = 2(length + breadth)
The measurement of all the possible rectangular frames which can be made out of it =
1 cm × 19 cm = 2 (length + breadth)
= 2 x (19 + 1)
= 40 cm
2 cm × 18 cm = 2 (length + breadth)
= 2 x (18 + 2)
= 40 cm
3 cm × 17 cm = 2 (length + breadth)
= 2 x (17 + 3)
= 40 cm
4 cm × 16 cm = 2 (length + breadth)
= 2 x (16 + 4)
= 40 cm
5 cm × 15 cm = 2 (length + breadth)
= 2 x (15 + 5)
= 40 cm
6 cm × 14 cm = 2 (length + breadth)
= 2 x (14 + 6)
= 40 cm
7 cm × 13 cm = 2 (length + breadth)
= 2 x (13 + 7)
= 40 cm
8 cm × 12 cm = 2 (length + breadth)
= 2 x (12 + 8)
= 40 cm
9 cm × 11 cm = 2 (length + breadth)
= 2 x (11 + 9)
= 40 cm
10 cm × 10 cm = 2 (length + breadth)
= 2 x (10 + 10)
= 40 cm
Alternative Solution,
The measurement for the possible rectangular frames is
1 cm × 19 cm, 2 cm × 18 cm
3 cm × 17 cm, 4 cm, × 16 cm, 5 cm × 15 cm, 6 cm × 14 cm,
7 cm × 13 cm, 8 cm × 12 cm, 9 cm × 11 cm, 10 cm × 10 cm
Question no – (36)
Solution :
Base of a tent is a regular hexagon of perimeter 60 cm.
Regular hexagon has 6 sides.
Perimeter of Regular hexagon = 6 x Length of side
Perimeter 60 cm is given.
60 cm = 6 x Length of side
Length of side of Regular hexagon
= 60/6
= 10 cm
Hence, Length of side of Regular hexagon is 10 cm.
Alternative Solution,
The length of each side of hexagon is,
= 60 cm/60
= 10 cm
∴ The length of each side of the base is 10 cm.
Question no – (37)
Solution :
There are 24 display boards each of length 1 m 50 cm and breadth 1 m.
There is a 100 m long aluminium strip, which is used to frame these boards.
We have to find the perimeter of each display board.
Display board is of rectangular shape.
Length of rectangular display boards 1.5 m and breadth is 1 m.
Length of rectangular display board = 1.5 m.
Breadth of rectangular display board= 1 m.
Perimeter of a rectangular display board = 2(length + breadth)
Perimeter of rectangular display board
= 2 x (1.5 m + 1 m)
Perimeter of rectangular display board
= 2 x (2.5 m)
= 5 m.
There are 24 display boards
= 5 x 24
= 120 cm
Perimeter of rectangular 24 display board = 120 cm
There is a 100 m long aluminium strip, which is used to frame these boards.
Perimeter of rectangular display board
= 2 x (2.5 m)
= 5 m.
Boards will be framed using this strip = we divide aluminium strip by 5 m
Boards will be framed using this strip
= 100 / 5
= 20 Boards.
20 Boards Are framed using 100 m long aluminium strip.
Remaining 4 board required
= 5 x 4
= 20 m Aluminium strip.
Length of the aluminium strip required for the remaining boards is 20 m.
Alternative Solution :
The perimeter of boards
= 2 (1.5 + 1) m
= 2 × 2.5 m
= 5 m
∴ 100m/5m = 20 boards will be framed with this strip.
Remaining boards = 4
Therefore 4 × 5 = 20 m strip required for the remaining board
= 20, 20 m.
Question no – (38)
Solution :
We have to find area of the rectangular display boards.
We know, Area of rectangular display boards = length x breadth
Length of rectangular display boards =1.5 m.
Breadth of rectangular display boards = 1 m.
= 1.5 m x 1 m = 1.5 sq. m
Area of rectangular display boards is 1.5 sq. m.
There are 24 display boards
Area of 24 display boards = 24 x 1.5 = 36 sq. m.
Length in m of the cloth used, if its breadth is 120 cm
Area of 24 display boards = length x breadth
Breadth of rectangular display boards = 1.2 m
Area of 24 display boards = length x 1.2
Area of 24 display boards = 36 sq. m.
36 sq. m. = length x 1.2
Length = 36 / 1.2
Length = 30 m
Therefore, 30 m length of the cloth used, if its breadth is 120 cm
Alternative Solution,
The area of the boards,
= 150 cm × 100 cm
= 1500 sq.cm
= 1.5 sq.cm
∴ 1.5 × 24 = 360 sq.m cloth is required to cover the boards.
If it’s breadth 120 cm the length of the cloth will be,
= 36 m × 10/12
= 30 m
Question no – (39)
Solution :
Length of outer boundary of the park shown in Fig. 6.16
We have to find perimeter of the park = sum of all outer sides.
Perimeter of the park
= 200 + 300 + 80 + 300 + 200 + 260
= 1340 m
Perimeter of the park is 1340 m.
Total cost of fencing it at the rate of Rs. 20 per metre
We multiplying 1340 m by Rs.20
Total cost of fencing it at the rate of Rs. 20 per metre
= 1340 x 20
Total cost of fencing it at the rate of Rs. 20 per metre = Rs.2680
There is a rectangular flower bed in the center of the park.
Area of rectangular flower bed
We know, Area of rectangular flower bed = length x breadth
Length of rectangular flower bed =100 m.
Breadth of rectangular flower bed = 80 m.
= 100 m x 80 m = 8000 sq. m
Area of rectangular flower bed is 8000 sq. m.
The cost of manuring the flower bed at the rate of Rs 50 per square metre
We multiplying 8000 sq. m.by Rs.50
The cost of manuring the flower bed at the rate of Rs 50 per square metre
= 8000 sq. m. x Rs.50
= Rs. 4, 00,000
The cost of manuring the flower bed at the rate of Rs 50 per square metre is Rs.4, 00,000
Alternative Solution,
The length of outer boundary of the park,
= (200 m + 300 m + 8 + 300 m + 200 m + 260 m + 200 m)
= 1340 m
∴ The total cost for fencing is,
= (1340m × 20)
= Rs. 26800
The area of the flower bed,
= (80 × 100)
= 8000 sq meter
The cost of reaming the flower bed,
= 8000 × 50
= Rs. 40000
∴ 1340 m; Rs. 26800; and Rs. 40000.
Question no – (40)
Solution :
We have to find perimeter of the park = sum of all outer sides.
Perimeter of the park
= 150 + 100 + 120 + 180 + 270 + 280
= 1100 m
Perimeter of the park is 1100 m.
Total cost of fencing the park shown in Fig. 6.17 is Rs 55000
The cost of fencing per metre,
We have to divide total cost by 1100 m.
The cost of fencing per metre = Rs 55000 /1100 m.
The cost of fencing per metre = Rs.50
The cost of fencing per metre is Rs.50
Alternative Solution,
The perimeter,
= (150 + 100 + 120 + 180 + 270 + 280) m
= 1100 m
∴ The cost of the fencing per metre,
= 55000/1100 = 50
= Rs 50
Question no – (41)
Solution :
(a) What is the perimeter of the rectangle ABCD?
In rectangle ABCD,
Side AB is 6 unit.
Side BC is 10 unit.
Side CD is 6 unit.
Side DA is 10 unit.
The perimeter of the rectangle ABCD = sum of (AB + BC + CD + DA)
The perimeter of the rectangle ABCD = 6 + 10 + 6 + 10 = 32 units.
(b) What is the area of the rectangle ABCD?
Area of the rectangle ABCD =
We know,
Area of rectangle ABCD = length x breadth
Length of rectangle ABCD =10 units.
Breadth of rectangle ABCD = 6 units.
= 10 units x 6 units = 60 sq. units
Area of rectangle ABCD is 60 sq. units.
(c) Divide this rectangle into ten parts of equal area by shading squares. (Two parts of equal area are shown here)
10 Equal parts. Each part is 6 units.
(d) Find the perimeter of each part which you have divided. Are they all equal?
Perimeter of each part is 10 units.
Out of 10 parts perimeter of 8 part is same but perimeter of 2 parts in different from we divided.
Hence, they all not equal.
Alternative Solution,
(a) The perimeter of the rectangle ABCD is
= 2 × (10 + 6)
= 2 × 16 = 32 units
(b) The area of the rectangle
= (10 × 6) sq unit
= 60 sq unit
Question no – (42)
Solution :
Rectangular wall MNOP of a kitchen is covered with square tiles of 15 cm length.
Area of square tiles,
We know, Area of square tiles = length x length
Length of square tiles = 15 cm
= 15 cm x 15 cm
= 225 sq. cm
Area of square tiles = 225 sq. cm
There are 28 Square tiles on wall MNOP.
The area of the wall,
We multiply 28 with 225 sq. cm
The area of the wall = 28 x 225
The area of the wall = 6300 sq.cm
Alternative Solution,
Area of the wall,
= {(15 × 4) × (15 × 7)} sq.cm
= (60 × 105) sq.cm
= 6300 sq.cm
∴ The area of the wall is 6300 sq.cm.
Question no – (43)
Solution :
Length of a rectangular field is 6 times its breadth. If the length of the field is 120 cm
Breadth of a rectangular field = we divide length of the field by 6.
Breadth of a rectangular field = 120 cm / 6
Breadth of a rectangular field = 20 cm
Length of rectangular field = 120 cm.
Breadth of rectangular field = 20 cm.
Perimeter of a rectangular field
= 2(length + breadth)
Perimeter of rectangular field,
= 2 x (120 cm + 20 cm)
Perimeter of rectangular field,
= 2 x (140 cm)
= 280 cm.
Alternative Solution,
The breadth = length/6 cm
= 120/6 cm
= 20 cm
Length = 120 cm
∴ The perimeter of the field,
= 2 × (20 + 120) cm
= 2 × 140 cm
= 280 cm.
Question no – (44)
Solution :
Anmol has a chart paper of measure 90 cm × 40 cm.
We know, Area of rectangular chart paper of Anmol = length x breadth
Length of rectangular chart paper = 90 cm.
Breadth of rectangular chart paper = 40 cm.
= 90 cm x 40 cm = 3600 sq. cm
Area of rectangular chart paper of Anmol is 3600 sq. cm.
Abhishek has one which measures 50cm × 70cm.
We know, Area of rectangular chart paper of Abhishek= length x breadth
Length of rectangular chart paper =70 cm.
Breadth of rectangular chart paper = 50 cm.
= 70 cm x 50 cm
= 3500 sq. cm
Area of rectangular chart paper of Abhishek is 3500 sq. cm.
Therefore, Area of rectangular chart paper of Anmol will cover more area on the table by 100 sq.cm.
Alternative Solution,
Anmols chart paper’s area,
= (90 cm × 40 cm)
= 3600 sq cm
Abhishek’s chart papers area,
= (50 cm × 70 cm)
= 35000 sq cm
∴ Anmol’s chart paper covers more area.
Question no – (45)
Solution :
A rectangular path of 60 m length and 3 m width is covered by square tiles of side 25 cm
First we have to find area of rectangular path and then area of square tiles.
Area of rectangular path,
We know, Area of rectangular path = length x breadth
Length of rectangular path = 60 m.
Breadth of rectangular path = 3 m.
= 60 m x 3 m
= 180 sq. m
Area of rectangular path is 180 sq. m.
We know, 1 sq.m = 10,000 sq.cm.
180 sq. m. = 180 x 10,000 sq.cm.
We know, Area of square tiles = length x length
Length of square tiles = 25 cm
= 25 cm x 25 cm
= 625 sq. cm
Area of square tiles = 625 sq. cm
Total no. of Square tiles = 180 x 10,000 sq.cm. / 625 sq. cm
Total no. of Square tiles = 2880.
Now, Width of rectangular path = 3 m = 300 cm.
Side of square tiles = 25 cm
Tiles will there be in one row along its width
= 300 cm / 25 cm
= 12 Tiles.
12 Tiles there in one row along its width.
Total no. of Square tiles = 2880.
No. of rows
= 2880 / 12
= 240
No. of rows is 240.
Alternative Solution,
There will be,
= 6000/25
= 240 tiles in the row.
There will be,
= 300/25
= 12 rows.
∴ Total tiles used,
= (240 × 12)
= 2880 tiles used to make this path.
Question no – (46)
Solution :
Square slabs each with side 90 cm.
Area of Square slabs,
We know, Area of square slabs = length x length
Length of square slabs = 90 cm
= 90 cm x 90 cm
= 8100 sq. cm
Area of square slabs = 8100 sq. cm
We know, 1 sq.m = 10,000 sq.cm.
81 sq. m. = 81 x 10,000 sq.cm.
Square slabs required = 81 x 10,000 sq.cm. / 8100 sq. cm
Square slabs required = 1 x 100
Square slabs required = 100.
100 Square slabs are required.
Alternative Solution,
Area of the slap
= (90 × 90) sq.cm
= 8100 sq.cm
Area of the floor,
= 81 sqm = 810000 sq.cm
= 810000/8100
= 100
∴ 100 square slabs are needed.
Question no – (47)
Solution :
The length of a rectangular field is 8 m and breadth is 2m.
A square field has the same perimeter as this rectangular field.
First we find perimeter of rectangular field,
Length of rectangular field = 8 m.
Breadth of rectangular field = 2 m.
Perimeter of a rectangular field = 2(length + breadth)
Perimeter of rectangular field = 2 x (8 m + 2 m)
Perimeter of rectangular field
= 2 x (10 m)
= 20 m.
A square field has the same perimeter as this rectangular field.
Perimeter of square field = 20 m.
Perimeter of square field = 4 x length
Length of square field = 20 m / 4
Length of square field = 5 m.
We know, Area of rectangular field = length x breadth
Length of rectangular field = 8 m.
Breadth of rectangular field = 2 m.
= 8 m x 2 m
= 16 sq. m
Area of rectangular field is 16 sq. m.
Now,
We know, Area of square field = length x length
Length of square field = 5 m
= 5 m x 5 m
= 25 sq. m
Area of square field = 25 sq. m
Area of square field has greater area.
Alternative Solution,
Area of rectangular field,
= 8 m × 2 m
= 16 sq.cm
Perimeter of rectangular field,
= 2 × (8 + 2) m
= 20 m
Perimeter of square field = 20 m
= 4 × side = 20 m
Side = 5 m
∴ Area of a square field,
= (5 × 5) sqm
= 25 sqm
∴ Square filed has greater area.
Question no – (48)
Solution :
Parmindar walks around a square park once and covers 800 m.
This is perimeter of Square Park.
Perimeter of Square Park = 800 m.
Perimeter of square Park = 4 x length
Length of square Park = 800 / 4 = 200 m.
Length of square Park = 200 m.
We know, Area of square Park = length x length
Length of square Park = 200 m
= 200 m x 200 m
= 40,000 sq. m
Area of square Park = 40,000 sq. m
Alternative Solution,
Side = 4 × side = 800
= 200 m
∴ Area of the square park,
= (200)2 sq. m
= 40000 sq. m
∴ The area of this park is 40000 sq. m.
Question no – (49)
Solution :
The side of a square is 5 cm.
We know, Area of square = length x length
Length of square = 5 cm
= 5 cm x 5 cm
= 25 sq. cm
Area of square = 25 sq. cm
If the square is doubled.
The side of a square is 10 cm.
We know, Area of square = length x length
Length of square = 10 cm
= 10 cm x 10 cm
= 100 sq. cm
Area of square = 100 sq. cm
4 times area increase, if the side of the square is doubled.
Alternative Solution,
Side = 5 cm
∴ Area = (5 × 5) sq. cm
= 25 sq.cm
If the side = 5 × 2 = 10 cm
∴ Area = (10 × 10) sq. m
= 100 sq.cm
∴ Therefore the area increase 4 times if the side of the square is doubled.
Question no – (50)
Solution :
Amita wants to make rectangular cards measuring 8cm × 5cm.
She has a square chart paper of side 60cm.
We find area of rectangular cards =
We know, Area of rectangular cards = length x breadth
Length of rectangular cards =8 cm
Breadth of rectangular cards = 5 cm
= 8 cm x 5 cm = 40 sq. cm
Area of rectangular cards is 40 sq. cm.
Now, we find area of square chart paper =
We know, Area of square chart paper = length x length
Length of square chart paper = 60 cm
= 60 cm x 60 cm
= 3600 sq. cm
Area of square chart paper = 3600 sq. cm
Complete cards can she make from this chart =
Area of square chart paper / Area of rectangular cards
Complete cards can she make from this chart = 3600 / 40
Complete cards can she make from this chart = 90
Alternative Solution,
Complete cards she can make from this chart,
= (12 × 7)
= 84.
∴ Chart paper will be left,
= (3600 – (84 × 40)}
= {3600 – 3360} cm²
= 240 cm²
Question no – (51)
Solution :
Page of the magazine is 15 cm × 24 cm.
We find area of rectangular Page of the magazine,
We know, Area of rectangular Page of the magazine = length x breadth
Length of rectangular Page of the magazine = 24 cm
Breadth of rectangular Page of the magazine = 15 cm
= 24 cm x 15 cm
= 360 sq. cm
Area of rectangular Page of the magazine is 360 sq. cm.
A company decided to order a half page advertisement.
Area of advertisement
= 360 sq. cm./2
= 180 sq. cm.
A magazine charges Rs 300 per 10 sq.cm area for advertising.
Amount will the company has to pay,
We multiply Rs 300 with 18
Amount will the company has to pay = Rs.300 x 18
∴ Amount will the company has to pay = Rs.5400
Alternative Solution,
The area of each page,
= 18 cm × 24cm
= 360 sq.cm
The area of half page = 180 sq. cm
∴ Company has to pay,
= 300/10 × 180
= 5400
∴ Company has to pay 5400 Rupees for it.
Question no – (52)
Solution :
The perimeter of a square garden is 48 m.
The perimeter of a square garden = 4 x length
Length of square garden = 48 / 4 = 12 m.
Now, we find area of square =
We know, Area of square garden = length x length
Length of square garden = 12 m
= 12 m x 12 m
= 144 sq. m
Area of square garden = 144 sq. m
A small flower bed covers 18 sq.m area inside this garden.
Area of the garden that is not covered by the flower bed = 144 sq. m – 18 sq.m.
Area of the garden that is not covered by the flower bed = 126 sq. m
A small flower bed covers 18 sq.m area
Area of square garden = 144 sq. m
What fractional part of the garden is covered by flower bed
= 18 / 144
= 1 /8
1 /8 fractional part of the garden is covered by flower bed.
Ratio of the area covered by the flower bed and the remaining area =
A small flower bed covers 18 sq.m area
Area of the garden that is not covered by the flower bed = 126 sq. m
Ratio of the area covered by the flower bed and the remaining area
= 18 / 126
= 1/ 7
∴ Ratio of the area covered by the flower bed and the remaining area is 1/ 7.
Alternative Solution,
Side = 4 × side = 48
= 12 m
Area = (12 × 12) sq. cm
= 144 sq. m
∴ The area not coved with the flower had is,
= (144 – 18) sq. m
= 126 sq. m
∴ The fraction part of the garden is coved by flower bed,
= 18/144
= 1/8
∴ The ratio of the area coved by flower bed and the remaining are,
= 18 : 126
= 1 : 7
Question no – (53)
Solution :
Perimeter of a square and a rectangle is same.
Side of the square is 15 cm.
Perimeter of square = 4 x length
Length of square = 15 cm
Perimeter of square = 4 x 15 = 60 cm
Perimeter of a square and a rectangle is same.
Perimeter of Rectangle = 60 cm.
Let, Length of rectangle = 18 cm
Perimeter of a rectangle = 2(length + breadth)
Perimeter of rectangle = 2 x (18 cm +breadth)
60 / 2 = (18 cm + breadth).
30 = 18 + breadth
Breadth = 30 – 18
Breadth = 12 cm
Area of rectangle,
We know, Area of rectangle = length x breadth
Length of rectangle =18 cm
Breadth of rectangle = 12 cm
= 18 cm x 12 cm = 216 sq. cm
Area of rectangle is 216 sq. cm.
Alternative Solution,
Side of square = 15 cm
Perimeter = 15 × 4 cm
= 60 cm
= 2 (ones side + 18) = 60
= one side + 18 = 30
= one side = 30 – 18
= 12
∴ The area of the rectangle
= 12 × 18
= 216 sq. cm
∴ The area of the rectangle is 216 sq. cm.
Question no – (54)
Solution :
A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2 cm.
Total area of the small squares is 28 square cm.
First we find area of small square =
We know, Area of square = length x length
Length of square = 2 cm
= 2 cm x 2 cm
= 4 sq. cm
Area of square = 4 sq. cm
Total area of the small squares is 28 sq. cm
No. of small squares = 28 sq. cm / 4 sq. cm
No. of small squares = 7.
Perimeter of square =
Perimeter of square = 4 x length
Length of square = 2 cm
Perimeter of square
= 4 x 2
= 8 cm
Total 7 small squares.
Perimeter of all 7 squares
= 7 x 8
= 56 cm.
Original length of the wire is the perimeter of all 7 squares.
Original length of the wire is 56 cm.
Alternative Solution :
Area of small square,
= (2)2 sq. m
= 4 cm
∴ Number of small square,
= 28/4
= 7
∴ Perimeter of small square,
= 4 × 2
= 8 cm
∴ The original length of the square,
= 8 × 7 cm
= 56 cm
Question no – (55)
Solution :
The total are of this park,
= {(150 × 100) + (270 × 180)})
= 15000 + 48600
= 63600 sq. m
∴ Packets of fertilizer required,
= 63600/300
= 212
Therefore, 212 packets of fertilizer are required for the whole park.
Question no – (56)
Solution :
Area of rectangular,
= 1600 sq. m
= 80 × breadth = 1600 m
breadth = 20 m
∴ The perimeter of the field,
= 2 × (80 + 20) m
= 2 × 100 m
= 200 m
Therefore, the perimeter of the field is 200 m.
Question no – (57)
Solution :
The area of each square on a chess board is 4 sq.cm
There are total 64 square on a chess board.
Area of 64 square on a chess board = 64 x 4
Area of 64 square on a chess board = 256 sq.cm
(a) At the beginning of game when all the chess men are put on the board, write area of the squares left unoccupied.
At the beginning of game when all the chess men are put on the board,
Area of the squares left unoccupied is half of the total area of chess board.
Area of the squares left unoccupied
= 256 sq.cm / 2
= 128 sq.cm
Area of the squares left unoccupied = 128 sq.cm
(b) Find the area of the squares occupied by chess men.
Area of the squares occupied by chess men is half of the total area of chess board.
Area of the squares occupied by chess men
= 256 sq.cm / 2
= 128 sq.cm
Area of the squares occupied by chess men = 128 sq.cm
Alternative Solution,
There are 64 square on a chess board
∴ The area of chess board,
= 4 × 64
= 256 sq. cm
(a) (4 × 8 × 4)
= 128 sq. cm left unoccupied.
(b) (4 × 8 × 4)
= 128 sq. m occupied by chess men.
Question no – (58)
Solution :
(a) Rectangle with a perimeter 36 cm.
Perimeter of a rectangle = 2(length + breadth)
17 cm × 1 cm = 2(length + breadth)
= 2 x (17 + 1)
= 36 cm
∴ Area of rectangle = length x breadth
17 x 1= 17 sq.cm
16 cm× 2 cm =2(length + breadth)
= 2 x (16 + 2)
= 36 cm
∴ Area of rectangle = length x breadth
16 x 2= 32 sq.cm
∴ 15 cm × 3 cm = 2(length + breadth)
= 2 x (15 + 3)
= 36 cm
∴ Area of rectangle = length x breadth
15 x 3 = 45 sq.cm
14 cm × 4 cm = 2(length + breadth)
= 2 x (14 + 4)
= 36 cm
∴ Area of rectangle = length x breadth
14 x 4 = 56 sq.cm
13 cm × 5 cm = 2(length + breadth)
= 2 x (13 + 5)
= 36 cm
∴ Area of rectangle = length x breadth
= 13 x 5
= 65 sq.cm
Alternative Solution,
Given, Perimeter = 36
= 2 × Dimension
∴ Dimension,
= (17 × 1, 16 × 2, 15 × 3, 14 × 4, 13 × 5, 12 × 6, 11 × 7, 10 × 8, 9 × 9)
∴ Areas,
= 17, 32, 45, 56, 65, 72, 77, 80, 81
(b) Rectangle with an area of 36 sq.cm
Area of rectangle = length x breadth
Area of rectangle
= 36 × 1
= 36 sq.cm
Perimeter of a rectangle = 2(length + breadth)
= 2 x (36 + 1)
= 74 cm.
Area of rectangle
= 18 × 2
= 36 sq.cm
Perimeter of a rectangle = 2(length + breadth)
= 2 x (18 + 2)
= 40 cm.
Area of rectangle
= 12 × 3
= 36 sq.cm
Perimeter of a rectangle = 2(length + breadth)
= 2 x (12 + 3)
= 30 cm.
Area of rectangle
= 9 × 4
= 36 sq.cm
Perimeter of a rectangle = 2(length + breadth)
= 2 x (9 + 4)
= 26 cm.
Area of rectangle
= 6 × 6
= 36 sq.cm
Perimeter of a rectangle = 2(length + breadth)
= 2 x (6 + 6)
= 24 cm.
Alternative Solution,
Dimension of the rectangle are,
= (36 × 1), (18 × 2), (9 × 4), (6 × 6), (12 × 3)
Perimeter of the rectangle in cm2 are :
= 74, 40, 30, 26, 24
Question no – (59)
Solution :
Figure – (i)
There are total 11 squares.
Area of fig. = 11 sq.cm
Perimeter of fig. = sum of all exterior sides.
Perimeter of fig. = Total 18 sides
Perimeter of fig. = 18 cm.
Figure – (ii)
There are total 13 squares.
Area of fig. = 13 sq.cm
Perimeter of fig. = sum of all exterior sides.
Perimeter of fig. = Total 28 sides
Perimeter of fig. = 28 cm.
Figure – (iii)
There are total 13 squares.
Area of fig. = 13 sq.cm
Perimeter of fig. = sum of all exterior sides.
Perimeter of fig. = Total 28 sides
Perimeter of fig. = 28 cm.
Alternative Solution,
In figure – (1)
The perimeter = 18 cm
∴ Area = 11 × 1 sq. cm
= 11 sq. cm
In figure – (2)
The perimeter = 28 cm
∴ Area = 13 sq. cm
In figure – (3)
The perimeter = 28 cm
∴ Area = 13 cm2
Question no – (60)
Solution :
The area of entire figure is 96 sq.cm
In fig. Total 24 small squares.
Area of each small squares = the area of entire figure / Total 24 small squares
Area of each small squares =96 sq.cm / 24
Area of each small squares = 4 sq.cm
Perimeter of fig. = sum of all exterior sides.
Perimeter of fig. = Total 34 sides
Perimeter of fig. = 34 cm.
Alternative Solution,
Side of small square,
= 96/24 cm
= 4 cm
The perimeter of the figure,
= 34 × 2 cm
= 68 cm
Therefore, the perimeter of the given figure will be 68 cm.
Next Chapter Solution :
👉 Unit – 7 Algebra 👈
