NCERT Exemplar Class 6 Maths Mensuration Solution: NCERT Exemplar Solution Class 6 Maths Chapter 6 Mensuration Full Explanation. NCERT Exemplar Class 6 Maths – Chapter 6 Mensuration. NCERT Exemplar Class 6 Maths Mensuration Solution by Expert.
NCERT Exemplar Class 6 Maths Mensuration Solution
In questions 1 to 6, out of the four options only one is correct. Write the correct answer.
Question 1:
Following figures are formed by joining six unit squares. Which figure has the smallest perimeter in Fig. 6.4?
(A) (ii)
(B) (iii)
(C) (iv)
(D) (i)
Solution: Answer: (D)
This figure is formed by joining six-unit squares and this figure has the smallest perimeters in fig. 6.4
Question 2:
A square shaped park ABCD of side 100m has two equal rectangular flower beds each of size 10m × 5m (Fig. 6.5). Length of the boundary of the remaining park is
(A) 360m
(B) 400m
(C) 340m
(D) 460m
Solution:
Answer: (B)
Flower bed each of length 10cm and breadth is 5m.
Now, Length of the boundary of the remaining park
= 90+5+10+95+90+5+10+95
= 400 m
Question 3:
The side of a square is 10cm. How many times will the new perimeter become if the side of the square is doubled?
(A) 2 times
(B) 4 times
(C) 6 times
(D) 8 times
Solution: Answer: (A)
Side of square = 10cm, perimeter of square = 4 x side = 4 x 10cm = 40cm
Side of the square X after doubling the length = 20cm
Perimeters of new square = 4 x 20cm = 80cm
If the side of square double, thus the new perimeter will also be double.
Question 4:
Length and breadth of a rectangular sheet of paper are 20cm and 10cm, 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?
(A) Perimeter remains same but area changes.
(B) Area remains the same but perimeter changes.
(C) Both area and perimeter are changing.
(D) Both area and perimeter remain the same.
Solution: Answer: (A)
The length and breadth of a rectangular sheet of paper are 20cm and 10cm. If we cut the rectangular piece from the sheet ten the parameter remains same but area change.
Question 5:
Two regular Hexagons of perimeter 30cm each are joined as shown in Fig. 6.7. The perimeter of the new figure is
(A) 65cm
(B) 60cm
(C) 55cm
(D) 50cm
Solution: Answer: (D)
Perimeter of hexagonal = 30cm
Each side of hexagonal = 30cm = 5cm
When two hexagonal are joined each side of each hexagonal overlaps side of new figure is 10.
New perimeter = 10 × 5
=50cm
Question 6:
In Fig. 6.8 which of the following is a regular polygon? All have equal side except (i)
(A) (i)
(B) (ii)
(C) (iii)
(D) (iv)
Solution: Answer: (B)
Square is the regular polygon.
Question 7:
Match the shapes (each sides measures 2cm) in column I with the corresponding perimeters in column II:
Solution:
[A] -[IV] = 28cm, perimeter of following diagram is 2 X 14 = 28. Number of sides is 14.
[B] -[I] = 16cm, perimeter of octagon. Side length = 2cm, 2 X 8 = 16cm.
[C] -[II] = 20cm, perimeter of following diagram. = 2cm X 10 = 20cm.
[D] -[III] = 24cm, perimeter of following diagram. = 2cm X 12 = 24cm
This matches the column base on perimeter of each figure.
In questions 9 to 13, fill in the blanks to make the statements true.
Question 8:
Perimeter of the shaded portion in Fig. 6.9 is
AB + _ + _ + _ + _ + _ + _ + HA
Solution: BM + MD + DE + EN + NG + GH
Question 9:
The amount of region enclosed by a plane closed figure is called its _________.
Solution: The amount of region enclosed by a plane closed figure is called its Area.
Question 10:
Area of a rectangle with length 5cm and breadth 3cm is _________.
Solution: Area of a rectangle with length 5cm and breadth 3cm is 16sq cm.
Question 11:
A rectangle and a square have the same perimeter (Fig. 6.10).
(a) The area of the rectangle is ______.
(b) The area of the square is ______.
Solution:
(a) The area of the rectangle is 12sq units.
(b) The area of the square is 16sq units.
Question 12:
(a) 1m = _________ cm.
(b) 1sqcm = _________ cm × 1cm.
(c) 1sqm = 1m × _________ m = 100cm × _________ cm.
(d) 1sqm = _________ sq. cm.
Solution:
(a) 1m = 100 cm.
(b) 1sqcm = 1 cm × 1cm.
(c) 1sqm = 1m × 1 m = 100cm × 100 cm.
(d) 1sqm = 10000 sq. cm.
In questions 14 to 20, state which of the statements are true and which are false.
Question 13:
If length of a rectangle is halved and breadth is doubled then the area of the rectangle obtained remains same.
Solution: The statement is true.
Question 14:
Area of a square is doubled if the side of the square is doubled.
Solution: False
Area of the original square = a²
Area of new square is = 4 a² = = (2a)²
Question 15:
Perimeter of a regular octagon of side 6cm is 36cm.
Solution: False
The perimeter of octagon is = side × 8 = 6 × 8 = 48
Question 16:
A farmer who wants to fence his field, must find the perimeter of the field.
Solution: The statement is true.
Question 17:
An engineer who plans to build a compound wall on all sides of a house must find the area of the compound.
Solution: False.
An engineer who plans to build a compound wall on all sides of a house must find the parameter of the compounds.
Question 18:
To find the cost of painting a wall we need to find the perimeter of the wall.
Solution: False.
To find the cost of painting a wall we need not to find the perimeter of the wall.
Question 19:
To find the cost of a frame of a picture, we need to find the perimeter of the picture.
Solution: The statement is true.
Question 20:
Four regular hexagons are drawn so as to form the design as shown in Fig. 6.11. If the perimeter of the design is 28cm, find the length of each side of the hexagon.
Solution:
Perimeter of given fig. = 28cm
Length of all side of hexagon are same
14 × side = 28cm
Side = 2cm
Question 21:
Perimeter of an isosceles triangle is 50cm. If one of the two equal sides is 18cm, find the third side.
Solution:
Length of two side is = 18cm and 18cm
Mean = 36cm Perimeter = 50cm
And the thirds side of an isosceles triangle.
Perimeter = 18 + 18 + Thirds side
50 = 36 + Thirds side
Thirds side = 14cm
Question 22:
Length of a rectangle is three times its breadth. Perimeter of the rectangle is 40cm. Find its length and width.
Solution:
Perimeter of rectangle is 40cm
Breadth we assume y and length is 3y.
Perimeter of rectangle = 2 (l + b)
= 2 (4y)
40cm = 8y
Y = 5cm
Length is 5 × 3 = 15cm and breadth is 5cm.
Question 23:
There is a rectangular lawn 10m long and 4m wide in front of Meena’s house (Fig. 6.12). It is fenced along the two smaller sides and one longer side leaving a gap of 1m for the entrance. Find the length of fencing.
Solution:
Length = 10m, Width is 4m
Perimeter of fencing = Two smaller size + one longer side – Gap
= 4 + 4 + 10 – 1
= 18 – 1
= 17m
Hence the length of fencing is 17m.
Question 24:
The region given in Fig. 6.13 is measured by taking as a unit. What is the area of the region?
Solution: Area of the region is 13sq units.
Question 25:
Tahir measured the distance around a square field as 200 rods (lathi). Later he found that the length of this rod was 140cm. Find the side of this field in metres.
Solution:
Length of rod is 140cm.
Total distance = number of rod × length of rod
= 200 × 40
= 28000 cm
Perimeter of square field = 4 × Length of the side
28000/4 = Length of the side of square field.
= 7000 cm, 100cm = 1m
Length of side of field is 70m.
Question 26:
The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of 6km. What is the length of the field?
Solution: Jamal jogged around it 4 times and covered distance 6km.
Perimeter of the rectangle = (Coverd distance)/(number of times of jamal jogged)
= 6/k km = 1.5km
We assume length and breadth of rectangle is 2x an x respectively.
Perimeter = 2 (2x + x)
1.5 = 2(3x)
= 6x/1.5
X = 0.25
length of the field is = 2 × 0.25 = 0.5km
= 0.5 × 1000 m
= 500 m
Question 27:
Three squares are joined together as shown in Fig. 6.14. Their sides are 4cm, 10cm and 3cm. Find the perimeter of the figure.
Solution:
Three squares are joined it means all square od side are same. Perimeter of the fig.
6 + 4 + 4 + 4 + 10 + 10 + 3 + 3 + 3 + 7 = 54cm
Question 28:
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. What is the perimeter of the figure?
Solution:
All triangle is equilateral Hence side of all triangle are same AB = 8 units means AC = BC = 8 units.
Perimeter of the figure = 1+1+1+1+1+1+1+1+1+4+2+2+4+4+2+2+4+4+4+2+2+4
= 45 units.
Question 29:
Length of a rectangular field is 250m and width is 150m. Anuradha runs around this field 3 times. How far did she run? How many times she should run around the field to cover a distance of 4km?
Solution: Length and width of all the rectangular field is 250m and 150m.
Perimeter of the rectangle = 2 x (250 + 150)
= 800m
Anuradha runs around thus field 3 times
= 3 x 800 = 2400m, 1km = 1000m
2km 400m = 2.4km
Number of times she should run around the to cover the distance of 4km = 4000/800
= 5 Times.
Question 30:
Bajinder runs ten times around a square track and covers 4km. Find the length of the track.
Solution:
Total distance = number of round × perimeter of square track.
4 = 10 × perimeter of square track
Perimeter of square track = 4/10
= 0.4 Km
Perimeter of square = 4 × length of the track
Length = Perimeter/4
= 0.4/4
= 0.1km, 1km = 1000m
Length of the track is 100m.
Question 31:
The lawn in front of Molly’s house is 12m× 8m, whereas the lawn in front of Dolly’s house is 15m×5m. A bamboo fencing is built around both the lawns. How much fencing is required for both?
Solution:
Perimeter of rectangle/Lawn = 2 (Length + breadth)
Perimeter of the lawn in front of angle of dolly house.
= 2 (15+5) = 40m
Perimeter of lawn in front of molly’s house
= 2 (12 + 8) = 40m
40m + 40m = 80m is facing is required for booth.
Question 32:
The perimeter of a regular pentagon is 1540cm. How long is its each side?
Solution:
Perimeter of a pentagon = 5 × Length of side
Perimeter of a pentagon = 1540 cm
Length of the side =1540/5
308cm long is its each side.
Question 33:
The perimeter of a triangle is 28cm. One of it’s sides is 8cm. Write all the sides of the possible isosceles triangles with these measurements.
Solution:
There are two possible cases to finding the length of the side of isosceles triangle.
8cm it may be the length of other side.
Perimeter = 8 + 8 + (Length of third side)
28 – 16 = Third side
Length of thirds side = 12
8, 8, 12 is length of side of isosceles triangle.
We find the length of side in another way.
Perimeter = y + y + 8
28 = 8 + 2y
28 – 8 = 2y
20/2 = y
Y = 10.
10, 10, 8 is the length of the side of the isosceles triangle.
Question 34:
The length of an aluminium strip is 40cm. If the lengths in cm are measured in natural numbers, write the measurement of all the possible rectangular frames which can be made out of it. (For example, a rectangular frame with 15cm length and 5cm breadth can be made from this strip.)
Solution:
Perimeter of frame = 2 (l + b)
40 = 2 (l + b)
20 = (l + b)
(19, 1), (18, 2), (17, 3), (16, 4), (15, 5), (14, 6), (13, 7), (12, 8), (11, 9).
These are the measurement of all the possible rectangular frame.
Question 35:
Base of a tent is a regular hexagon of perimeter 60cm. What is the length of each side of the base?
Solution:
Pediments of regular hexagon = 6 × side
60 = 6 × side
Side = 60/6 = 10cm
Length of each side of the base is 10cm.
Question 36:
In an exhibition hall, there are 24 display boards each of length 1m 50cm and breadth 1m. There is a 100m long aluminium strip, which is used to frame these boards. How many boards will be framed using this strip? Find also the length of the aluminium strip required for the remaining boards.
Solution:
Perimeter of a board = 2(l + b)
= 2 × (1.5 + 1) m
= 5m
Number of boards that can be framed = (Length of strip)/perimeter = 100/5 = 20. 20 is the number of boards.
Length of strip is required for remaining boards
= (Total boards – number of boards that can be formed) × perimeter
= (24 – 20) × 5m
= 20m is the length of the aluminium strip required for remining boards.
Question 37:
In the above question, how many square metres of cloth is required to cover all the display boards? What will be the length in m of the cloth used, if its breadth is 120cm?
Solution:
Breadth of the cloth 120cm
Breadth of the cloth 120/100 m = 1.2m
Length of board and breadth is 1m 50cm = 1.5m and 1m
Area of all the display board = 24 × l × b
= 24 × 1.5 × 1m = 36m²
Area of the board = Length of the cloth × Breadth of the cloth
Length of the cloth = Area/Breadth = (36m²)/1.2m = 36m/1.2 = 30m
Question 38:
What is the length of outer boundary of the park shown in Fig. 6.16? 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 center of the park. Find the cost of manuring the flower bed at the rate of Rs 50 per square metre.
Solution:
Perimeter of the figure = 300+300+200+200+80+260
= 1340 m
Total cost = perimeter × rate
=1340 × 20
= Rs 26, 800
Area of flower bed which is rectangular in shape = l × b
= 100m × 80m
= 8000 m²
Cost for manufacturing flower bed
= Area of flower bed × Cost
= 8000 × 50
= 400000
Question 39:
Total cost of fencing the park shown in Fig. 6.17 is Rs 55000. Find the cost of fencing per metre.
Solution:
Perimeter of the figure
= 270 + 280 + 150 + 100 + 120 + 180 m
= 1100 m
Cost of facing = (Total cost of fencing)/(perimeter of the fig,)
= 55000/1100
= Rs 50
Rs 50 is the cost of fencing per meter.
Question 40:
In Fig. 6.18 each square is of unit length
(a) What is the perimeter of the rectangle ABCD?
(b) What is the area of the rectangle ABCD?
(c) Divide this rectangle into ten parts of equal area by shading squares. (Two parts of equal area are shown here)
(d) Find the perimeter of each part which you have divided. Are they all equal?
Solution:
Length = 10 units, breadth = 6 units
(a) Perimeter of rectangle
ABCD = 2 (l + b)
= 2 (10 + 6)
= 32 units
(b) Area of rectangle i.e. ABCD
= I + b = 10 × 6 = 60 Unit²
(c) This rectangle is divided into 10 equal parts of Unit²
Question 41:
Rectangular wall MNOP of a kitchen is covered with square tiles of 15 cm length (Fig. 6.19). Find the area of the wall.
Solution:
Rectangular wall MNOP of a kitchen is covered with square tiles of 15cm length
Number of squares along MN = 7
Number of squares along MP = 4
Length of rectangle = 7 × 15
= 105cm
Breadth of rectangle = 4 × 15
= 60cm
Area of rectangle = 105 × 60 cm²
= 6300 cm²
Is the area of the wall.
Question 42:
Length of a rectangular field is 6 times its breadth. If the length of the field is 120cm, find the breadth and perimeter of the field.
Solution:
We assume breadth of rectangle is x and he length is 6x and the length of rectangle is given
6x = 120cm
X = 20cm
Breadth = 20cm
Perimeter of rectangular field = 2 (l + b)
= 2 (120 + 20) cm
= 2 (140) cm
Perimeter = 280 cm
Question 43:
Anmol has a chart paper of measure 90cm × 40cm, whereas Abhishek has one which measures 50cm × 70cm. Which will cover more area on the table and by how much?
Solution:
Length = 50cm and breadth = 70cm
Area of Abhishek charts = l × b
= 50 × 70 cm²
= 3500 cm²
Length = 90cm and breadth = 40cm
Area of Anmol chart = l × b
= 90 × 40 cm²
= 3600 cm²
3600 cm² – 3500 cm² = 100 cm²
Difference in area = 100 cm²
Question 44:
A rectangular path of 60m length and 3m width is covered by square tiles of side 25cm. How many tiles will there be in one row along its width? How many such rows will be there? Find the number of tiles used to make this path?
Solution:
Area of path = l × b
= 60m × 3m
= 180m²
Area of rectangular path = 1800000 cm²
1 m² = 10000 cm²
Area of tile = 25 × 25 cm²
= 625 cm²
(Area of rectangular path )/(Area of tile) = (1800000 cm²)/(625 cm²)
= 2880 is the number of tiles which are required.
(Length of width)/(Length of side of square) = (8 × 100 cm)/25cm
= 12 is the number of tiles along the width of rectangle.
(Total tiles)/(Number of tiles in one row) = 2880/12
= 240 is the number of rows.
Question 45:
How many square slabs each with side 90cm are needed to cover a floor of area 81sqm.
Solution:
Area of square slab = 90 × 90 cm²
= 8100 cm²
Area of floor = 810000 cm²
1m² = 10000 cm²
= 810000/8100 = 100 is the number of square slabs is required.
Question 46:
The length of a rectangular field is 8m and breadth is 2m. If a square field has the same perimeter as this rectangular field, find which field has the greater area.
Solution:
Perimeter = 2 (l + b)
Of rectangular = 2 (8 + 2) m = 20m field
Area of rectangular field = l × b
= 8 × 2
= 16
Perimeter of square field = 4 × side
= 4x
We assume side = x
As square field has the same perimeter as its rectangular field
4x = 20
X = 20/4
X = 5
Area of square field = 5 × 5
= 25m²
Question 47:
Parmindar walks around a square park once and covers 800m. What will be the area of this park?
Solution:
Perimeter = 4 × side
Side = (Perimeter)/4 m
= 800/4 m
= 200m Perimeter 800m
Area of park = 200 × 200
= 40000 m²
Question 48:
The side of a square is 5cm. How many times does the area increase, if the side of the square is doubled?
Solution:
Side = 5 cm
Area of square = 5 × 5 cm²
= 25 cm²
New side = 10 cm
Area of new square = 10 × 10 cm²
= 100 cm²
Area of a square become 4 time if its side is double.
Question 49:
Amita wants to make rectangular cards measuring 8cm × 5cm. She has a square chart paper of side 60cm. How many complete cards can she make from this chart? What area of the chart paper will be left?
Solution:
Total number of cards = 7 × 12 = 84 complete cards she can make from this chart square charts of length = 60 cm
Area of sheet = 60 × 60 cm²
= 3600 cm²
Area of one card = 8 × 5 cm² = 40 cm²
Area of 84 rectangular cards = 84 × 40 cm²
= 3360 cm²
(3600 – 3360) cm² = 240 cm² is the area of charts will be left.
Question 50:
A magazine charges Rs 300 per 10sqcm area for advertising. A company decided to order a half page advertisement. If each page of the magazine is 15cm × 24cm, what amount will the company has to pay for it?
Solution:
(Length × Breadth)/2 = (15 × 24)/2 cm²
= (360)/2 cm²
= 180 cm² is the area for advertisement.
Magazine charge Rs30 per 10sqcm area for advertisement.
Rs (300)/10 = Rs 30
= 180 × 30 = Rs 5400 is the total cost for advertisement.
Question 51:
The perimeter of a square garden is 48m. A small flower bed covers 18sqm area inside this garden. What is the area of the garden that is not covered by the flower bed? What fractional part of the garden is covered by flower bed? Find the ratio of the area covered by the flower bed and the remaining area.
Solution:
Perimeter of square garden is 48m
Perimeter = 4 × side
48 = 4 × side
Side = 12m
Area of square garden = 12² m²
= 144 m²
A small floor bed = 18 m²
144 m² – 18 m² = 12 m² is the area of garden not covered by flower bed.
Question 52:
Perimeter of a square and a rectangle is same. If a side of the square is 15cm and one side of the rectangle is 18cm, find the area of the rectangle.
Solution:
Perimeter of the rectangle = 2 (18 + x)
Length = 18cm and
Breadth we assume to be x cm
Perimeter of square = 4 × 15 cm
Length of side of square = 15cm
As, perimeter of square and rectangle is same
60 = 2 (18 + x)
30 = 18 + x
X = 12 cm is the area of rectangle.
Question 53:
A wire is cut into several small pieces. Each of the small pieces is bent into a square of side 2cm. If the total area of the small squares is 28 square cm, what was the original length of the wire?
Solution:
Length of side = 2 cm
Area of square = 2 × 2 cm²
= 4 cm²
Perimeter of square = 4 × 2 cm
= 8 cm
Number of square × perimeter = 7 × 8 cm
= 56cm
56cm is the original length of the wire
Question 54:
Divide the park shown in Fig. 6.17 of question 40 into two rectangles. Find the total area of this park. If one packet of fertilizer is used for 300sqm, how many packets of fertilizer are required for the whole park?
Solution:
Area of ABCD= l × b
= 150 × 100 m²
= 15000 m²
Area of AGFE = l × b
= 270 × 180 m²
= 48600 m²
Total area of the park = 15000 + 48600 m²
= 63600 m²
(63600)/300 = 212 is the number of packets of the fertilizer used as the number of packets of the fertilizer used for 300 m² = 1.
Question 55:
The area of a rectangular field is 1600sqm. If the length of the field is 80m, find the perimeter of the field.
Solution:
Length of the field = 80m
Area of the rectangle filed = 1600 m²
Area of the rectangle = l × b
1600m² = 80m × b
Breadth = b = 20m
Perimeter of the rectangle = 2(l + b)
= 2 (20 + 80)
Perimeter = 200 m
Question 56:
The area of each square on a chess board is 4sqcm. Find the area of the board.
(a) At the beginning of game when all the chess men are put on the board, write area of the squares left unoccupied.
(b) Find the area of the squares occupied by chess men.
Solution:
The area of each square on chessboard is 4 cm²
Area of the chessboard = 64 × 4 cm²
= 256 cm²
There are 64 squares on chessboard.
(a) At the beginning of the game when all the chess men are put on the board.
= 32 × area of square
= 32 × 4 cm² is the area of the square left unoccupied.
(b) 32 × area of square
= 128 cm² is the area of the square occupied by chess men.
Question 57:
(a) Find all the possible dimensions (in natural numbers) of a rectangle with a perimeter 36cm and find their areas.
(b) Find all the possible dimensions (in natural numbers) of a rectangle with an area of 36sqcm, and find their perimeters.
Solution:
Perimeter of the rectangle = 36cm
Perimeter = 2 (l + b)
36 = 2 (l + b)
l + b = 18 cm
(17, 1), (16, 2) , (15, 3), (14, 4), (13, 5), (12, 6), (11, 7), (10, 8) and (19, 9)
These all-possible combination of length and breadth.
Area of the rectangle = 17 × 1 = 17
Area = 16 × 2 = 32
(b) Area of the rectangle = 36 cm² and perimeter of rectangle = 2 (l + b)
Question 58:
Find the area and Perimeter of each of the following figures, if area of each small square is 1sqcm.
Solution:
Fig (i) Area of the figure = 11 × 1 cm²
= 11 cm²
There are ii square and area of each square is 1 cm²
Perimeter – 18 × 1 cm = 18cm
Fig (ii) Area of the fig = 13 × 1 cm² = 13 cm²
There are 13 square each square is 1sqcm
Perimeter = 28 × 1 cm = 28cm
(iii) There are 13 square and area of each square 1sqcm
Area = (13 × 1) cm² = 13 cm²
Perimeter = 28 × 1 cm = 28 cm
Question 59:
What is the area of each small square in the Fig. 6.21 if the area of entire figure is 96sqcm. Find the perimeter of the figure.
Solution:
Area of the fig = 96 cm²
There are 24 number of squares
(96)/24 cm² = 4 cm² is the area of one small square
Area of square = 4 cm²
Side = √4 cm²
Side = 2 cm
Number of outside edges = 34
Perimeter = 34 × 2cm = 68 cm
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