NCERT Exemplar Solutions: Mensuration | Mathematics (Maths) Class 6 PDF Download

Figure (i) has the smallest perimeter.
Perimeter of a closed figure is the distance covered in one round along the boundary of the figure.
We know that, perimeter of figure = numbers of sides × length of each sides
Consider the given figures,
(i) figure (i) has 10 units and each side have length 1 cm
So, perimeter = 10 × 1
= 10 cm
(ii) figure (ii) has 12 units and each side have length 1 cm
So, perimeter = 12 × 1
= 12 cm
(iii) figure (iii) has 14 sides and each side have length 1 cm
So, perimeter = 14 × 1
= 14 cm
(iv) figure (iv) has 14 sides and each side have length 1 cm
So, perimeter = 14 × 1
= 14 cm
By comparing all the perimeters of given figures, figure (i) has smallest perimeter.

Ans: (b)
We know that, Length of the boundary = Perimeter of boundary
Given, two rectangular flower beds each of size 10m × 5m
So, perimeter of remaining park = 90 + 5 + 10 + 95 + 90 + 5 + 10 + 95
= 400 m

Ans: (a)
As we know that, perimeter of square = side × 4
= 10 × 4
= 40 cm
Given, side of square = 10 cm
The side of the square is doubled = 10 + 10
= 20 cm
Therefore, the new perimeter becomes 2 times the side of the square is doubled.

Ans: (a)
(A) Perimeter remains same but area changes.
We know that, area of big rectangle = length × breadth
= 10 × 20
= 200 cm²
Area of small rectangle = 5 × 2
= 10 cm²
Perimeter of rectangle = 2(length + breadth)
= 2(20 + 10)
= 2 × 30
= 60 cm
Then, perimeter of new figure = 20 + 8 + 5 + 2 + 15 + 10
= 60
Area of new figure = Area of big rectangle – Area of new figure
= 200 – 10
= 190 cm²
By comparing all the results, Perimeter remains same but area changes.

Ans: (d)
From the question it is given that, perimeter of each hexagon = 30 cm
We know hat, hexagon has 6 sides.
So, length of each side of hexagon = 30/6
= 5 cm
Now, consider two hexagons are joined together.
Then, perimeter of new figure = number of sides × length of each side
= 10 × 5
= 50 cm

Ans: (b)
A closed figure in which all sides and angles are equal is called a regular polygon.

Given, each side measures 2cm.
Figure (A) has 14 sides.
Perimeter of figure = number of sides × length of each side
= 14 × 2
= 28 cm
Figure (B) has 8 sides.
Perimeter of figure = number of side × length of each side
= 8 × 2
= 16 cm
Figure (C) has 10 sides.
Perimeter of figure = number of sides × length of each side
= 10 × 2
= 20 cm
Figure (D) has 12 sides.
Perimeter of figure = number of sides × length of each side
= 12 × 2
= 24 cm

(A) Perimeter of a rectangle = 2 × (length + breadth)
= 2 × (6 + 4)
= 2 × 10
= 20
(B) Perimeter of a square = 4 × length of its side
= 4 × 5
= 20
(C) Perimeter of an equilateral triangle = 3 × length of a side
= 3 × 6
= 18
(D) perimeter of isosceles triangle = sum of length of all sides
= 4 + 4 + 2
= 10

We know that perimeter of a figure = sum of length of all sides
= AB + BM + MD + DE + EN + NG + GH + HA

The amount of region enclosed by a plane closed figure is called its area.

Area of a rectangle with length 5cm and breadth 3cm is 15 cm.
Area of rectangle = length × breadth
= 5 cm × 3 cm
= 15 cm

The area of the rectangle is 12 sq. units.
From figure, length of rectangle is 6 and breadth is 2.
Then, area of rectangle = length × breadth
= 6 × 2
= 12 sq. units
Perimeter of rectangle = 2 × (length + breadth)
= 2 × (6 + 2)
= 2 × 8
= 16 units

The area of the square is 16 sq. Units.
From the question it is given that, rectangle and a square have the same perimeter.
So, length of each side of a square = 16/4
= 4 units.
Area of square = side × side
= 4 × 4
= 16 sq. Units

1 m = 100 cm.

1 sqcm = 1 cm × 1cm.

1 sqm = 1 m × 1 m
= 100 cm × 100 cm.

1 sqm = 10000 sqcm.
We know that, 1 m = 100 cm
Then, 1 sqm = 100 × 100
= 10000 sqcm

True.

We know that, area of rectangle = length × breadth
As per the condition given in the question, length of a rectangle is halved = l/2
Breadth is doubled = 2b
Then, new area = l/2 × 2b
= l × b.
Therefore, if length of a rectangle is halved and breadth is doubled then the area of the rectangle obtained remains same.

False

We know that, area of square = side × side
As per the condition given in the question, side of the square is doubled = 2 × side
Then, new area = 2side × 2side
= 4 side²
Therefore, if the side of the square is doubled then the area of the square obtained is 4 times the old area.

False.

Perimeter of regular octagon = number of sides × length of reach sides
= 8 × 6
= 48 cm

True.

False.

An engineer who plans to build a compound wall on all sides of a house must find the perimeter of the compound.

False.

To find the cost of painting a wall we need to find the area of the wall.

True.

From the question it is given that, the perimeter of the design is 28cm.
Given design has 14 sides, so length of each side = 28/14 = 2 cm
In the given figure, 4 regular hexagons are joined.
Therefore, the length of each side of the hexagon = 2 cm

From the question it is given that,
Perimeter of rectangle = 40 cm
Length of a rectangle is three times its breadth = 3b
We know that, perimeter of rectangle = 2 × (length + breadth)
40 = 2 × (3b + b)
40 = 2 × 4b
40 = 8b
40/8 = b
b = 5 cm
Therefore, width of rectangle = 5 cm
Length of rectangle = 3b
= 3 × 5
= 15 cm

From the question it is given that,
Length of rectangular lawn = 10m
Width of rectangular lawn = 4m
Then,
Perimeter of fencing = length of fencing
Perimeter of fencing = two smaller sides + one longer side – gap in the longer side
= 4 + 4 + 10 – 1
= 18 – 1
= 17 m
Therefore, length of fencing is 17 m.

From the question it is given that, a region measured by taking rectangle is as one unit.
Given figure contain 13 rectangles.
Therefore, area of the region = 13 × 1
= 13 sq. Units

From the question it is given that,
Tahir measured the distance around a square field as 200 rods
Length of the rod = 140 cm
So, total distance of square field = 200 × 140
= 28000 cm
Then, length of one side of square field = 28000/4
= 7000 cm
We know that, 1 m = 100 cm
Therefore, side of field = 7000/100
= 70 m

From the question it is given that,
The length of a rectangular field is twice its breadth = 2b
Perimeter of rectangular field = 6km/4
= 1.5 km
We know that, perimeter of rectangle = 2 × (length + breadth)
1.5 = 2 × (2b + b)
1.5 = 2 × 3b
1.5 = 6b
b = 1.5/6
b = 0.25 km
length of rectangle field = 2 × 0.25 = 0.5km
= 0.5 × 1000
= 500 m

We know that, perimeter of figure = sum of all sides
= AB + BC + CD + DE + EF + FG + GH + HI + JI +JA
= 4 + 4 + (10 – 4) + 10 + (10 – 3) + 3 + 3 + 3 + 10 + 4
= 4 + 4 + 6 + 10 + 7 + 3 + 3 + 3 + 10 + 4
= 54 cm

45 units

Perimeter of given figure = sum of length of each side
= BN + NM + ML + LK + KJ + JI + IA + AH + HV + VG + GF + FE + ED + DC + UC + UT + TS + SR + RQ + PQ + BP
= 4 + 2 + 1 + 1 + 1 + 2 + 4 + 4 + 2 + 1 + 1 + 1 + 2 + 4 + 4 + 2 + 1 + 1 + 1 + 2 + 4
= 45 units

From the question it is given that,
Length of a rectangular field is 250 m
Width of rectangular field = 150 m
Perimeter of rectangle = 2 × (length + breadth)
= 2 × (250 + 150)
= 2 × 400
= 800 m
Given that, Anuradha runs around this field 3 times = 3 × 800
= 2400 m
We know that, 1 km = 1000 m
So, 2400/1000
= 2.4 km i.e. 2 km 400 m
Number of times Anuradha should run around the field to cover a distance of 4 km = 4000/800
= 5 times
Therefore, 5 times Anuradha should run around the field to cover a distance of 4 km.

Let a be the side of the square track.
Perimeter of the square track = 4a
Bajinder runs ten times around the square track and covers 4 km, i.e.,
10 × 4a = 4 km
40a = 4 × 1000 m [∵ 1 km = 1000 m]
a = (4 × 1000) / 40 m = 100 m
Length of the track = 4a = 4 × 100 m = 400 m.

Given, size of lawn in front of Molly’s house = 12 m × 8 m
Size of lawn in front of Dolly’s house = 15 m × 5 m
Now, fencing required for both = 2(12 + 8) + 2(15 + 5)
= 40 + 40 = 80m.

Given, perimeter of pentagon = 1540 cm
Sides in pentagon = 5
5 × side = 1540
Now, length of each side = 1540 / 5 = 308 cm.

Given, Perimeter of a triangle = 28 cm
One of its sides = 8 cm
Let two equal sides of the isosceles triangle be a.
Now we have, a + a + 8 = 28
2a + 8 = 28
2a = 20
a = 10 cm
All sides of the isosceles triangle are 8 cm, 10 cm, and 10 cm.

Given length of aluminium strip = 40 cm
This implies, we have the perimeter of the rectangular frame = 40 cm
Now 2(l + b) = 40
l + b = 20
From this equation, we can suggest values of l and b as 19×1 cm, 18×2 cm, 16×4 cm, and many more.

Given perimeter of hexagon = 60 cm
No. of sides in hexagon = 6
Now, length of each side of hexagon = 60 / 6 = 10 cm.

Given, Total number of display boards = 24
Length of strip = 100 m
Length of each display board = 1.5 m
Breadth of one display board = 1 m
Perimeter of one display board = 2(l + b) = 2 (1.5 + 1) = 5 m
Now, number of boards which will be framed = 100 / 5 = 20
Thus, number of boards remained unframed = 24 – 20 = 4
Length of strip required for remaining boards = 4 × 5 = 20 m.

As given, Length of display board = 1m 50 cm = 1.5 m
Breadth of display board = 1m
Area of 24 display boards = 24(l × b) = 24(1.5 × 1) = 24 × 1.5 = 36 sqm
Therefore, 36 sqm of cloth is required to cover all the display boards.
Now, breadth of cloth = 120 cm = 1.2m
Length of cloth = 36 / 1.2 = 30 m.

The length of the outer boundary of the park = (200 + 300 + 80 + 300 + 200 + 260) m = 1340 m
Cost of fencing the park at the rate of Rs. 20 per metre = Rs. (20 × 1340) = Rs 26800
Area of the flower bed = (100 × 80) = 8000 sqm
Cost of manuring the flower bed at the rate of Rs. 50 per square metre = Rs. (50 × 8000) = Rs. 400000.

Given, perimeter of the given figure = 150 + 100 + 120 + 180 + 270 + 280 = 1100m
Total cost of fencing = Rs. 55000
Cost of fencing per metre = 55000 / 1100 = Rs. 50.

(a) Perimeter of rectangle ABCD = 2(10 + 6) = 120 units.
(b) Area of rectangle ABCD = 10 × 6 = 60 square units.
(c) When the rectangle is divided into 10 equal parts, the area of each part is 6 square units.
(d) Perimeter of each part = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12 units.
And the perimeter of all parts is equal.

Given, length of square tile = 15 cm
No. of tiles in given figure = 28
Now, area of each tile = 15 × 15 = 225 sq cm
This implies area of 28 tiles = 28 × 225 = 6300 square cm.
Thus, area of wall = 6300 square cm.

Given, Length of a rectangular field = 6 times its breadth.
This implies breadth of the field = Length / 6.
Length of the field = 120 cm.
Thus, breadth = 120 / 6 = 20 cm.
Now, perimeter of the field = 2(l + b) = 2(120 + 20) = 280 cm.

Area of chart paper of Anmol = 90 × 40 = 3600 square cm
Also, area of chart paper of Abhishek = 50 × 70 = 3500 square cm
Thus, chart paper of Anmol covers more area on the table by = 3600 – 3500 = 100 square cm.

Given, length and width of rectangular path = 60 m and 3 m
Also, side of square tile = 25 cm
Thus, no. of tiles in a row = 3 / 0.25 = 12
Now, no. of rows = 60 / 0.25 = 240
Hence, total no. of tiles = 12 × 240 = 2880.

Area of each square slab = 90 × 90 = 8100 square cm
Also, area of floor = 81 square m = 81 × 10000 = 810000 square cm
No. of square slabs required = 810000 / 8100 = 100.

Given, Perimeter of rectangular field = Perimeter of square field
This implies, 2(8 m + 2 m) = 4 × side
Side of square field = 20 / 4 = 5 m
Now, area of rectangular field = 8 × 2 = 16 m²
Area of square field = 5 × 5 = 25 m²
So, the area of the square field is greater than the area of the rectangular field.

Given, perimeter of the park = 800 m
Thus, if the side of the square is a, then 4 × a = 800
a = 200 m
Now, area of the park = 200 × 200 = 40000 m².

Given, side of square = 5 cm
Area of square = 5 × 5 = 25 cm²
Now, side of square becomes = 5 × 2 = 10 cm
Now, area of square = 10 × 10 = 100 cm²
Area increases by = 100 – 25 = 75 cm².

Given, area of the square chart = 60 × 60 = 3600 cm²
Now, area of each card = 8 × 5 = 40 cm²
This implies, No. of cards that can be made = 3600 / 40 = 90 cards.

Area of one page of magazine = 15 × 24 = 360 cm²
Thus, area of the half-page of the magazine = 360 / 2 = 180 cm²
Also given, cost of advertising for 10 sq cm = Rs. 300
This implies cost of advertising for 180 sq cm = (300 / 10) × 180 = Rs. 5400.

Given, perimeter of the square garden = 48 m
This implies, 4 × side = 48 m
Side = 48 / 4 = 12 m
Now, area of the square garden = 12 × 12 = 144 m²
Also, area of the small flower bed = 18 m²
So, the area of the garden that is not covered by the flower bed = 144 – 18 = 126 m².
Now, the fractional part of the garden covered by the flower bed = 18 / 144 = 1 / 8.
In addition, the ratio of the area covered by the flower bed and the remaining area = 18 / 126 = 1 / 7 = 1:7.

Given, Perimeter of a square = Perimeter of a rectangle
Also, side of square = 15 cm, side of rectangle = 18 cm
Thus, 4 × 15 = 2(18 + b)
b = 30 – 18 = 12 cm
Now, area of the rectangle = 18 × 12 = 216 cm².

Let the number of small pieces in which the wire is cut be x.
Also, given side of each square-shaped piece = 2 cm.
So, Area of each piece = 2 × 2 = 4 cm².
Given total area of the small squares = 28 cm².
Thus, 4x = 28, x = 7.
So, the number of pieces = 7.
It will give the perimeter of each piece with side 2 cm = 4 × 2 = 8 cm.
Therefore, the total length of the wire = 7 × 8 = 56 cm.

Total area of the park = 15000 + (270 x 180) = 15000 + 48600 = 63600 m²
Now one packet of fertilizer is used for area = 300 m²
Thus, total packets of fertilizer required = 63600 / 300 = 121

Given, area of a rectangular field = 1600 m²
And length of field = 80m
Now area of rectangular field is l x b
This implies, 80 x b = 1600
b = 1600 / 80 = 20 m
Now, perimeter of rectangular field = 2(l + b) = 2(80 + 20) = 200 m

Given, area of one square on a chess board = 4 sq cm
Now there are 64 squares on a chess board.
This implies area of chess board = 64 × 4 cm² = 256 cm²
(a) At the beginning of the game when all the chess men are put on the board, write the area of the squares left unoccupied.
At the beginning of the game, 32 squares are left unoccupied.
Thus, area of the squares left unoccupied = 32 x 4 cm² = 128 cm²
(b) Find the area of the squares occupied by chess men.
Area occupied by chess men = 32 x 4 cm² = 128 cm²

(a) Perimeter of the rectangle = 36 cm
2(length + breadth) = 36 cm
Length + breadth = 36/2 = 18 cm
All the possible dimensions (in natural numbers) of a rectangle with a perimeter of 36 cm are:
17 x 1 cm, 16 x 2 cm, 15 x 3 cm, 14 x 4 cm, 13 x 5 cm, 12 x 6 cm, 11 x 7 cm, 10 x 8 cm, 9 x 9 cm respectively.
Their corresponding areas are 17 cm², 32 cm², 45 cm², 56 cm², 65 cm², 72 cm², 77 cm², 80 cm², and 81 cm².
(b) The area of the rectangle = 36 sq cm 
length × breadth = 36 sq cm

(i) Area of the given figure = 11 x 1 cm² = 11 cm²
Perimeter of the given figure = 18 x 1 = 18 cm

(ii) Area of the given figure = 13 x 1 cm² = 13 cm²
Perimeter of the given figure = 28 x 1 = 28 cm

(iii) Area of the given figure = 13 x 1 cm² = 13 cm²
Perimeter of the given figure = 28 x 1 = 28 cm

Total number of small squares in the given figure = 24
Area of the total figure = 96 cm²
So, the area of each small square = 96 / 24 = 4 cm²
Now, the side of each small square = 2 cm
Perimeter of the figure = 34 x 2 = 68 cm