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 NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

NCERT class 7 maths chapter 10 mensuration topic 10.2 perimeter

Question:3 Meera went to a park 150 m long and 80 m wide. She took one complete round on its boundary. What is the distance covered by her?

Answer: The park is 150m long and 80 m wide. The parks are of the shape of the rectangle. The distance covered by Meera = perimeter of park = 2 (150+80)= 460 m

Question:4 Find the perimeter of the following figures:

Perimeter = AB + BC + CD + DA
= __+__+__+__
= ______

Perimeter = AB + BC + CD + DA
= __ + __ + __+ __
= ______

Perimeter = AB + BC + CD + DE + EF + FG + GH +HI + IJ + JK + KL + LA
= __ + __ +__ + __ + __ + __ + __ + __ +__+ __ + __ + __
= ______

Perimeter = AB + BC + CD + DE + EF+ FA
= __ + __ + __ + __ + __ + __
= ______

Answer: Find the perimeter of the following figures:

Perimeter = AB + BC + CD + DA
= 40+10+40+10
= 100

Perimeter = AB + BC + CD + DA
=5+5+5+5=20

Perimeter = AB + BC + CD + DE + EF + FG + GH +HI + IJ + JK + KL + LA
=1+3+3+1+3+3+1+3+3+1+3+3=28 cm

Perimeter = AB + BC + CD + DE + EF+ FA
=100 + 120 + 90 + 45+ 60 +80=495 cm

NCERT class 7 maths chapter 10 mensuration topic 10.2.1 perimeter of a rectangle

Question: Find the perimeter of the following rectangles:

Answer:

Length of rectangle

Breadth of rectangle

Perimeter by adding all sides

perimeter by 2\times(length + breadth)

25 cm

12 cm

=25+12+25+12=74 cm

\\2\times(25+12)\\2\times37=74\ cm

0.5 cm

0.25cm

=0.5+0.25+0.5+0.25=1.5cm

\\2\times(0.25+0.5)\\2\times0.75=1.5\ cm

18 cm

15 cm

=18+15+18+15=66 cm

\\2\times(18+15)\\2\times33=66\ cm

10.5 cm

8.5 cm

=10.5+8.5+10.5+8.5=38 cm

\\2\times(10.5+8.5)\\2\times19=38\ cm

NCERT class 6 maths chapter 10 mensuration exercise: 10.1

Question: 1 Find the perimeter of each of the following figures :

Answer: The perimeter of the following figures-
Perimeter = sum of all sides

Therefore,
(a) Perimeter = 4 cm + 2cm +1 cm +5cm = 12 cm

(b) Perimeter = 23 cm + 35 cm + 40cm + 35cm = 133 cm

(c) Required perimeter = 15cm + 15 cm + 15 cm +15 cm = 60 cm

(d) Required perimeter= (4 + 4 + 4 + 4 + 4) cm = 20 cm

(e) Perimeter = 4cm + 0.5 cm +2.5cm +2.5cm + 0.5cm +4cm +1cm = 15cm

(f) Perimeter = Sum of all sides = 4\times(4cm) + 4\times(1 cm) + 8\times(3 cm) + 4\times(2cm) = (16 + 4 + 24 + 8)cm =52 cm

Question: 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?

Answer: We have,
Length = 40 cm and,
Breadth = 10 cm

So, the length of tape required = perimeter of the rectangular box
= 2 (L + B)
= 2 \times 50 cm
= 100 cm

Hence 100 cm tape is required.

Question: 3 A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Answer: We have table-top measures 2 m 25 cm by 1 m 50 cm.
Therefore, the Perimeter of the table-top = sum of all sides
= 2 \times (L+B)
= 2 \times ((2m + 0.25m) + (1m + 0.5m))
= 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?

Answer: We have,
Length and breadth of a photograph is 32 cm and 21 cm respectively.

So, According to question,
Length of wooden strip required to frame the photograph = 2 [L + B]
= 2 [32 + 21] cm
= 2 \times 53 cm = 106 cm

Hence the required length of the strip is 106 cm or 1.06 m

Question: 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?

Answer: We have,
A rectangular piece of land measures 0.7 km by 0.5 km and side is to be fenced with 4 rows of wires.

As per the question,

Required length of wires
= 4 \times 2 [ 0.7 km + 0.5 km]
= 8 \times 1.2 km
= 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.

Answer: (a) The perimeter of a triangle = Sum of all the three sides
= 3cm + 4cm + 5cm
= 12 cm

(b) The perimeter of a eqilateral triangle
= Sum of all the sides of the triangle
= (9cm +9cm +9cm)
= 27cm

(c) Perimeter of a given isosceles triangle
= Sum of all sides
= 8cm + 8cm +6cm
= 22 cm

Question: 7 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Answer: We all know that,
Perimeter of a triangle = sum of all the sides of the triangle
=10cm + 14cm +15cm
= 39 cm

Question: 8 Find the perimeter of a regular hexagon with each side measuring 8 m.

Answer: It is known that,
A regular hexagon has six equal sides.
So, the perimeter of the regular hexagon
= 6 \times side
= 6 \times 8m = 48m

Question: 9 Find the side of the square whose perimeter is 20 m.

Answer: Given that,
The perimeter of the square is 20m
According to question,
The perimeter of square = 4 \times side
\therefore\ side = \frac{20}{4}=5\ m

Question: 10 The perimeter of a regular pentagon is 100 cm. How long is its each side?

Answer: Given,
The perimeter of a regular pentagon is 100cm and,
no. of sides in regular pentagon is five (5)
therefore, the length of each side
=\frac{100}{5}=20cm

Question: 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?

Answer: Given that,

Length of the string is 30 cm
(a)For square,
Number of sides = 4
Therefore, Length of one side = 30 /4 = 7.5 cm

(b)For equilateral triangle,
Number of sides = 3
Therefore, length of sides = 30/3 = 10cm

(c) For a regular hexagon,
Number of sides = 6
Therefore, Length of one side = 30/6 = 5cm

Question: 12 Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?

Answer: Given that,
The two sides of the triangle are 12 cm and 14 cm. let the third side be A.
And the perimeter of the triangle is 36cm.

Therefore, the Perimeter of the triangle
= sum of all sides of the triangle
= 12cm +14cm + A = 36cm
= A + 26cm =36cm
= A = 10cm

Question: 13 Find the cost of fencing a square park of side 250 m at the rate of rupees 20 per metre.

Answer: Given that,
Length of the square park = 250 m

Cost of fencing = rs. 20 per meter

Therefore, the Perimeter of the square park =
= 4 \times 250
=1000 m
= 1km

So, total cost = 20 \times 1000 = 20,000 Rs

Question: 14 Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of rupees 12 per metre.

Answer: Given that,
The length of the rectangular park is 15 m and the breadth is 125m.
Cost of fencing rate = Rs 12 per meter

Therefore,
Perimeter of the park
=2 [L + B]
= 2 [175 +125]
=2 \times 300m
=600m

So, the total cost of fencing = 600 \times 12 = Rs. 7,200

Question: 15 Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?

Answer: Given that,
Length of the square park = 75m
Length and breadth of the rectangular park is 60 m and 45m respectively

Therefore,
The perimeter of the square park
= 4 \times 75
=300 m

Also, the Perimeter of the rectangular park
= 2 \times ( 60 + 45)
=2 \times 105m
=210m

Hence Bulbul covers the less distance.

Question: 16 What is the perimeter of each of the following figures? What do you infer from the answers?

Answer: (a) The perimeter of the square
= Sum of all the four side = 4 \times side
= 4 \times 25
= 100 cm

(b)The perimeter of the rectangle
= SUm of all the sides = 2 \times [length + breadth )
= 2 \times (20 + 30)
= 2 \times 50
= 100

(c)The perimeter of the rectangle
= Sum of all the sides = 2 \times [length + breadth ]
= 2 \times (40 + 10)
= 2 \times 50
= 100 cm

(d)The perimeter of the triangle
= Sum of all three sides of the triangle
= 30cm + 30cm + 40cm
=100cm

We conclude all the given figures have the same perimeter.

Question:1 7 Avneet buys 9 square paving slabs, each with a side of \frac{1}{2}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?

Answer: Given that,
Length of each slab = 1/2 m
So, the length of the square = 3/2\ m

 


(a) The first figure is a square
So, the perimeter of the square = 4 \times side
4 \times 3/2 m = 6\ m

(b) The perimeter of the 2nd figure
= Sum of all the sides
(4\times \frac{1}{2} + 8 \times 1)
= 10 m

(c) Figure (ii) has a greater perimeter than fig (i)

(d) Arrange all the slab in horizontally, it forms a rectangle whose dimensions are \frac{9}{2}\ by\ \frac{1}{2}

NCERT class 6 maths chapter 10 mensuration exercise 10.2

Question: 1 Find the areas of the following figures by counting square:

Answer: Area of 1 square = 1sq. unit

(a) Number of total squares = 9
Therefore, the area of 9 squares = 9 sq. units

(b) In fig (b) the total number of square = 5
Therefore, are = 5 sq units

(c) In fig (c) the number of squares = 4
So, the area is 4 sq. units

(d) In fig (d) number of squares = 8
Therefore, area = 8 sq. units

(e) Total number of square = 10
therefore, area = 10 sq. units

(f) total number of squares = 4
Therefore, area = 4 sq. units

(g) Total number of square = 6
therefore, area = 6 sq. units

(h) Total number of square = 5
therefore, area = 5 sq. units

(i) Total number of square = 9
therefore, area = 9 sq. units

(j) Total number of square = 4
therefore, are = 4 sq. units

(k) Total number of square = 5
therefore, area = 5 sq. units

(l) Total number of full squares = 4
Squares more than half = 3
half squares = 2
Therefore, the total area = 8 sq. units

(m) Count more than half squares as 1
Now, the total number of squares = 14
Therefore, the total area = 14 sq. unit

(n) Count more than half squares as 1
Now, the total number of squares = 18
Therefore, the total area = 18 sq. unit

NCERT class 6 maths chapter 10 mensuration topic 10.3.1 area of a rectangle

Question:1 Find the area of the floor of your classroom.

Answer: If floor length is 3m and breadth is 2m then

floor \ area=length\times breadth=3\times2=6m^2

Question:2 Find the area of any one door in your house.

Answer: Suppose the length of door is 2m and breadth is 0.5 m

then the area of door = length\times breadth=2\times0.5=1m^2

NCERT class 6 maths chapter 10 mensuration 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

Answer: We know that area of a rectangle = length \times breadth
Therefore,
(a) Here length is 3cm and the breadth of the rectangle is 4cm
So, the area of the rectangle
3 \times 4 = 12\ cm^2
(b) Here length is 12m and the breadth of the rectangle is 21m
So, the area of the rectangle
12 \times 21 = 252\ m^2
(c) Here length is 3km and the breadth of the rectangle is 3km
So, the area of the rectangle ?
2 \times 3 = 6\ km^2
(d) Here length is 2m and the breadth of the rectangle is 70cm = 70/100 = 0.7 m
So, the area of the rectangle
2 \times 0.7= 1.4\ m^2

Question: 2 Find the areas of the squares whose sides are :

(a) 10 cm (b) 14 cm (c) 5 m

Answer: It is known that area of a square = (side)^2

(a) The length of a side of a square = 10cm
So, the area = (10)^2 =100cm^2

(b)The length of a side of a square = 14cm
So, the area = (14)^2 =196cm^2

(c) The length of a side of a square = 5m
So, the area = (5)^2 =25m^2

Question: 3 The length and 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?

Answer: We know that area of a rectangle = length \times breadth
Therefore,
(a) Here length is 9m and the breadth of the rectangle is 6m
So, the area of the rectangle
9\times 6 = 54m^2

(b) Here length is 17 m and the breadth of the rectangle is 3m
So, the area of the rectangle
17\times 3 = 51m^2

(c) Here length is 4m and the breadth of the rectangle is 14m
So, the area of the rectangle
4\times 14 = 56m^2

Hence rectangle C has the greatest area and rectangle B has the least area.

Question: 4 The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.

Answer: Given that,
Area of the rectangular garden = 300 sq. m
and the length = 50 m

Therefore, the Width of the garden = \frac{Area}{length}
300/50 = 6\ m

Hence the width of the garden is 6 m

Question: 5 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of rupees 8 per hundred sq m.?

Answer: Given that,
Length of the rectangular plot = 500m
and the width = 200 m
fencing cost = Rs 8 per hundred sq. meter

Therefore, the area of the plot
500\times 200 = 100,000 \ m^2

Now, Total rate of tiling of 100,1000 m 2 = \frac{8}{100}\times 100,000
= Rs 8000

Question: 6 A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?

Answer: Given that,
Length of the table-top = 2m
and it's breadth = 1.5 m

Therefore, the area of the tabletop
= length \times breadth
= 2 \times 1.5
=3\ m^2

Question: 7 A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?

Answer: Given that,
Length of the room = 4 m
And it's breadth = 3.5 m
Area of the room = = length \times breadth
= 4 \times 3.5
14 m^2

Hence the area of carpet needed is 14 m^2

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.

Answer: Given that,
Length of the floor = 5 m
and its width = 4 m
Therefore, the area of the floor
= length \times breadth
= 5\times 4 = 20\ m^2

And, side of the square carpet = 3m
Therefore, the area of the square carpet = (side)^2
= 3m \times 3m = 9m 2

Thus, the area of the floor which is not covered with carpet = 20 m^2-9m^2= 11m^2

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?

Answer: Given that,
Length and breadth of the land are 5m and 4 m.
Side of the square flower bed = 1m
Therefore the area of the bed = 1 m^2

So, Area of 5 square flower bed = 5 m^2

Area of the and = length \times breadth
5m\times 4m = 20 m^2

Thus, the area of the remaining part of the land = Area of land - area of the flower bed
= 20 m^2 - 5 m^2 = 15 m^2

Question:1 0 By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Answer: (a) Split the given figures into small rectangular parts as shown here;

Now, the area of the rectangle (I) = l\times b = 4 \times 3 = 12 cm^2
the area of the rectangle (II) = l\times b = 3 \times 2= 6 cm^2
the area of the rectangle (III) = l\times b = 4 \times 1 = 4 cm^2
the area of the rectangle (IV) = l\times b = 3 \times 2 = 6 cm^2

So, the total area of the figure = sum area of all rectangles
= (12 + 6 + 4 + 6) cm^2
= 28 cm^2

(b) Split the given figure into smalle rectangles as shown here;

the area of the rectangle (I) = l\times b = 3 \times 1 = 3 cm^2
the area of the rectangle (II) = l\times b = 3 \times 1= 3 cm^2
the area of the rectangle (III) = l\times b = 3 \times 1 = 3 cm^2 ?

Therefore, the total area of the figure = sum of the area of all the small rectangles
3+3+3
9cm^2

Question:11 Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Answer: (a) Split the given figure into a small rectangle, as shown here;

Now, the area of the rectangle (I) = l\times b = 12 \times 2 = 24cm^2
the area of the rectangle (II) = l\times b = 8 \times 2= 16cm^2

Therefore, Total area = 24cm^2 + 16cm^2 = 40cm^2

(b) Split the given figure into a small rectangle, as shown here;

Now, the area of the rectangle (I) = l\times b = 7 \times 7 = 49 cm^2
the area of the rectangle (II) = l\times b = 21 \times 7= 147 cm^2
Now, the area of the rectangle (III) = l\times b = 7 \times 7 = 49 cm^2

Therefore, Total area = 49cm^2 + 147cm^2 + 49cm^2 = 245 cm^2

(c) Split the given figure into a small rectangle, as shown here;

Now, the area of the rectangle (I) = l\times b = 2 \times 1 = 2 cm^2
the area of the rectangle (II) = l\times b = 5 \times 1= 5 cm^2
Now, the area of the rectangle (III) = l\times b = 2 \times 1 = 2 cm^2

Therefore, Total area = 2 cm^2 + 5cm^2 + 2cm^2 = 9cm^2

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.

Answer: Given that,
the length and breadth of the tiles are 12 cm and 5 cm

Therefore, the area of the tiles
= length \times breadth
= 12 cm \times 5 cm
= 60 cm 2

Now, (a) For 100 cm and 144 cm rectangular region,
Area of region = length \times breadth
= 14,400 cm 2

Therefore, the number of tiles needed = \frac{14,400}{60}=240 tiles

(b) For 70 cm and 36 cm rectangular region,
Area of region = length \times breadth
= 70 cm \times 36 cm
= 2520 m 2

Required number of tiles = \frac{2520}{60}=42

Chapter No. Chapter Name
Chapter 1 NCERT Solutions for class 6 maths chapter 1 Knowing Our Numbers
Chapter 2 NCERT solutions for class 6 maths chapter 2 Whole Numbers
Chapter 3 NCERT solutions for class 6 maths chapter 3 Playing with Numbers
Chapter 4 NCERT Solutions for class 6 maths chapter 4 Basic Geometrical Ideas
Chapter 5 NCERT solutions for class 6 maths chapter 5 Understanding Elementary Shapes
Chapter 6 NCERT solutions for class 6 maths chapter 6 Integers
Chapter 7 NCERT Solutions for class 6 maths chapter 7 Fractions
Chapter 8 NCERT solutions for class 6 maths chapter 8 Decimals
Chapter 9 NCERT solutions for class 6 maths chapter 9 Data Handling
Chapter 10 NCERT solutions for class 6 maths chapter 10 Mensuration
Chapter 11 NCERT Solutions for class 6 maths chapter 11 Algebra
Chapter 12 NCERT solutions for class 6 maths chapter 12 Ratio and Proportion
Chapter 13 NCERT solutions for class 6 maths chapter 13 Symmetry
Chapter 14 NCERT Solutions for class 6 maths chapter 14 Practical Geometry

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
The CBSE NCERT Solutions for Class 6 Maths Chapter 10 Mensuration are very useful for students to prepare for final exams. Click to download.
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