NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration: It is a process of measurement which is needed for different geometric shapes both 2D and 3D shapes. In CBSE NCERT solutions for class 6 maths chapter 10 mensuration, you will get many questions related to calculating the perimeter and area of a rectangle, square and a triangle. You must go through these NCERT solutions to get a better idea of how to solve a particular problem. NCERT Class 6 solution for chapter 10 is more focused on the measurement of 2-D shapes. As per the NCERT syllabus for Class 6 Maths, there are 30 questions given in the 3 exercises of this chapter. In solutions of NCERT for class 6 maths chapter 10 mensuration, there are some problems where you have to find areas of figures by counting the number of squares and sum of their area is the total area of that figure. Refer to the NCERT Books for Class 6 Maths for more clarity of the concepts. All these questions are explained in NCERT solutions for class 6. It will help you to understand the concept of NCERT in a more easy way.
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The main topics of the Solutions for NCERT Class 6 Maths Chapter 10 Mensuration are:
10.1 Introduction
10.2 Perimeter
10.2.1 Perimeter of a rectangle
10.2.2 Perimeter of regular shapes
10.3 Area
10.3.1 Area of a rectangle
10.3.2 Area of a square
NCERT class 7 maths chapter 10 mensuration topic 10.2 perimeter
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+ 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 |
25 cm | 12 cm | =25+12+25+12=74 cm | |
0.5 cm | 0.25cm | =0.5+0.25+0.5+0.25=1.5cm | |
18 cm | 15 cm | =18+15+18+15=66 cm | |
10.5 cm | 8.5 cm | =10.5+8.5+10.5+8.5=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 (4cm) + 4 (1 cm) + 8 (3 cm) + 4 (2cm) = (16 + 4 + 24 + 8)cm =52 cm
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 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 (L+B)
= 2 ((2m + 0.25m) + (1m + 0.5m))
= 7.5 m
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 53 cm = 106 cm
Hence the required length of the strip is 106 cm or 1.06 m
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 2 [ 0.7 km + 0.5 km]
= 8 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 side
= 6 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 side
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
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 =
= 4 250
=1000 m
= 1km
So, total cost = 20 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 300m
=600m
So, the total cost of fencing = 600 12 = Rs. 7,200
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 75
=300 m
Also, the Perimeter of the rectangular park
= 2 ( 60 + 45)
=2 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 side
= 4 25
= 100 cm
(b)The perimeter of the rectangle
= SUm of all the sides = 2 [length + breadth )
= 2 (20 + 30)
= 2 50
= 100
(c)The perimeter of the rectangle
= Sum of all the sides = 2 [length + breadth ]
= 2 (40 + 10)
= 2 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 . He lays them in the form of a square.
(a) What is the perimeter of his arrangement ?
(c) Which has greater perimeter?
Answer: Given that,
Length of each slab =
So, the length of the square =
(a) The first figure is a square
So, the perimeter of the square = 4 side
=
(b) The perimeter of the 2nd figure
= Sum of all the sides
=
= 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
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
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 =
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 breadth
Therefore,
(a) Here length is 3cm and the breadth of the rectangle is 4cm
So, the area of the rectangle
=
(b) Here length is 12m and the breadth of the rectangle is 21m
So, the area of the rectangle
=
(c) Here length is 3km and the breadth of the rectangle is 3km
So, the area of the rectangle ?
=
(d) Here length is 2m and the breadth of the rectangle is 70cm =
So, the area of the rectangle
=
Question: 2 Find the areas of the squares whose sides are :
Answer: It is known that area of a square =
(a) The length of a side of a square = 10cm
So, the area =
(b)The length of a side of a square = 14cm
So, the area =
(c) The length of a side of a square = 5m
So, the area =
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 breadth
Therefore,
(a) Here length is 9m and the breadth of the rectangle is 6m
So, the area of the rectangle
=
(b) Here length is 17 m and the breadth of the rectangle is 3m
So, the area of the rectangle
=
(c) Here length is 4m and the breadth of the rectangle is 14m
So, the area of the rectangle
=
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 =
=
Hence the width of the garden is 6 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
=
Now, Total rate of tiling of 100,1000 m ^{ 2 } =
= 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 breadth
Answer: Given that,
Length of the room = 4 m
And it's breadth = 3.5 m
Area of the room = = length breadth
=
Hence the area of carpet needed is
Answer: Given that,
Length of the floor = 5 m
and its width = 4 m
Therefore, the area of the floor
= length breadth
And, side of the square carpet = 3m
Therefore, the area of the square carpet =
= 3m 3m = 9m ^{ 2 }
Thus, the area of the floor which is not covered with carpet =
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
So, Area of 5 square flower bed = 5
Area of the and = length breadth
=
Thus, the area of the remaining part of the land = Area of land - area of the flower bed
= 20 - 5 = 15
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) = = 4 3 = 12
the area of the rectangle (II) = = 3 2= 6
the area of the rectangle (III) = = 4 1 = 4
the area of the rectangle (IV) = = 3 2 = 6
So, the total area of the figure = sum area of all rectangles
= (12 + 6 + 4 + 6)
= 28
(b) Split the given figure into smalle rectangles as shown here;
the area of the rectangle (I) = = 3 1 = 3
the area of the rectangle (II) = = 3 1= 3
the area of the rectangle (III) = = 3 1 = 3 ?
Therefore, the total area of the figure = sum of the area of all the small rectangles
=
=
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) = = 12 2 =
the area of the rectangle (II) = = 8 2=
Therefore, Total area =
(b) Split the given figure into a small rectangle, as shown here;
Now, the area of the rectangle (I) = = 7 7 =
the area of the rectangle (II) = = 21 7=
Now, the area of the rectangle (III) = = 7 7 =
Therefore, Total area =
(c) Split the given figure into a small rectangle, as shown here;
Now, the area of the rectangle (I) = = 2 1 =
the area of the rectangle (II) = = 5 1=
Now, the area of the rectangle (III) = = 2 1 =
Therefore, Total area =
(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 breadth
= 12 cm 5 cm
= 60 cm ^{ 2 }
Now, (a) For 100 cm and 144 cm rectangular region,
Area of region = length breadth
= 14,400 cm ^{ 2 }
Therefore, the number of tiles needed = tiles
(b) For 70 cm and 36 cm rectangular region,
Area of region = length breadth
= 70 cm 36 cm
= 2520 m ^{ 2 }
Required number of tiles =
NCERT Solutions for Class 6 Mathematics Chapter wise
Chapters No. | Chapters Name |
Chapter - 1 | NCERT Solutions for class 6 maths chapter 1 Knowing Our Numbers |
Chapter - 2 | |
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 | |
Chapter - 7 | |
Chapter - 8 | |
Chapter - 9 | |
Chapter -10 | |
Chapter -11 | |
Chapter -12 | NCERT solutions for class 6 maths chapter 12 Ratio and Proportion |
Chapter -13 | |
Chapter -14 | NCERT Solutions for class 6 maths chapter 14 Practical Geometry |
NCERT Solutions for Class 6 Subject wise
Some important formulas of class 6 maths chapter 10 mensuration
- Perimeter of rectangle:
Perimeter = 2 × (length + breadth) Area of the rectangle with length l and breadth b is the product of length and breadth = lb. The perimeter= l+b+l+b=2(l+b)
- Perimeter of a square:
Perimeter = 4 × length of its side
- The perimeter of a triangle:
Perimeter = (a+b+c)
a- First side of a triangle
b- Second side of a triangle
c- Third side of a triangle
For an equilateral triangle, all the sides are equal so its perimeter is 3l.
l- side of an equilateral triangle
- Area of a rectangle:
Area = length(l) × breadth (b)
- Area of a square:
Area = (side x side) A triangle has three sides.
You can derive all these formulas from the basic concepts of this chapter. There many questions in CBSE NCERT solutions for class 6 maths chapter 10 mensuration where you will find the complex geometry, to solve that you have to use your basic concepts of the geometry.
Happy Reading!!!
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