CBSE Important Questions Class 6 Maths Chapter 10
Important Questions Class 6 Maths Chapter 10 – Mensuration
Maths is an important subject taught in school. We need Maths in our everyday life because it helps us to solve different real-life problems. Chapter 10 of Class 6 Maths deals with mensuration, a part of geometry. It covers the calculation of the length, area, or perimeter of different geometric shapes.
Mensuration is a relatively new concept to Class 6 students. They have become familiar with different geometric shapes in previous chapters. In this chapter, they will learn about how to calculate areas, perimeters, lengths or widths of different geometric shapes. The chapter mainly concentrates on rectangles, and students will learn the perimeter and area calculation of rectangles.
Extramarks is a leading company that provides all the important study materials related to CBSE and NCERT. Our experts have made the Important Questions Class 6 Maths Chapter 10 to help students practice and collect the questions from different sources. They have taken help from the textbook exercises, CBSE sample papers, NCERT exemplar, important reference books and CBSE past years’ question papers. Students will upgrade their level of preparation if they solve these questions thoroughly.
Extramarks is a leading company that provides all the study materials related to CBSE and NCERT. You can register on our official website of Extramarks and download these materials. We provide CBSE syllabus, NCERT books, CBSE past years’ question papers, NCERT solutions, NCERT exemplar solutions, CBSE extra questions, CBSE revision notes, NCERT important questions, CBSE sample papers, vital formulas and many more.
CBSE Important Questions for Class 6 Maths |
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Sr No | Chapter No | Chapter Name |
1 | Chapter 1 | Knowing Our Numbers |
2 | Chapter 2 | Whole Numbers |
3 | Chapter 3 | Playing with Numbers |
4 | Chapter 4 | Basic Geometrical Ideas |
5 | Chapter 5 | Understanding Elementary Shapes |
6 | Chapter 6 | Integers |
7 | Chapter 7 | Fractions |
8 | Chapter 8 | Decimals |
9 | Chapter 9 | Data Handling |
10 | Chapter 10 | Mensuration |
11 | Chapter 11 | Algebra |
12 | Chapter 12 | Ratio and Proportion |
13 | Chapter 13 | Symmetry |
14 | Chapter 14 | Practical Geometry |
Important Questions Class 6 Maths Chapter 10 – With Solutions
The experts of Extramarks believe in continuous practice for better marks. They have made this question series to help students in solving questions. The experts have collected the questions from different sources like textbook exercises, CBSE sample papers, NCERT exemplars and important reference books. Apart from this, they have also included a few questions from CBSE past years’ question papers so that students may know which kind of questions come in exams. They also gave the solutions in Important Questions Class 6 Maths Chapter 10. The important questions are-
Question 1. The lid of a rectangular box is of sides 40 cm by 10 cm and is covered all around with a sellotape. What is the total length of the sellotape required to cover the rectangle?
Answer 1:
The length of the required tape is equal to the Perimeter of the rectangle.
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (40 + 10) cm
= 2 (50) cm
= 100 cm
∴ Hence the required length of tape is 100 cm.
Question 2. A tabletop measures about 2 m 25 cm(length) by 1 m 50 cm (breadth). What is the total Perimeter of the tabletop?
Answer 2:
The length of the tabletop is = 2 m 25 cm, which then becomes 2.25 m
The Breadth of a tabletop is = 1 m 50 cm, which then becomes 1.50 m
Perimeter of tabletop will be = 2 (Length + Breadth)
= 2 (2.25 + 1.50)
= 2 (3.75)
= 2 × 3.75
= 7.5 m
∴ So the Perimeter of the table top is 7.5 m
Question 3. What is the total length of the wooden strip which is required to frame a photograph of length and Breadth of 32 cm and 21 cm, respectively?
Answer 3:
The required length of the wooden strip will be equal to the Perimeter of the photograph.
= 2 (Length + Breadth)
= 2 (32 + 21)
= 2 (53)
= 2 × 53
= 106 cm
∴ The required length of the wooden strip will be 106 cm.
Question 4. A rectangular piece of land measures about 0.7 km by 0.5 km. Each side of the rectangular piece is to be fenced with four rows of wires. What is the total length of the wire needed?
Answer 4:
The Perimeter of the field is equal to 2 (Length + Breadth)
= 2 (0.7 + 0.5)
= 2 (1.2)
= 2 × 1.2
= 2.4 km
We know that each side is to be fenced with four rows, so = 4 × 2.4
= 9.6 km
∴ Hence the total length of the required wire is 9.6 km.
Question5 . Find the total Perimeter of each of the following shapes given below:
(a) A normal triangle of sides 3 cm, 4 cm and 5 cm
(b) An equilateral triangle on each side of 9 cm
(c) An isosceles triangle with equal sides of 8 cm each and the third side of 6 cm.
Answer 5:
(a) Perimeter of triangle will be = 3 + 4 + 5
= 12 cm
(b) Perimeter of an equilateral triangle will be = 3 × side
= 3 × 9
= 27 cm
(c) Perimeter of isosceles triangle will be = 8 + 8 + 6
= 22 cm
Question 6. Find the total Perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Answer 6:
Perimeter of the triangle will be = 10 + 14 + 15
= 39 cm
∴ Hence the Perimeter of the triangle will be 39 cm.
Question 7. Find the total Perimeter of a regular hexagon with each side measuring almost 8 m.
Answer 7:
The final Perimeter of the hexagon will be = 6 × 8
= 48 m
∴ Hence Perimeter of a regular hexagon is 48 m
Question 8. Find the side of the square whose final Perimeter is 20 m.
Answer 8:
The Perimeter of the square = 4 × side
20 = 4 × side
Per Side = 20 / 4
One Side = 5 m
∴ The length of the side of the square is 5 m
Question 9. The Perimeter of a regular pentagon is found to be 100 cm. How long is each side?
Answer 9:
Perimeter of regular pentagon = 100 cm
5 × side = 100 cm
Side = 100 / 5
Side = 20 cm
∴ The side of the pentagon is 20 cm.
Question 10. A piece of string is 30 cm long. What will be the length of each side of the string if the string is used to form the following figures:
(i) a square?
(ii) an equilateral triangle?
(iii) a regular hexagon?
Answer 10:
(i) Perimeter of square = 30 cm
4 × side = 30
Side = 30 / 4
Side = 7.5 cm
(ii) Perimeter of an equilateral triangle is 30 cm
3 × side = 30
Per Side = 30 / 3
One Side = 10 cm
(iii) Perimeter of a regular hexagon is 30 cm
6 × side = 30
Per Side = 30 / 6
One Side = 5 cm
Question 11. The two sides of a triangle are 12 cm and 14 cm, respectively. The Perimeter of the triangle is found to be 36 cm. What is the length of the third side?
Answer 11
Let x cm be the length of the third side.
The Perimeter of the triangle becomes 36 cm.
12 + 14 + x = 36
26 + x = 36
x = 36 – 26
x = 10 cm
∴ The length of the third side is 10 cm.
Question 12. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Answer 12:
The side of the square is 250 m
The Perimeter of the square will be four × sides.
= 4 × 250
= 1000 m
The cost of fencing = ₹ 20 per m
The cost of fencing for 1000 m is = ₹ 20 × 1000
= ₹ 20,000
Question 13. Find the total cost of fencing a rectangular park of length 175 cm and Breadth as 125 m at the rate of ₹ 12 per metre.
Answer 13:
Length = 175 cm
Breadth = 125 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (175 + 125)
= 2 (300)
= 2 × 300
= 600 m
Cost of fencing = 12 × 600
= 7200
∴ The cost of fencing is ₹ 7,200
Question 14. Find the Area of the given following figures by counting the square:
Answer 14.
(a) The figure contains only nine squares which are fully filled squares. Hence, the Area of the following figure will be 9 square units.
(b) The figure contains only five squares which are fully filled squares. Hence, the Area of the following figure will be 5 square units.
(c) The figure contains two squares which are fully-filled squares and four half-filled squares. Hence, the Area of the following figure will be 4 square units.
(d) The figure contains only eight squares which are fully filled squares. Hence, the Area of the following figure will be 8 square units.
(e) The figure contains only ten squares which are fully filled squares. Hence, the Area of the following figure will be 10 square units.
(f) The figure contains only two squares which are fully-filled squares and four half-filled squares. Hence, the Area of the following figure will be 4 square units.
(g) The figure contains four squares which are fully-filled squares and four half-filled squares. Hence, the Area of the following figure will be 6 square units.
(h) The figure contains five squares which are fully filled squares. Hence, the Area of the following figure will be 5 square units.
(i) The figure contains nine squares which are fully filled squares. Hence, the Area of the Area of the figure will be 9 square units.
(j) The figure contains two squares which are fully-filled squares and four half-filled squares. Hence, the Area of the Area of the figure will be 4 square units.
(k) The figure contains four squares which are fully-filled squares and two half-filled squares. Hence, the Area of the figure will be 5 square units.
(l) From the given figure, we observe
Covered Area Number Area estimate (square units)
Fully filled squares 2 2
Half-filled squares – –
More than half-filled squares 6 6
Less than half-filled squares 6 0
Therefore total area = 2 + 6
= 8 square units.
Question 15. Find the Area of the rectangle whose sides are given below:
(i) 3 cm and 4 cm
(ii) 12 m and 21 m
(iii) 2 km and 3 km
(iv) 2 m and 70 cm
Answer 15:
We know the formula for the Area of the rectangle is,
Area of rectangle = Length multiplied by Breadth
(i) Here, l = 3 cm and b = 4 cm
So the Area = l × b = 3 × 4
= 12 cmsq
(ii) Here, l = 12 m and b = 21 m
So the Area = l × b = 12 × 21
= 252 m sq
(iii) Here, l = 2 km and b = 3 km
So the Area = l × b = 2 × 3
= 6 km sq
(iv) Here l = 2 m and b = 70 cm = 0.70 m
So the Area = l × b = 2 × 0.70
= 1.40 m sq
Question 16. Find the Area of the squares whose sides are given below:
(i) 10 cm
(ii) 14 cm
(iii) 5 m
Answer 16-
(i) We know that, Area of square = side2
= 102
= 100 cm2
(ii) We know that, Area of square = side2
= 142
= 196 cm2
(iii) We know that, Area of square = side2
= 52
=25 cm2
Question 17. The length and Breadth of the three rectangles are as given below:
(i) 9 m and 6 m
(ii) 17 m and 3 m
(iii) 4 m and 14 m
Which one of the above-given statements has the largest Area, and which one has the smallest?
Answer 17:
(i) Area of rectangle = l × b
= 9 × 6
= 54 m2
(ii) Area of rectangle = l × b
= 17 × 3
= 51 m2
(iii) Area of rectangle = l × b
= 4 × 14
= 56 m2
The Area of the rectangle is 56 m2, So (c) is the largest Area, and the Area of the rectangle is 51 m2, so (b) is the smallest Area.
Question 18. How many tiles of length and breadth 12 cm and 5 cm, respectively, will be needed to fit in the following rectangular region whose length and Breadth are found to be?
(i) 100 cm and 144 cm
(ii) 70 cm and 36 cm
Answer 18:
(i) Area of rectangle = 100 × 144
= 14400 cm
Area of one tile = 5 × 12
= 60 cm2
The number of tiles = (Area of rectangle) divided by (Area of one tile)
= 14400 / 60
= 240
Hence, 240 tiles are needed.
(ii) Area of rectangle = 70 × 36
= 2520 cm2
Area of one tile = 5 × 12
= 60 cm2
The number of tiles = (Area of rectangle) divided by (Area of one tile)
= 2520 / 60
= 42
Hence, 42 tiles are needed.
Question 19. The side of a square is 10cm. How many times will the new Perimeter be if the side of the square is doubled?
(i) 2 times (ii) 4 times (iii) 6 times (iv) 8 times
Answer 19:-
(i) 2 times
As we know, the Perimeter of the square is = side × 4
= 10 × 4 cm
= 40 cm
Given the side of the square = 10 cm.
The side of the square is then doubled = 10 + 10
= 20 cm
Therefore, the new Perimeter becomes two times as the side of the square is doubled.
Question 20. The Breadth and length of a rectangular sheet of paper are 10cm and 20cm, respectively. A rectangular piece is cut from the sheet, as shown in the figure given below. Which of the following statements is correct for the present remaining sheet?
(i) Perimeter remains the same but the area changes.
(ii) Area remains the same but the perimeter changes.
(iii) Both Area and Perimeter are changing accordingly
(iv) Both Area and Perimeter remain the same.
Answer 20:-
(i) Perimeter remains the same but the area changes.
We know that the Area of the big rectangle is length × Breadth.
= 10 × 20
= 200 cm2
Then the Area of the small rectangle is equal to 5 × 2
= 10 cm2
The Perimeter of the rectangle becomes 2(length + Breadth)
= 2(20 + 10)
= 2 × 30
= 60 cm
Then the perimeter of the new figure becomes = 20 + 8 + 5 + 2 + 15 + 10
= 60 cm
Area of new figure = Area of the big rectangle – Area of a new figure
= 200 – 10
= 190 cm sq
By comparing all the results, we find out that the Perimeter remains the same but the area changes.
Question 21.
A rectangle and a square have an equal or the same Perimeter. So find the following statements.
(a) The area of the rectangle is _________.
Answer 21:-
The Area of the rectangle becomes 12 sq. units.
From the given figure, the length of a rectangle is six, and the Breadth is 2.
So the total area of the rectangle will be length × Breadth.
= 6 × 2
= 12 sq. units
The Perimeter of the rectangle is two × (length + Breadth)
= 2 × (6 + 2)
= 2 × 8
= 16 units
(b) The area of the square is _________.
Answer:-
The Area of the square becomes 16 sq. Units.
From the above question, it is given that a rectangle and a square have the same Perimeter.
So, the length of each side of a square will be = 16/4
= 4 units.
Area of square = side × side
= 4 × 4
= 16 sq. Units
Question 22. (a) 1m = _________ cm.
Answer:-
1m = 100 cm.
(b) 1sqcm = _________ cm × 1cm.
1sqcm = 1 cm × 1cm.
(c) 1sqm = 1m × _________ m = 100cm × _________ cm.
1sqm = 1m × 1 m
= 100cm × 100 cm.
(d) 1sqm = _________ sqcm.
1sqm = 10000 sqm.
We know that 1 m = 100 cm.
Then, 1 sqm = 100 × 100
= 10000sqcm
Question 23. In the below questions, state which of the statements are true and which of them are false.
- a) If the length of a rectangle is halved and the Breadth is doubled, then the Area of the rectangle remains the same.
Answer:-
True.
We know that the area of a rectangle is equal to the length × Breadth.
As per the conditions given in the question, the length of a rectangle is halved. So, length = l/2
Breadth is doubled So, breadth = 2b
Then, the new Area becomes = l/2 × 2b
= l × b.
Therefore, if the length of the rectangle is halved and the Breadth gets doubled, then the Area of the rectangle obtained remains the same.
- b) Area of a square gets doubled if the side of the square is doubled.
Answer:-
False
We know that the Area of the square gets as side × side.
As per the conditions given in the question, the side of the square gets doubled = 2 × side.
Then, the new Area becomes = 2side × 2side
= 4 side2
Therefore, when the side of the square is doubled, then the Area of the square obtained becomes four times of the old Area.
c). Perimeter of a regular octagon of side 6cm is 36cm.
Answer:-
The statement is False.
The Perimeter of regular octagon is = number of sides × length of reach sides.
= 8 × 6
= 48 cm
d). A farmer who wants to fence his field must find the Perimeter of the field.
Answer:-
True.
e). An engineer who plans to build a compound wall on all sides of a house must find the Area of the following compound.
Answer:-
The statement is False.
An engineer who plans to build a compound wall on all sides of the house must find the Perimeter of the compound.
- To find the total cost of painting a wall, we need to find the Perimeter of the wall.
Answer:-
The statement is False.
To find the cost of painting on a wall, we will need to find the total area of the wall.
- To find the cost of a frame of a picture, we need to find the Perimeter of the picture.
Solution:-
True.
Question 24. The Perimeter of an isosceles triangle is found to be 50cm. If one of the two equal sides is 18 centimeters, find the length of the third side.
Answer 24:-
We know that in an isosceles triangle, there are two equal sides.
From the above question, it is given that one of the two equal sides of the isosceles triangle is 18cm.
And the Perimeter of an isosceles triangle becomes 50cm.
By this, we know that Perimeter of an isosceles triangle is equal to the sum of all sides of the triangle.
Let us assume the third side to be x.
Then,
50 = 18 + 18 + x
50 = 36 + x
X = 50 – 36
X = 14 cm
Hence, the length of the third side of an isosceles triangle is 14 cm.
Question 25. The length of a rectangle is three times its Breadth, and the Perimeter of the rectangle is 40cm. Find its length and width.
Answer 25:-
From the above question, it is given that the,
The perimeter of a rectangle is given as 40 cm.
The length of a rectangle is said to be three times its Breadth So, = 3b
We know that Perimeter of the rectangle is equal to 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
Question 26. The length of a rectangular field is twice its Breadth. Jamal then jogged around it four times and covered a distance of 6km. What is the total length of the field?
Answer 26:-
From the above question, it is stated that,
The length of a rectangular field is said to be twice its Breadth, which means the length will be 2b
The perimeter of rectangular field is 6km/4
= 1.5 km
We know that Perimeter of the rectangle is equal to 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
Benefits of Solving Important Questions Class 6 Maths Chapter 10
Practice is very important to score better in exams. Many students fear maths because they don’t understand the concepts clearly and get confused. So, students must practice questions as much as possible for better exam preparation. The practice habit is very important because it helps students in many ways. It not only helps them to score better but also clear concepts and boosts confidence. Solving the Class 6 Maths Chapter 10 Important Questions will have several benefits. These are-
- Sometimes, the textbook exercise is not enough for practice. Students must take help from other sources for regular practice, but it may be a tiring job for them. Hence, the experts in Extramarks have made this question series to help students. They collected the questions from several sources, such as the CBSE sample papers, the textbook exercise, NCERT exemplar and important reference books. They have included several questions from CBSE past years’ question papers too. Thus, the Important Questions Class 6 Maths Chapter 10 will help students in regular practice and assist in their preparation.
- Apart from the questions, the experts have solved the questions too. They have provided a step-by-step process for explaining each question in this question series. Students may follow the solutions if they cannot solve the questions. Experienced professionals have further checked the answers to ensure the best quality of content for students. Students can further their answers with the answers given by professionals. Thus, the Maths Class 6 Chapter 10 Important Questions will help them solve those questions and generate interest in the subject.
- Practice is very much important for better preparation for exams. Practice will help students to gain much confidence in the subject. It will also help those students who have a fear of Maths. Thus, the Important Questions Class 6 Maths Chapter 10 will help the students in various ways. It will boost their preparation, upgrade their preparation and help them score higher marks in exams.
Extramarks is a leading company that will help students with all the necessary study materials. You may register on the official website of Extramarks and download these study materials. We provide CBSE syllabus, CBSE sample papers, CBSE past years’ question papers, CBSE extra questions, CBSE revision notes, NCERT books, NCERT solutions, NCERT Exemplar, NCERT exemplar, vital formulas and many more. Like the Chapter 10 Class 6 Maths Important Questions, you will also find important questions for other chapters. Links to the study materials are given below-
- NCERT books
- Important questions
- CBSE Revision Notes
- CBSE syllabus
- CBSE sample papers
- CBSE past years’ question papers
- Important formulas
- CBSE extra questions
Q.1 A floor is 10 m long and 8 m wide. A square carpet of side 4.5 m is laid on it. Find the area of floor uncovered by the carpet.
$\begin{array}{l}\text{Area of rectangular floor = 10m \xd7 8m}\\ {\text{=80m}}^{\text{2}}\\ \text{Area of square carpet = 4.5m\xd74.5m}\\ {\text{=20.25 m}}^{\text{2}}\\ {\text{Area of uncovered floor =80m}}^{\text{2}}{\text{-20.25 m}}^{\text{2}}\\ {\text{=59.75m}}^{\text{2}}\\ {\text{Thus, area of uncovered floor is 59.75 m}}^{}\end{array}$
Marks:4
Ans
$\begin{array}{l}\text{Area of rectangular floor = 10m \u2014 8m}\\ {\text{=80m}}^{\text{2}}\\ \text{Area of square carpet = 4.5m\u20144.5m}\\ {\text{=20.25 m}}^{\text{2}}\\ {\text{Area of uncovered floor =80m}}^{\text{2}}{\text{-20.25 m}}^{\text{2}}\\ {\text{=59.75m}}^{\text{2}}\\ {\text{Thus, area of uncovered floor is 59.75 m}}^{\text{2}}\text{.}\end{array}$
Q.2 A rectangular grassy lawn measuring 48 m by 35 m is to be surrounded externally by a path, which is 2.5 m wide. Find the cost of the leveling the path at the rate of 4.50 per sq m.
Let KLMN be a grassy lawn surrounded by the path which is 2.5 m wide.
Area of lawn = 48 ×35
= 1680 m^{2}
Length of lawn including the path = 48 + 2 × 2.5
= 53 m
Breadth of lawn including the path = 35 + 2 × 2.5
= 40 m
Total area of lawn including the path = 53 × 40
= 2120 m^{2}
Area of path = Total area Area of the grassy lawn
= 2120 m^{2 }– 1680 m^{2}
= 440 m^{2}
Cost of leveling the path = 440 × 4.50
Marks:5
Ans
Let KLMN be a grassy lawn surrounded by the path which is 2.5 m wide.
Area of lawn = 48 —35
= 1680 m^{2}
Length of lawn including the path = 48 + 2 — 2.5
= 53 m
Breadth of lawn including the path = 35 + 2 — 2.5
= 40 m
Total area of lawn including the path = 53 — 40
= 2120 m^{2}
Area of path = Total area €“ Area of the grassy lawn
= 2120 m^{2 }– 1680 m^{2}
= 440 m^{2}
Cost of leveling the path = 440 — 4.50
= 1980
Q.3 A marble tile measures 25 cm by 20 cm. To cover a wall of size 4 m by 3 m, the number of required tiles is:
A. 340
B. 270
C. 240
D. 120
Marks:1
Ans
The measure of marble tile = 25 cm by 20 cm
So, area of marble tile = 25 cm × 20 cm
= 500 cm^{2}
Wall size = 4m by 3m
= 4 × 100 cm by 3 × 100 cm
So, wall size = 400 cm by 300 cm
Area of the wall = 400 cm × 300 cm
= 120000 cm^{2}
So, number of tiles = Area of wall/Area of tiles
= 120000/500
= 240
Q.4 Area of a square is 196 cm^{2}. What is its perimeter
A. 14 cm
B. 28 cm
C. 42 cm
D. 56 cm
Marks:1
Ans
Q.5 The perimeter of 12 m wide rectangular field is 104 m. What is its area
A. 240 m^{2}
B. 120 m^{2}
C. 480 m^{2}
D. 160 m^{2}
Marks:1
Ans
FAQs (Frequently Asked Questions)
1. What is Class 6 Maths Chapter 10 about?
Chapter 10 of Class 6 Maths is about mensuration. Mensuration is a branch of Geometry that deals with the length, area and volume calculation of different geometric shapes. Students will learn area calculation and calculation of the perimeter of rectangles. Thus, the chapter mainly concentrates on the rectangle. Students can learn the concepts if they follow the textbook carefully. They can take help from the Important Questions Class 6 Maths Chapter 10 prepared by the experts of Extramarks.
2. How can the Important Questions Class 6 Maths Chapter 10 help students?
The experts of Extramarks have made this question series to help students. They collected the questions from different sources so that students would practice the questions regularly. They have taken help from the textbook exercises, NCERT exemplar, CBSE sample papers and other important reference books. They have taken help from the CBSE past years’ question papers to include questions that have come before in exams. Thus, the Important Questions Class 6 Maths Chapter 10 will help students better prepare. Apart from this, they have solved the questions, too, so that students can follow their solutions. So, it will help me to score better in exams.