Benefits of RD Sharma Solutions for Class 6 Maths

• The chapter-wise and exercise-wise solutions can be easily downloaded by the students, based on their requirement.
• To make learning more interesting, each solution is provided with a pictorial representation to improve analysing skills among students.
• The solutions improve speed in solving problems more accurately, as per the prescribed CBSE syllabus.
• The students can self analyse their areas of weaknesses and can work on them to score good marks.

Is RD Sharma Solutions the right book for CBSE Class 6 students?

Yes RD Sharma is the right book for CBSE Class 6 Students as it provides lots of questions to practice. The presentation of each topic in the book is described uniquely, which means the topics covered in the book are not stretched beyond their necessary length. They yet vividly describe all the variations of a particular concept, applied to different questions.

Name the chapters present in RD Sharma Class 6 Maths Textbook.

There are 23 chapters are there in RD Sharma Class 6 Maths Textbook. Viz, Knowing our Numbers, Playing with numbers, Whole Numbers, Operations on Whole Numbers, Negative Numbers and Integers, Fractions, Decimals, Introduction to Algebra, Ratio, Proportion and Unitary Method, Basic Geometrical Concepts, Angles, Triangles, Quadrilaterals, Circles, Pair of Lines and Transversal, Understanding Three Dimensional Shapes, Symmetry, Basic Geometrical Tools, Geometrical Constructions, Mensuration, Data Handling – I (Presentation of Data), Data Handling – II (Pictographs) and Data Handling – III (Bar Graphs).

How many exercises does the RD Sharma Solutions provide for Class 6 students?

Exercises of RD Sharma Textbook vary according to the concepts covered in the chapters. Some chapters have 6 and some have only one. It depends on the topic of the lesson.

How many four digit numbers are there in all?

Solution:

We know that the 10 digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

0 cannot be used in thousands place so only nine digits can be used.

10 digits can be used in hundreds, tens and units place

The number of four digit numbers = 9 × 10 × 10 × 10 = 9000

Therefore, 9000 four digit numbers are there in all.

2. Write the smallest and the largest six digit numbers. How many numbers are between these two.

Solution:

We know that the smallest digit is 0 which cannot be used in the highest place value.

So 1 which is the second smallest digit can be used in the highest place value

The required smallest six digit number is 100000

We know that the largest digit is 9 which can be used in any place

The required largest six digit number is 999999

So we get the difference = 999999 – 100000 = 899999

Therefore, the smallest six digit number is 100000, the largest six digit number is 999999 and 899999 numbers are between these two numbers.

3. How many 8-digit numbers are there in all?

Solution:

We know that the 10 digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

0 cannot be used in the highest place value and 9 can be used in the highest place value

So the 10 digits can be used in the remaining places of 8 digit numbers

The total number of 8 digit numbers = 9 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 90000000

Therefore, 90000000 eight digit numbers are there in all.

4. Write 10075302 in words and rearrange the digits to get the smallest and the largest numbers.

Solution:

The given number 10075302 can be written as one crore seventy five thousand three hundred and two.

To get smallest 8 digit number using 0, 1, 2, 3, 5 and 7

We use 1 which is the smallest digit in the highest place and largest digit 7 at the units place

Further we put 5 in the tens place, 3 in the hundreds place and 2 in thousands place

So the required smallest number is 10002357

To get largest 8 digit number using 0, 1, 2, 3, 5 and 7

We use 7 which is the largest digit in the highest place value, 5 in a place after highest place, 3 as the next one, 2 as the smallest digit and then 1.

So the required largest number is 75321000.

5. What is the smallest 3-digit number with unique digits?

Solution:

102 is the smallest 3-digit number with unique digits.

6. What is the largest 5-digit number with unique digits?

Solution:

98765 is the largest 5-digit number with unique digits.

7. Write the smallest 3-digit number which does not change if the digits are in reverse order.

Solution:

101 is the smallest 3-digit number which does not change if the digits are in reverse order.

8. Find the difference between the number 279 and that obtained on reversing its digits.

Solution:

The reverse of 279 is 972

Difference between both the numbers = 972 – 279 = 693

Therefore, the difference between the number 279 and that obtained on reversing its digits is 693.

9. Form the largest and smallest 4-digit numbers using each of digits 7, 1, 0, 5 only once.

Solution:

The largest 4 digit number = 7510

Smallest 4 digit number = 1057

Therefore, the largest and smallest 4-digit numbers using each of digits 7, 1, 0, 5 only once is 7510 and 1057.

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1. Find the HCF of the following numbers using prime factorization method:

(i) 144, 198

(ii) 81, 117

(iii) 84, 98

(iv) 225, 450

(v) 170, 238

(vi) 504, 980

(vii) 150, 140, 210

(viii) 84, 120, 138

(ix) 106, 159, 265

Solution:

(i) 144, 198

We know that the prime factorization of 144 = 2 × 2 × 2 × 2 × 3 × 3

The same way prime factorization of 198 = 2 × 3 × 3 × 11

Hence, HCF of 144, 198 is 2 × 3 × 3 = 18

(ii) 81, 117

We know that prime factorization of 81 = 3 × 3 × 3 × 3

The same way prime factorization of 117 = 3 × 3 × 13

Hence, HCF of 81, 117 = 3 × 3 = 9

(iii) 84, 98

We know that prime factorization of 84 = 2 × 2 × 3 × 7

The same way prime factorization of 98 = 2 × 7 × 7

Hence, HCF of 84, 98 = 2 × 7 = 14

(iv) 225, 450

We know that prime factorization of 225 = 3 × 3 × 5 × 5

The same way prime factorization of 450 = 2 × 3 × 3 × 5 × 5

Hence, HCF of 225, 450 = 3 × 3 × 5 × 5 = 225

(v) 170, 238

We know that prime factorization of 170 = 2 × 5 × 17

The same way prime factorization of 238 = 2 × 7 × 17

Hence, HCF of 170, 238 = 2 × 17 = 34

(vi) 504, 980

We know that the prime factorization of 504 = 2 × 2 × 2 × 3 × 3 × 7

The same way prime factorization of 980 = 2 × 2 × 5 × 7 × 7

Hence, HCF of 504, 980 = 2 × 2 × 7 = 28

(vii) 150, 140, 210

We know that prime factorization of 150 = 2 × 3 × 5 × 5

The same way prime factorization of 140 = 2 × 2 × 5 × 7

Prime factorization of 210 = 2 × 3 × 5 × 7

Hence, HCF of 150, 140, 210 = 2 × 5 = 10

(viii) 84, 120, 138

We know that prime factorization of 84 = 2 × 2 × 3 × 7

The same way prime factorization of 120 = 2 × 2 × 2 × 3 × 5

Prime factorization of 138 = 2 × 3 × 23

Hence, HCF of 84, 120, 138 = 2 × 3 = 6

(ix) 106, 159, 265

We know that prime factorization of 106 = 2 × 53

The same way prime factorization of 159 = 3 × 53

Prime factorization of 265 = 5 × 53

Hence, HCF of 106, 159, 265 = 53

2. What is the HCF of two consecutive

(i) Numbers

(ii) even numbers

(iii) odd numbers

Solution:

(i) We know that the common factor of two consecutive numbers is 1.

Hence, HCF of two consecutive numbers is 1.

(ii) We know that the common factors of two consecutive even numbers are 1 and 2.

Hence, HCF of two consecutive even numbers is 2.

(iii) We know that the common factors of two consecutive odd numbers is 1.

Hence, HCF of two consecutive odd numbers is 1.

3. HCF of co-prime numbers 4 and 15 was found as follows:

4 = 2 × 2 and 15 = 3 × 5

Since there is no common prime factor. So, HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Solution:

No. It is not correct.

The HCF of two co-prime numbers is 1.

We know that 4 and 15 are co-prime numbers having common factor 1.

Therefore, HCF of 4 and 15 is 1.

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1. Write down the smallest natural number.

Solution:

1 is the smallest natural number.

2. Write down the smallest whole number.

Solution:

0 is the smallest whole number.

3. Write down, if possible, the largest natural number.

Solution:

The largest natural number does not exist.

4. Write down, if possible, the largest whole number.

Solution:

The largest whole number does not exist.

5. Are all natural numbers also whole numbers?

Solution:

Yes. All natural numbers are also whole numbers.

6. Are all whole numbers also natural numbers?

Solution:

No. All whole numbers are not natural numbers.

7. Give successor of each of the following whole numbers:

(i) 1000909

(ii) 2340900

(iii) 7039999

Solution:

(i) The successor of 1000909 is 1000910.

(ii) The successor of 2340900 is 2340901.

(iii) The successor of 7039999 is 7040000.

8. Write down the predecessor of each of the following whole numbers:

(i) 10000

(ii) 807000

(iii) 7005000

Solution:

(i) The predecessor of 10000 is 9999.

(ii) The predecessor of 807000 is 806999.

(iii) The predecessor of 7005000 is 7004999.

4. What is the difference between the largest number of five digits and the smallest number of six digits?

Solution:

99999 is the largest number of five digits

100000 is the largest number of six digits

Difference = 100000 – 99999 = 1

Therefore, 1 is the difference between the largest number of five digits and smallest number of six digits.

5. Find the difference between the largest number of 4 digits and the smallest number of 7 digits.

Solution:

9999 is the largest number of 4 digits

1000000 is the smallest number of 6 digits

Difference = 1000000 – 9999 = 990001

Therefore, 990001 is the difference between the largest number of 4 digits and the smallest number of 7 digits.

6. Rohit deposited Rs 125000 in his savings bank account. Later he withdrew Rs 35425 from it. How much money was left in his account?

Solution:

Money deposited in savings bank account = Rs 125000

Money withdrawn = Rs 35425

So the money which is left out in his account = 125000 – 35425 = Rs 89575

Hence, Rs 89575 is left in his account.

7. The population of a town is 96209. If the number of men is 29642 and that of women is 29167, determine the number of children.

Solution:

Population of a town = 96209

No. of men = 29642

No. of women = 29167

Total number of men and women = 29642 + 29167 = 58809

So the number of children = Population of a town – Total number of men and women

Number of children = 96209 – 58809 = 37400

Hence, there are 37400 children.

8. The digits of 6 and 9 of the number 36490 are interchanged. Find the difference between the original number and the new number.

Solution:

It is given that

Original Number = 39460

Number after interchanging 6 and 9 = 36490

Difference between them = 39460 – 36490 = 2790

Therefore, the difference between the original number and new number is 2970.

1. Write the opposite of each of the following:

(i) Increase in population

(ii) Depositing money in a bank

(iii) Earning money

(iv) Going North

(v) Gaining a weight of 4kg

(vi) A loss of Rs 1000

(vii) 25

(viii) – 15

Solution:

(i) The opposite of Increase in population is Decrease in population.

(ii) The opposite of Depositing money in a bank is Withdrawing money from a bank.

(iii) The opposite of earning money is Spending money.

(iv) The opposite of Going North is Going South.

(v) The opposite of gaining a weight of 4kg is losing a weight of 4kg.

(vi) The opposite of a loss of Rs 1000 is a gain of Rs 1000.

(vii) The opposite of 25 is – 25.

(viii) The opposite of – 15 is 15.

2. Indicate the following by using integers:

(i) 25o above zero

(ii) 5o below zero

(iii) A profit of Rs 800

(iv) A deposit of Rs 2500

(v) 3km above sea level

(vi) 2km below level

Solution:

(i) 25o above zero is + 25o.

(ii) 5o below zero is – 5o.

(iii) A profit of Rs 800 is + 800.

(iv) A deposit of Rs 2500 is + 2500.

(v) 3km above sea level is + 3.

(vi) 2km below level is – 2.

1. Write each of the following divisions as fractions:

(i) 6 ÷ 3

(ii) 25 ÷ 5

(iii) 125 ÷ 50

(iv) 55 ÷ 11

Solution:

(i) The division 6 ÷ 3 can be written as 6/3.

(ii) The division 25 ÷ 5 can be written as 25/5.

(iii) The division 125 ÷ 50 can be written as 125/50.

(iv) The division 55 ÷ 11 can be written as 55/11.

2. Write each of the following fractions as divisions:

(i) 9/7

(ii) 3/11

(iii) 90/63

(iv) 1/5

Solution:

(i) The fraction 9/7 can be written as 9 ÷ 7.

(ii) The fraction 3/11 can be written as 3 ÷ 11.

(iii) The fraction 90/63 can be written as 90 ÷ 63.

(iv) The fraction 1/5 can be written as 1 ÷ 5.

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1. Choose the decimal (s) from the brackets which is (are) not equivalent to the given decimals:

(i) 0.8 (0.80, 0.85, 0.800, 0.08)

(ii) 25.1 (25.01, 25.10, 25.100, 25.001)

(iii) 45.05 (45.050, 45.005, 45.500, 45.0500)

Solution:

(i) 0.8 (0.80, 0.85, 0.800, 0.08)

We know that 0.85 and 0.08 are not equivalent to the given decimal

For 0.85, 5 is in the hundredth place and for 0.8 the hundredth value is 0.

For 0.08, 0 is in the tenth place and for 0.8 the tenth value is 8.

(ii) 25.1 (25.01, 25.10, 25.100, 25.001)

For 25.01, 0 is in the tenth place and for 25.001 the tenth value is 0.

(iii) 45.05 (45.050, 45.005, 45.500, 45.0500)

For 45.005, 0 is in the hundredth place and for 45.05 the hundredth value is 5

For 45.500, 5 is in the tenth place and for 45.05 the tenth value is 0.

2. Which of the following are like decimals:

(i) 0.34, 0.07, 5.35, 24.70

(ii) 45.05, 4.505, 20.55, 20.5

(iii) 8.80, 17.08, 8.94, 0.27

(iv) 4.50, 16.80, 0.700, 7.08

Solution:

(i) 0.34, 0.07, 5.35, 24.70

The given values are like decimals because equal number of digits are present after the decimal point.

(ii) 45.05, 4.505, 20.55, 20.5

The given values are unlike decimals because different number of digits are present after the decimal point.

(iii) 8.80, 17.08, 8.94, 0.27

The given values are like decimals because equal number of digits are present after the decimal point.

(iv) 4.50, 16.80, 0.700, 7.08

The given values are unlike decimals because different number of digits are present after the decimal point.

3. Which of the following statements are correct?

(i) 8.05 and 7.95 are like decimals.

(ii) 0.95, 0.306, 7.10 are unlike decimals.

(iii) 3.70 and 3.7 are like decimals.

(iv) 13.59, 1.359, 135.9 are like decimals.

(v) 5.60, 3.04, 0.45 are like decimals.

Solution:

(i) Correct because the two decimals have same number of digits present after decimal point.

(ii) Correct because the three decimals have different number of digits present after decimal point.

(iii) Incorrect because the two decimals have different number of digits present after decimal point.

(iv) Incorrect because the three decimals have different number of digits present after decimal point.

(v) Correct because the three decimals have same number of digits present after decimal point.

4. Convert each of the following sets of unlike decimals to like decimal:

(i) 7.8, 7.85

(ii) 2.02, 3.2

(iii) 0.6, 5.8, 12.765

(iv) 5.296, 5.2, 5.29

(v) 4.3294, 43.29, 432.94

Solution:

(i) 7.8, 7.85

In the given values we know that 7.85 contains 2 digits after decimal point so by changing 7.8 as 7.80

Hence, 7.80 and 7.85 are like decimals.

(ii) 2.02, 3.2

In the given values we know that 2.02 contains 2 digits after decimal point so by changing 3.2 as 3.20

Hence, 2.02 and 3.20 are like decimals.

(iii) 0.6, 5.8, 12.765

In the given values we know that 12.765 contains 3 digits after decimal point so by changing 0.6 as 0.600 and 5.8 as 5.800

Hence, 0.600, 5.800 and 12.765 are like decimals.

(iv) 5.296, 5.2, 5.29

In the given values we know that 5.296 contains 3 digits after decimal point so by changing 5.2 as 5.200 and 5.29 as 5.290

Hence, 5.296, 5.200 and 5.290 are like decimals.

(v) 4.3294, 43.29, 432.94

In the given values we know that 4.3294 contains 4 digits after decimal point so by changing 43.29 as 43.2900 and 432.94 as 432.9400

Hence, 4.3294, 43.2900 and 432.9400 are like decimals.

1. Write each of the following products in exponential form:

(i) a × a × a × a × …….. 15 times

(ii) 8 × b × b × b × a × a × a × a

(iii) 5 × a × a × a × b × b × c × c × c

(iv) 7 × a × a × a …….. 8 times × b × b × b × …… 5 times

(v) 4 × a × a × …… 5 times × b × b × ……. 12 times × c × c …… 15 times

Solution:

(i) a × a × a × a × …….. 15 times is written in exponential form as a15.

(ii) 8 × b × b × b × a × a × a × a is written in exponential form as 8a4b3.

(iii) 5 × a × a × a × b × b × c × c × c is written in exponential form as 5a3b2c3.

(iv) 7 × a × a × a …….. 8 times × b × b × b × …… 5 times is written in exponential form as 7a8b5.

(v) 4 × a × a × …… 5 times × b × b × ……. 12 times × c × c …… 15 times is written in exponential form as 4a5b12c15.

2. Write each of the following in the product form:

(i) a2 b5

(ii) 8x3

(iii) 7a3b4

(iv) 15 a9b8c6

(v) 30x4y4z5

(vi) 43p10q5r15

(vii) 17p12q20

Solution:

(i) a2 b5 is written in the product form as a × a × b × b × b × b × b.

(ii) 8x3 is written in the product form as 8 × x × x × x.

(iii) 7a3b4 is written in the product form as 7 × a × a × a × b × b × b × b.

(iv) 15 a9b8c6 is written in the product form as 15 × a × a …… 9 times × b × b × … 8 times × c × c × ….. 6 times.

(v) 30x4y4z5 is written in the product form as 30 × x × x × x × x × y × y × y × y × z × z × z × z × z.

(vi) 43p10q5r15 is written in the product form as 43 × p × p …. 10 times × q × q …. 5 times × r × r × …. 15 times.

(vii) 17p12q20 is written in the product form as 17 × p × p …. 12 times × q × q × ….. 20 times.

3. Write down each of the following in exponential form:

(i) 4a3 × 6ab2 × c2

(ii) 5xy × 3x2y × 7y2

(iii) a3 × 3ab2 × 2a2b2

Solution:

(i) 4a3 × 6ab2 × c2 is written in exponential form as 24a4b2c2.

(ii) 5xy × 3x2y × 7y2 is written in exponential form as 105x3y4.

(iii) a3 × 3ab2 × 2a2b2 is written in exponential form as 6a6b4.

4. The number of bacteria in a culture is x now. It becomes square of itself after one week. What will be its number after two weeks?

Solution:

Number of bacteria in a culture = x

It is given that

Number of bacteria becomes square of itself in one week = x2

So the number of bacteria after two weeks = (x2)2 = x4

Hence, the number of bacteria after two weeks is x4.

5. The area of a rectangle is given by the product of its length and breadth. The length of a rectangle is two-third of its breadth. Find its area if its breadth is x cm.

Solution:

It is given that

Area of rectangle = l × b

Breadth = x cm

Length = (2/3) x cm

So the area of the rectangle = (2/3) x × x = 2/3 x2 cm2

Hence, the area of rectangle is (2/3) x2 cm2.

6. If there are x rows of chairs and each row contains x2 chairs. Determine the total number of chairs.

Solution:

Number of rows of chairs = x

Each row contains = x2 chairs

So the total number of chairs = number of rows of chairs × chairs in each row

We get

Total number of chairs = x × x2 = x3

Hence, the total number of chairs is x3.

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6. You are standing in a class-room facing north. In what direction are you facing after making a quarter turn?

Solution:

By making a quarter turn (90o), I will be facing towards east if I turn to my right hand and if I turn to my left hand, I will be facing towards west.

7. A bicycle wheel makes four and a half turns. Find the number of right angles through which it turns.

Solution:

We know that the wheel of a bicycle covers 360o in one turn.

It can be written as

360/90 = 4 right angles

We know that in four and half turns the wheel turns by 4 (4.5) = 18 right angles

Hence, the number of right angles through which it turns is 18.

8. Look at your watch face. Through how many right angles does the minute-hand moves between 8: 00 O’ clock and 10: 30 O’ clock?

Solution:

We know that the time interval between 8: 00 O’ clock and 10: 30 O’ clock is two and half hours

The minute hand turns 360o in 1 hour

360/90 = 4 right angles

So in two and half hours the minute hand turns by 2.5 (4) = 10 right angles.

Hence, the minute hand turns by 10 right angles.

9. If a bicycle wheel has 48 spokes, then find the angle between a pair of adjacent spokes.

Solution:

The central angle in a bicycle is 360o which consists of 48 spokes.

So the angle between a pair of adjacent spokes = 360/48 = 7.5o

Hence, the angle between a pair of adjacent spokes is 7.5o.

Types of Angles

• Acute Angle: An angle whose measure is less than 90° is called an acute angle.
• Right Angle: An angle whose measure is 90° is called right angle.
• Obtuse Angle: An angle whose measure is greater than 90° but less than 180° is called an obtuse angle.
• Straight Angle: 
• Reflex Angle:
• Zero Angle:

What is the measure of right angle Class 6?

Answer: An acute angle measures between 0° and 90°; an obtuse angle measures between 90° and 180°; a straight angle measures 180°; a right angle measures 90°; a zero angle measures 0° and a complete angle measures 360°.

What type of angle is 26 degrees?

acute angle-an angle between 0 and 90 degrees. right angle-an 90 degree angle. obtuse angle-an angle between 90 and 180 degrees.

 

(i) Triangle – A plane figure formed by three non-parallel line segments is called a triangle.

(ii) Parts or elements of a triangle – The three sides, three angles and three vertices of a triangles are called parts or elements of a triangle.

(iii) Scalene triangle – A triangle in which all the sides are different is called a scalene triangle.

(iv) Isosceles triangle – A triangle whose two sides are equal is called an isosceles triangle.

(v) Equilateral triangle – A triangle whose all sides are equal to one another is called an equilateral triangle.

(vi) Acute triangle – A triangle whose all the angles are acute is called acute-angled triangle or an acute triangle.

(vii) Right triangle – A triangle whose one angle is a right angle is called a right-angled triangle or a right triangle.

(viii) Obtuse triangle – A triangle whose one angle is obtuse is called an obtuse-angled triangle or an obtuse triangle.

(ix) Interior of a triangle – The region which lies inside the triangle is called the interior of a triangle.

(x) Exterior of a triangle – The region which lies outside the triangle is called the exterior of a triangle.

What is the correct name for a quadrilateral?

Parallelogram is the proper name of the quadrilateral.

Is Triangle a quadrilateral?

A quadrilateral is a polygon. In fact it is a 4-sided polygon, just like a triangle is a 3-sided polygon, a pentagon is a 5-sided polygon, and so on.

What makes a shape a quadrilateral?

A quadrilateral is a polygon that has exactly four sides. (This also means that a quadrilateral has exactly four vertices, and exactly four angles.)

What is a quadrilateral 3rd grade?

A quadrilateral is a 4-sided two-dimensional shape. There are 4 types of quadrilaterals: parallelograms, trapezoids, rectangles, and rhombuses. A rectangle has 4 right angles. A rhombus is a quadrilateral with all 4 sides of the same length.

1. Explain the following:

(i) Circle

(ii) Radius

(iii) Centre

(iv) Diameter

(v) Chord

(vi) Interior of a circle.

Solution:

(i) Circle – A circle is a set of all those points in a plane whose distance from a fixed point remains constant.

(ii) Radius – The radius of a circle is the distance between the all the points of the circle to its centre.

(iii) Centre – The centre of a circle is a fixed point which is at a constant distance from all the points.

(iv) Diameter – A line segment passing through the centre of a circle, and having its end-points on the circle is called a diameter of the circle.

(v) Chord – A line segment with its end-points lying on a circle is called the chord of the circle.

(vi) Interior of a circle – The part of a plane inside the circle consisting of all the points is called the interior of a circle.

What is diameter of a circle class 6?

Diameter of a circle is a chord of the circle, not radius. It is the longest chord of the circle. A perpendicular drawn from the centre of the circle to the chord, bisects the chord. It is a line passing through the circle that intersects the circle at two points.

State the properties of a transversal.

The properties of a transversal are that first one being over here, the vertically opposite angles are equal. Further, the corresponding angles are equal and the interior angles which form on the same side of the transversal are supplementary. Finally, the alternate angles are equal.

Name any four objects from your environment, which have the form of

(i) a cuboid (ii) a cube

Solution:

(i) The objects which have the form of a cuboid are a compass box, a lunch box, a duster and a book.

(ii) The objects which have the form of a cube are a chalk box, a disc, a tissue box and a cubical cabin.

How do you describe a three-dimensional shape?

In geometry, a three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.

Why is it important to learn about 3D shapes?

Learning shapes not only helps children identify and organize visual information, it helps them learn skills in other curriculum areas including reading, math, and science. ... Learning shapes also helps children understand other signs and symbols.

What is line of symmetry class 6?

When a figure is folded along a line such a way that the two parts exactly fit on top of each other, then the figure is said to have a line symmetry. Line of symmetry is the line which divides figure into two identical parts and these are mirror image of each others.

Why is symmetry important?

Symmetry is a fundamental part of geometry, nature, and shapes. It creates patterns that help us organize our world conceptually. We see symmetry every day but often don't realize it. Geometry software is a very important tool in developing and testing individual ideas in the classroom.

Which triangle has 2 lines of symmetry?

The division of triangles into scalene, isosceles, and equilateral can be thought of in terms of lines of symmetry. A scalene triangle is a triangle with no lines of symmetry while an isosceles triangle has at least one line of symmetry and an equilateral triangle has three lines of symmetry.

What are the steps in writing a construction class 6?

Steps for construction:

1. Draw a line BX and take a point A, such that AB is equal to 4.5 cm.
2. Draw ∠ABP = 60° with the help of protractor.
3. With A as the centre and a radius of 5 cm, draw an arc cutting PB at C.
4. Draw AC.
5. Now, draw ∠BCY = 60°.
6. Then, draw ∠ABW, such that ∠ABW is equal to∠CAX, which cut the ray CY at D

Introduction to Mensuration

Mensuration

• Mensuration is the branch of mathematics that deals with the measurement of length, area or volume of various geometric shapes.

Shapes

• A shape is the form of an object.
• Examples of two-dimensional shapes are square, rectangle and triangle, and of three-dimensional shapes are cube, cuboid and sphere.

What is the difference between mensuration and geometry?

Mensuration refers to the calculation of various parameters of shapes like the perimeter, area, volume, etc. whereas, Geometry deals with the study of properties and relations of points and lines of various shapes.

What are 2D and 3D Mensuration?

2D mensuration deals with the calculation of various parameters like area and perimeter of 2-dimensional shapes like square, rectangle, circle, triangles, etc.

3D mensuration is concerned with the study and calculation of surface area, lateral surface area, and volume of 3-dimensional figures like a cube, sphere, cuboid, cone, cylinder, etc.

Types of data

Data handling method can be performed based on the types of data. The data is classified into two types, such as:

• Qualitative Data
• Quantitative Data

Qualitative data gives descriptive information of something whereas quantitative data gives numerical information about something. Here, the quantitative data is further divided into two. They are discrete data and continuous data. The discrete data can take only certain values such as whole numbers. The continuous data can take a value within the provided range.

How to Represent Data?

The data can be usually represented in any one of the following ways. They are:

• Bar Graph
• Line Graphs
• Pictographs
• Histograms
• Stem and Leaf Plot
• Dot Plots
• Frequency Distribution
• Cumulative Tables and Graphs