Perimeter and Area – Complete Guide For Class 6th Math Chapter 6

Welcome to iPrep, your Learning Super App. Our learning resources for the chapter, Perimeter and Area in Mathematics for Class 6th are designed to ensure that you grasp this concept with clarity and perfection. Whether you’re studying for an upcoming exam or strengthening your concepts, our engaging animated videos, practice questions and notes offer you the best of integrated learning with interesting explanations and examples. 

The chapter Perimeter and Area introduces students to the fundamental concepts of measuring the boundaries and enclosed spaces of various shapes. The perimeter is defined as the total length around a figure, such as a rectangle, square, or any polygon, calculated by summing the lengths of all its sides. The area, on the other hand, measures the space within a closed figure and is typically expressed in square units. Through practical examples and activities, students develop a solid understanding of how to apply these concepts to solve real-world problems, laying the groundwork for more advanced geometric studies.

Perimeter and Area

Perimeter

The perimeter of any closed plane figure is the total distance covered along its boundary when you trace around it once. 

For polygons, which are closed plane figures made up of straight line segments, the perimeter is simply the sum of the lengths of all its sides.

Perimeter Formulas for Common Shapes

Perimeter of Rectangle

As stated in the chapter perimeter and area, the perimeter of a rectangle is calculated by adding together the lengths of all four sides. A simpler formula is to double the sum of the rectangle’s length and breadth:

Perimeter of a rectangle = 2 × (length + breadth)

For example, if a rectangle has a length of 12 cm and a breadth of 8 cm, its perimeter would be 40 cm.

an image depicting a rectangle from class 6 math chapter 6 - Perimeter and Area

Perimeter of rectangle = Sum of the lengths of its four sides 

                                            = AB + BC + CD + DA

                                            = AB + BC + AB + BC

                                            = 2 × AB + 2 × BC 

                                            = 2 × (AB + BC) 

                                            = 2 × (12 cm + 8 cm)

                                            = 2 × (20 cm) = 40 cm.

From this example, we see that the perimeter of a rectangle = length + breadth + length + breadth. 

The perimeter of a rectangle = 2 × (length + breadth). 

The perimeter of a rectangle is twice the sum of its length and breadth.

Perimeter of Square

As mentioned in the chapter perimeter and area, Since all four sides of a square are equal, the perimeter is four times the length of one side:

The perimeter of a square = 4 × length of a side

For instance, a square with a side length of 1 meter would have a perimeter of 4 meters.

image 179

The perimeter of the square = sum of the lengths of all four sides of the square 

                                               = 1 m + 1 m + 1 m + 1 m = 4 m. 

Now, we know that all four sides of a square are equal in length. 

Therefore, in place of adding the lengths of each side, we can simply multiply the length of one side by 4. 

Thus, the length of the tape required = 4 × 1 m = 4 m. 

From this example, we see that the Perimeter of a square = 4 × length of a side. 

The perimeter of a square is quadruple the length of its side.

The Perimeter of a Triangle

The chapter perimeter and area also cover The perimeter of a triangle, which is the sum of the lengths of its three sides. 

For example, Consider a triangle having three given sides of lengths 4 cm, 5 cm and 7 cm. Find its perimeter. 

image 182

The perimeter of the triangle = 4 cm + 5 cm + 7 cm = 16 cm. 

The perimeter of a triangle is = sum of the lengths of its three sides

Practical Examples of Common Shapes

Example 1: Akshi wants to put lace around a rectangular tablecloth that is 3 meters long and 2 meters wide. 

Solution: Length of the rectangular table cover = 3 m. 

The breadth of the rectangular table cover = 2 m. 

Akshi wants to put lace all around the tablecloth. 

Therefore, the length of the lace required will be the perimeter of the rectangular tablecloth. 

Now, the perimeter of the rectangular tablecloth = 2 × (length + breadth) = 2 × (3 m + 2 m) = 2 × 5 m = 10 m. 

Hence, the length of the lace required is 10 m.

Example 2: Usha takes three rounds of a square park with each side measuring 75 meters. 

image 183

Solution

Perimeter of the square park = 4 × length of a side = 4 × 75 m = 300 m. Distance covered by Usha in one round = 300 m. 

Therefore, the total distance traveled by Usha in three rounds = 3 × 300 m = 900 m.

Deep Dive Activity

Understanding Perimeters through Real-Life Applications

In races, tracks often share a common finish line. 

Imagine two square tracks: the inner track has sides of 100 meters, while the outer track has sides of 150 meters. 

If the race is 350 meters long, where should the starting positions be to ensure both runners finish at the same point? This type of problem helps students apply perimeter concepts to real-world situations.

Explore and Verify

To better understand perimeter, try this activity: 

Take a piece of paper, cut it into different shapes, and estimate the perimeter. Then, measure the actual perimeter with a scale or tape measure to see how close your estimation was.

Perimeter of Regular Polygons

Regular polygons are closed figures with all sides and angles equal, like equilateral triangles or regular pentagons. 

The perimeter of a regular polygon is the length of one side multiplied by the number of sides.

Perimeter of Equilateral Triangle

We know that for any triangle its perimeter is sum of all three sides.  

Using this understanding, we can find the perimeter of an equilateral triangle. 

image 182

Perimeter of an equilateral triangle = AB + BC + AC 

                                                                    = AB + AB + AB (all sides are equal)

                                                                   = 3 times length of one side.

Perimeter of an equilateral triangle = 3 × length of a side

Understanding Area

The chapter perimeter and area covers a deep understanding of the concept of Area. This involves-

Area

The area of a closed figure is the amount of region enclosed within its boundaries. This concept was introduced in earlier grades using square grid paper.

Area of a Rectangle

Area = length × width

Area of a Square

Area = side length × side length

Example 1: 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.

Solution

Length of the floor = 5 m. 

Width of the floor = 4 m. 

Area of the floor = length × width = 5 m × 4 m = 20 sq m. 

Length of the square carpet = 3 m. 

Area of the carpet = length × length = 3 m × 3 m = 9 sq m. 

Hence, the area of the floor laid with carpet is 9 sq m. 

Area of the floor that is not carpeted is =  area of the floor – the area of the floor laid with carpet 

             = 20 sq m – 9 sq m 

             = 11 sq m.

Example 2: Four square flower beds each of side 4 m are in four corners on a piece of land 12 m long and 10 m wide. Find the area of the remaining part of the land.  

Solution

Length of the land (l) = 12 m. 

Width of land (w) = 10 m. 

Area of the whole land = l × w = 12 m × 10 m = 120 sq m. 

The sidelength of each of the four square flower beds is (s) = 4 m. 

Area of one flower bed = s × s = 4 m × 4 m = 16 sq m. 

Hence, the area of the four flower beds = 4 × 16 sq m = 64 sq m. 

Area of the remaining part of the land = area of the complete land – the area of all four flower beds 

                             = 120 sq m – 64 sq m 

                             = 56 sq m.

Estimating Area Using Graph Paper

The chapter perimeter and area also covers estimating area using graph paper. You can estimate the area of any simple closed shape using graph paper by counting the full squares and appropriately accounting for partial squares.

Look at the figures below and guess which one of them has a larger area.

image 184

We can estimate the area of any simple closed shape by using a sheet of squared paper or graph paper where every square measures 1 unit × 1 unit or 1 square unit. 

To estimate the area, we can trace the shape onto a piece of transparent paper and place the same on a piece of squared or graph paper and then follow the below conventions— 

1. The area of one full small square of the squared or graph paper is taken as 1 sq unit. 

2. Ignore portions of the area that are less than half a square. 

3. If more than half of a square is in a region, just count it as 1 sq unit. 

Area of a Triangle

The chapter Perimeter and area covers the area of a triangle. It includes-

Understanding Triangle Area

Activity: Cut a rectangle along its diagonal to form two triangles. 

Notice that these triangles have the same area, indicating that the area of each triangle is half the area of the rectangle.

image 181

Formula: Area of a Triangle: For any triangle formed by drawing a diagonal in a rectangle, the area of the triangle is half the area of the rectangle.

Find the Area of triangle ABE?

Solution

The  area of triangle BAD is half of the area of the rectangle ABCD.

Area of triangle ABE = Area of triangle AEF + Area of triangle BEF.

 Here, the area of triangle AEF = half of the area of rectangle AFED. 

Similarly, the area of triangle BEF = half of the area of rectangle BFEC. 

Thus, the area of triangle ABE = half of the area of rectangle AFED +half of the area of rectangle BFEC.

= half of the sum of the areas of the rectangles AFED and BFEC 

= half of the area of the rectangle ABCD.  

Perimeter and Area Relationship

Perimeter: The sum of all the sides of a polygon.

Area: The measure of the enclosed space.

Key Insight: Two shapes can have the same area but different perimeters, or the same perimeter but different areas.

Let’s Conclude

In summary, CBSE Class 6th Math, Chapter 6 – Perimeter and Area equips students with essential skills to measure and understand both the boundaries and the spaces within various shapes. The chapter on Perimeter and Area provides clear explanations of how to calculate the perimeter of rectangles, squares, and triangles, alongside real-life applications and practical examples. Similarly, the section on Area delves into methods for finding the area of rectangles, squares, and triangles, enhancing students’ ability to apply these concepts in everyday situations.

Through engaging activities, such as estimating perimeter and areas with graph paper and practical problems like measuring lace around a tablecloth or calculating distances in a park, students gain a comprehensive grasp of the concepts. The relationship between perimeter and area further reinforces the understanding that while two shapes can share the same area, their perimeters might differ, and vice versa. By mastering these principles, students are well-prepared for more advanced mathematical concepts and applications.

The chapter Perimeter and Area is a foundational element of geometry that helps bridge the gap between basic and more complex geometric studies. Remember, the chapter Perimeter and Area not only builds mathematical skills but also encourages students to appreciate the practical uses of geometry in the real world.

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Prime Time – Complete Guide For Class 6 Math Chapter 5

Welcome to iPrep, your Learning Super App. Our learning resources for the chapter, Prime Time in Mathematics for Class 6th are designed to ensure that you grasp this concept with clarity and perfection. Whether you’re studying for an upcoming exam or strengthening your concepts, our engaging animated videos, practice questions and notes offer you the best of integrated learning with interesting explanations and examples. 

The chapter Prime Time introduces students to the fundamental concepts of prime and composite numbers. It explores the idea that a prime number has exactly two distinct factors: 1 and the number itself, while a composite number has more than two factors. The chapter also covers the concept of divisibility, helping students understand how to determine if one number is divisible by another. Additionally, students learn about the importance of prime factorization, which involves breaking down a number into its prime factors, a crucial skill that lays the foundation for more advanced topics in mathematics.

Prime Time

Common Multiples and Common Factors

Let’s start the chapter of prime time with a fun game (Idli-Vada Game) that you can play with your friends! Imagine you’re sitting in a circle with other children, and the game is all about numbers.

  • One child starts by saying “1”.
  • The second player says “2”, and so on.
  • But when it’s the turn of multiples of 3 (like 3, 6, 9…), the player should say “idli” instead of the number.
  • When it’s the turn of multiples of 5 (like 5, 10, 15…), the player should say “vada” instead of the number.
  • If a number is a multiple of both 3 and 5 (like 15), the player should say “idli-vada”!

If a player makes a mistake, they are out of the game. The game continues in rounds until only one player remains.

Key Questions:

  • For which numbers should players say “idli”? These would be 3, 6, 9, 12, 15, 18, and so on.
  • For which numbers should players say “vada”? These would be 5, 10, 15, 20, 25, and so on.
  • Which is the first number for which players should say “idli-vada”? The answer is 15, as it is a multiple of both 3 and 5. Such numbers are called Common Multiples.

Jump Jackpot: A Game of Multiples

Let’s understand the chapter prime time further with a game. In this game, Jumpy and Grumpy are playing a treasure hunt game. Grumpy places a treasure on a number, say 24, and Jumpy has to jump on multiples of a chosen jump size, starting from 0, to reach the treasure.

For example, if Jumpy chooses a jump size of 4, he will jump on 4 → 8 → 12 → 16 → 20 → 24, successfully landing on 24.

Other successful jump sizes for 24 include 2, 3, 6, 8, 12, and 24.

But what happens if there are two treasures, placed on different numbers like 14 and 36? Jumpy needs to choose a jump size that lands on both numbers.

  • Factors of 14: 1, 2, 7, 14
  • Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

Common Factors of 14 and 36 are 1 and 2. So, Jumpy should choose a jump size of 1 or 2 to land on both treasures. 

Common Factors or Divisors are the numbers that can divide both numbers exactly.

Perfect Numbers

As stated in the chapter prime time, a number for which the sum of all its factors is equal to twice the number is called a Perfect Number

The number 28 is an example of a perfect number. Its factors are 1, 2, 4, 7, 14, and 28. Now let’s explore another very important topic of the chapter Prime Time, named perfect numbers.

Prime Numbers

Definition: Numbers with only two factors are called Prime Numbers.

Example:

Guna and Anshu are arranging figs (anjeer) in boxes.

  • Guna wants to put 12 figs in each box.
  • Anshu wants to put 7 figs in each box.

Guna can arrange the 12 figs in multiple ways:

  • 1 row of 12 figs
  • 2 rows of 6 figs
  • 3 rows of 4 figs
  • 4 rows of 3 figs
  • 6 rows of 2 figs

However, Anshu can only arrange the 7 figs in one way:

  • 1 row of 7 figs

This difference is because 12 has more than two factors, while 7 has only two factors—1 and 7. Now let’s explore another topic of the chapter prime time named Prime vs. Composite Numbers.

Prime vs. Composite Numbers

  • Prime Numbers have only two factors: 1 and the number itself. Examples include 2, 3, 5, 7, 11, 13, 17, 19.
  • Composite Numbers have more than two factors. Examples include 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20.

Note: The number 1 is neither a prime nor a composite number.

The Sieve of Eratosthenes: Finding Prime Numbers

image 178

Here’s an ancient method to find prime numbers:

  1. Cross out 1 because it is neither prime nor composite.
  2. Circle 2, then cross out all multiples of 2.
  3. Circle 3, then cross out all multiples of 3.
  4. Circle 5, then cross out all multiples of 5.

Continue this process until all numbers are either circled (primes) or crossed out (composites). This method is called the Sieve of Eratosthenes. Now let’s explore the another topic of prime time names twin primes.

Twin Primes

According to the chapter Prime Time, twin primes are pairs of prime numbers with a difference of 2. Examples include 3 and 5, 17 and 19. 

Co-prime Numbers: Safe Keeping Treasures

In the treasure hunt game, Jumpy has to reach two treasures placed on different numbers using the same jump size, but with a new rule: the jump size of 1 is not allowed.

To ensure Jumpy cannot reach both treasures, Grumpy should place them on numbers that have no common factors other than 1, called Co-prime Numbers.

For example:

  • 12 and 26 are not co-prime because they have a common factor of 2.
  • 4 and 9 are co-prime because they have no common factors other than 1.

Prime Factorisation: Breaking Down Numbers

As stated in the chapter Prime Time, when checking if two numbers are co-prime, a systematic approach called Prime Factorisation is used.

For example:

  • Prime Factorisation of 56: 56 = 2 × 2 × 2 × 7
  • Prime Factorisation of 63: 63 = 3 × 3 × 7

These factors are all primes, and the product of these prime factors gives the original number.

Key Points:

  • Every number greater than 1 has a prime factorization. 
  • The idea is the same: keep breaking the composite numbers into factors till only primes are left. 
  • The number 1 does not have any prime factorization. It is not divisible by any prime number.

Here, you see four different ways to get a prime factorization of 36. Observe that in all four cases, we get two 2s and two 3s. Multiply back to see that you get 36 in all four cases.

An illustration of prime factorization from the chapter prime time from class 6 math

Does the order matter?

When multiplying numbers, we can do so in any order. The end result is the same. That is why, when two 2s and two 3s are multiplied in any order, we get 36.

Thus, the order does not matter. Usually, we write the prime numbers in increasing order. For example, 225 = 3 × 3 × 5 × 5 or 30 = 2 × 3 × 5. 

Prime Factorization of a Product of Two Numbers

When we perform prime factorization, we start by expressing a number as a product of two factors. For example, 72 can be written as 12 × 6. We then find the prime factorization of each factor:

  • 12 = 2 × 2 × 3
  • 6 = 2 × 3

By combining these, the prime factorization of 72 is:

  • 72 = 2 × 2 × 3 × 2 × 3
  • This can also be written as 2 × 2 × 2 × 3 × 3.

Always remember to multiply the factors to verify that you get the original number, 72, in this case. 

Notice how each prime factor appears a specific number of times in the factorization of 72, compared to how they appear in the factorization of 12 and 6 individually.

Using Prime Factorization to Check Co-prime Numbers

Co-prime numbers are pairs of numbers that have no common prime factors. Let’s explore this with an example:

  • 56 = 2 × 2 × 2 × 7
  • 63 = 3 × 3 × 7

Since 7 is a common prime factor in both 56 and 63, they are not co-prime.

Consider another pair:

  • 80 = 2 × 2 × 2 × 2 × 5
  • 63 = 3 × 3 × 7

Here, there are no common prime factors, so 80 and 63 are co-prime.

Additional Examples:

  • 40 = 2 × 2 × 2 × 5
  • 231 = 3 × 7 × 11

Since they have no common prime factors, 40 and 231 are co-prime.

  • 242 = 2 × 11 × 11
  • 195 = 3 × 5 × 13

Again, no common prime factors exist, so 242 and 195 are co-prime.

Using Prime Factorization to Check Divisibility

If one number is divisible by another, the prime factorization of the second number will be a part of the prime factorization of the first. Let’s look at an example:

Example 1: Is 168 divisible by 12?

  • 168 = 2 × 2 × 2 × 3 × 7
  • 12 = 2 × 2 × 3

Since the prime factors of 12 are included in the prime factorization of 168, 168 is divisible by 12.

Example 2: Is 75 divisible by 21?

  • 75 = 3 × 5 × 5
  • 21 = 3 × 7

Since 7 is a prime factor of 21 but not of 75, 75 is not divisible by 21.

Example 3: Is 42 divisible by 12?

  • 42 = 2 × 3 × 7
  • 12 = 2 × 2 × 3

Since factor 2 appears twice in 12 but only once in 42, 42 is not divisible by 12.

Divisibility Tests: Simplifying Factorization for Large Numbers

As stated in the chapter Prime Time, finding factors of smaller numbers is straightforward, but what about larger numbers? 

Let’s take 8560 as an example. Does it have any factors from 2 to 10? 

We can easily determine this without long division by applying simple divisibility rules.

  1. Divisibility by 10

       To check if a number is divisible by 10, observe the last digit:

  • Numbers ending in 0 are divisible by 10.

             Example: Is 8560 divisible by 10?

                                Yes, because it ends in 0.

  1. Divisibility by 5

             Similarly, a number is divisible by 5 if it ends in 0 or 5.

            Example: Is 8560 divisible by 5?

                   Yes, because it ends in 0.

  1. Divisibility by 2

              A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.

              Example: Is 8560 divisible by 2?

                 Yes, because it ends in 0.

  1. Divisibility by 4

Checking divisibility by 4 involves looking at the last two digits. If they form a number divisible by 4, then the whole number is divisible by 4.

              Example: Is 8536 divisible by 4?

                 Yes, because 36 is divisible by 4.

  1. Divisibility by 8

For divisibility by 8, check the last three digits. If they form a number divisible by 8, then the entire number is divisible by 8.

             Example: Is 8560 divisible by 8?

                  No, because 560 is not divisible by 8.

Special Numbers and Fun with Prime Numbers

As stated in the chapter Prime Time, special numbers often have unique characteristics. 

Let’s explore some examples:

Consider the numbers 9, 16, 25, 43. Each has something special:

  • 9 is the only single-digit number.
  • 16 is the only even number and a multiple of 4.
  • 25 is the only multiple of 5.
  • 43 is the only prime number and not a perfect square.

A Prime Puzzle

Lastly, try your hand at a prime number puzzle! It’s a fun way to engage with primes and their properties.

Rules:  Fill the grid with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column.

Example:

75
42
102
1703063X

Solution

55375
23742
1723102
1703063X

In conclusion, Chapter 5 of CBSE Class 6th Math, “Prime Time,” provides a foundational understanding of prime and composite numbers, along with essential concepts such as divisibility, common factors, and prime factorization. By exploring the distinction between prime and composite numbers, students can build a strong mathematical base. Prime Time is filled with interactive games and activities that make learning these topics engaging and fun. Whether it’s understanding prime factorization or playing with common multiples, the lessons in Prime Time are crucial for grasping higher-level math concepts. Be sure to explore all the resources iPrep offers for Prime Time to master these essential math skills!

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