NCERT Solutions for Class 6 Maths Chapter 6: Perimeter and Area

NCERT Solutions for Chapter 6 Class 6 Maths, "Perimeter and Area," covers the following exercises:

Exercise 6.1: Perimeter

This exercise introduces the concept of perimeter, guiding students on how to calculate the total distance around different shapes. It helps them understand practical applications of perimeter in real-life scenarios.

Exercise 6.2: Area

In this exercise, students learn about the area as the amount of space inside a shape. They practise using formulas to calculate the area of squares and rectangles, enhancing their understanding of measurement.

Exercise 6.3: Area of a Triangle

This exercise focuses on calculating the area of triangles. Students learn and practice applying the specific formula, which helps reinforce their knowledge of different geometric shapes.

Access NCERT Solutions for Class 6 Maths Chapter 6 Perimeter and Area

Exercise 6.1

A- Figure it Out 

1. Find the missing terms: 

a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?. 

b. Perimeter of a square = 20 cm; side of a length = ?. 

c. Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?. 

Ans:

a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length =?

Formula: Perimeter = 2 × (length + breadth)

14 = 2 × (length + 2)

14 = 2 × (length + 2)

14 ÷ 2 = length + 2

7 = length + 2

Length = 7 - 2 = 5 cm

b. Perimeter of a square = 20 cm; side length = ?

Formula: Perimeter = 4 × side

20 = 4 × side

Side = 20 ÷ 4 = 5 cm

c. Perimeter of a rectangle = 12 m; length = 3 m; breadth =?

Formula: Perimeter = 2 × (length + breadth)

12 = 2 × (3 + breadth)

12 ÷ 2 = 3 + breadth

6 = 3 + breadth

Breadth = 6 - 3 = 3 m

Final answers: a. Length = 5 cm b. Side = 5 cm c. Breadth = 3 m

2. A rectangle having sidelengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square? 

Ans: To solve this problem:

1. Find the perimeter of the rectangle:

The formula for the perimeter of a rectangle is:

\[     \text{Perimeter} = 2 \times (\text{length} + \text{breadth})     \]

For the given rectangle with side lengths of 5 cm and 3 cm:

\[     \text{Perimeter} = 2 \times (5 + 3) = 2 \times 8 = 16 \, \text{cm}  \]

So, the total length of the wire is 16 cm.

2. When the wire is bent to form a square, the perimeter of the square will also be 16 cm.

3. Find the side length of the square:

The formula for the perimeter of a square is:

\[     \text{Perimeter} = 4 \times \text{side}  \]

Given that the perimeter is 16 cm:

\[ 16 = 4 \times \text{side}     \]

Solving for side:

\[     \text{side} = 16 \div 4 = 4 \, \text{cm}  \]

Therefore, the length of each side of the square will be 4 cm.

3. Find the length of the third side of a triangle having a perimeter of 55 cm and having two sides of length 20 cm and 14 cm, respectively. 

Ans: The perimeter of the triangle is the sum of all three sides.

Given two sides are 20 cm and 14 cm, let the third side be \( x \).

\[\text{Perimeter} = 20 + 14 + x\]

\[55 = 34 + x\]

\[x = 55 - 34 = 21 \, \text{cm}\]

The length of the third side is 21 cm.

4. What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs `40 per metre? 

Ans: First, find the perimeter of the park.

The perimeter of a rectangle is given by:

\[\text{Perimeter} = 2 \times (\text{length} + \text{breadth}) = 2 \times (150 + 120) = 2 \times 270 = 540 \, \text{m}\]

 Now, calculate the cost of fencing:

\[\text{Cost} = \text{Perimeter} \times \text{Cost per metre} = 540 \times 40 = 21,600 \, \text{₹}\]

The cost of fencing the park is ₹21,600.

5. A piece of string is 36 cm long. What will be the length of each side, if it is used to form: 

a. A square, 

b. A triangle with all sides of equal length, and 

c. A hexagon (a six sided closed figure) with sides of equal length? 

Ans:

a. A square:

The perimeter of the square is 36 cm, so each side of the square is: 

\[\text{Side length} = \frac{\text{Perimeter}}{4} = \frac{36}{4} = 9 \, \text{cm}\]

b. A triangle with all sides of equal length:

The perimeter of the triangle is 36 cm, so each side of the equilateral triangle is:

 \[\text{Side length} = \frac{\text{Perimeter}}{3} = \frac{36}{3} = 12 \, \text{cm}\]

c. A hexagon (six-sided closed figure) with sides of equal length:

The perimeter of the hexagon is 36 cm, so each side of the regular hexagon is:

\[\text{Side length} = \frac{\text{Perimeter}}{6} = \frac{36}{6} = 6 \, \text{cm}\]

a. Side length of the square = 9 cm

b. Side length of the triangle = 12 cm

c. Side length of the hexagon = 6 cm

6. A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed?

Ans: To find the total length of rope needed to fence the rectangular field with 3 rounds of rope, follow these steps:

1. Find the perimeter of the rectangular field:

The formula for the perimeter of a rectangle is:

\[     \text{Perimeter} = 2 \times (\text{length} + \text{breadth})  \]

Given the length is 230 m and the breadth is 160 m:

\[     \text{Perimeter} = 2 \times (230 + 160) = 2 \times 390 = 780 \, \text{m}  \]

2. Calculate the total length of rope for 3 rounds:

Since the farmer wants 3 rounds of rope, the total length of rope required is:

\[     \text{Total rope length} = 3 \times \text{Perimeter} = 3 \times 780 = 2340 \, \text{m} \]

So, the total length of rope needed is 2340 m.

B- Figure it Out

Each track is a rectangle. Akshi’s track has length 70 m and breadth 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in one round.

1. Find out the total distance Akshi has covered in 5 rounds. 

2. Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance? 

3. Think and mark the positions as directed— 

a. Mark ‘A’ at the point where Akshi will be after she ran 250 m. 

b. Mark ‘B’ at the point where Akshi will be after she ran 500 m. 

c. Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘C’. 

d. Mark ‘X’ at the point where Toshi will be after she ran 250 m. 

e. Mark ‘Y’ at the point where Toshi will be after she ran 500 m. 

f. Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘Z’

Ans:

1. Total distance Akshi covered in 5 rounds:

Each round is 220 meters (m).

Total distance = 5 × 220 m = 1100 m.

2. Total distance Toshi covered in 7 rounds:

   - Each round is 200 meters (m).

   - Total distance = 7 × 200 m = 1400 m.

3. Who ran the longer distance?

   - Toshi ran the longer distance, covering 1400 m, which is more than Akshi’s 1100 m.

For marking points A, B, C, X, and Y as instructed:

A: Mark the point where Akshi is after 250 m.

B: Mark the point where Akshi is after 500 m.

C: Mark the point where Akshi completes 1000 m.

X: Mark the point where Toshi is after 250 m.

Y: Mark the point where Toshi is after 500 m.

Exercise 6.2

Figure it Out

1. The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden? 

Ans: Given:

Length of the garden $ l = 25 \, \text{m} $

Area of the garden $ A = 300 \, \text{sq m} $

Formula:

\[A = l \times w\]

 Where:

 - $ A $ is the area,

 - $ l $ is the length,

 - $ w $ is the width.

 Solution:

 \[300 = 25 \times w\]

 Divide both sides by 25:

 \[w = \frac{300}{25} = 12 \, \text{m}\]

 The width of the garden is 12 meters.

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

Ans: Given:

Length of the plot $ l = 500 \, \text{m} $

Width of the plot $ w = 200 \, \text{m} $

Rate of tiling $ \text{₹} 8 $ per 100 square meters.

Formula:

The area of the plot is:

\[A = l \times w\]

Total cost of tiling is:

\[\text{Cost} = \left( \frac{A}{100} \right) \times \text{Rate per 100 sq m}\]

Solution:

Calculate the area of the plot:

\[A = 500 \times 200 = 100,000 \, \text{sq m}\]

Now calculate the cost:

\[\text{Cost} = \left( \frac{100,000}{100} \right) \times 8 = 1000 \times 8 = \text{₹} 8000\]

The cost of tiling the plot is ₹ 8000.

3. A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove? 

Ans: Given:

Length of the grove $ l = 100 \, \text{m} $

Width of the grove $ w = 50 \, \text{m} $

Area required for each coconut tree $ = 25 \, \text{sq m} $

Formula:

The area of the grove is:

\[A = l \times w\]

The number of trees that can be planted is:

 \[\text{Number of trees} = \frac{\text{Total area of grove}}{\text{Area per tree}}\]

Solution:

Calculate the area of the grove:

\[A = 100 \times 50 = 5000 \, \text{sq m}\]

Now calculate the number of trees:

\[\text{Number of trees} = \frac{5000}{25} = 200\]

The maximum number of trees that can be planted is 200.

4. By splitting the following figures into rectangles, find their areas (all measures are given in metres).

Ans:

To solve these problems, we will split each figure into smaller rectangles, calculate the area of each rectangle, and then sum the areas to get the total area for each figure.

(a) Figure a

We can split the figure into three smaller rectangles as follows:

1. Rectangle 1 (top-right rectangle):

Length = 3 meters

Width = 1 meter

Area = \( 3 \times 1 = 3 \, \text{sq meters} \)

2. Rectangle 2 (middle rectangle):

Length = 4 meters

Width = 2 meters

Area = \( 4 \times 2 = 8 \, \text{sq meters} \)

3. Rectangle 3 (bottom-left rectangle):

Length = 3 meters

Width = 3 meters

Area = \( 3 \times 3 = 9 \, \text{sq meters} \)

Total area for figure (a):

\[\text{Total area} = 3 + 8 + 9 = 20 \, \text{sq meters}\]

(b) Figure b

We can split this figure into two rectangles as follows:

1. Rectangle 1 (outer rectangle):

Length = 5 meters

Width = 3 meters

Area = \( 5 \times 3 = 15 \, \text{sq meters} \)

2. Rectangle 2 (inner cut-out rectangle):

Length = 3 meters

Width = 2 meters

Area = \( 3 \times 2 = 6 \, \text{sq meters} \)

Total area for figure (b):

\[\text{Total area} = \text{Area of outer rectangle} - \text{Area of inner cut-out}\]

\[\text{Total area} = 15 - 6 = 9 \, \text{sq meters}\]

Total area for figure (a): 20 sq meters

Total area for figure (b): 9 sq meters

Figure it Out

Cut out the tangram pieces given at the end of your textbook.

1. Explore and figure out how many pieces have the same area. 

Ans: Shapes C, E, and F have the same area. These are all triangles of equal size.

2. How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D and E? 

Ans: Shape D is twice the area of Shape C (or E). Shapes C and E have the same area, while Shape D is exactly double their area.

3. Which shape has more area: Shape D or F? Give reasons for your answer. 

Ans: Shape D has more area than Shape F. This is because Shape D is twice the area of Shape C (or E), while Shape F has the same area as Shape C and E.

4. Which shape has more area: Shape F or G? Give reasons for your answer. 

Ans: Shape G has more area than Shape F. Shape G is a square, and its area is larger than the area of Shape F, which is one of the smaller triangles.

5. What is the area of Shape A as compared to Shape G? Is it twice as big? Four times as big? 

Hint: In the tangram pieces, by placing the shapes over each other, we can find out that Shapes A and B have the same area, Shapes C and E have the same area. You would have also figured out that Shape D can be exactly covered using Shapes C and E, which means Shape D has twice the area of Shape C or shape E, etc. 

Ans: Shape A is exactly four times the area of Shape G. This can be deduced by the relative size of the pieces and the way they fit together.

6. Can you now figure out the area of the big square formed with all seven pieces in terms of the area of Shape C? 

Ans: The big square formed by all seven pieces has an area that is 8 times the area of Shape C. This is because the combined areas of the tangram pieces form a large square, and when you calculate the pieces' relative sizes, you find that the total area is 8 times the area of one Shape C.

7. Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the area of Shape C now? Give reasons for your answer. 

Ans: When the 7 pieces are arranged to form a rectangle, the area of the rectangle will still be 8 times the area of Shape C. The total area of the pieces remains constant regardless of the shape they form (whether a square or rectangle).

8. Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Give an explanation for your answer.

Ans: The perimeters of the square and the rectangle formed by the 7 pieces will be different. Although the areas remain the same, the shape of the perimeter changes when you rearrange the pieces into different arrangements, changing the overall perimeter.

Exercise 6.3

1. Find the areas of the figures below by dividing them into rectangles and triangles.

Ans:

(a) 

Total area of the figure = 20 + 4 = 24 sq. units

(b) 

Total area of the figure = 25 + 4 = 29 sq. units

(c)

Total area of the figure = 36 + 1 + 8 = 45 sq. units

(d)

Total area of the figure = 13 + 3 = 16 sq. units

(e)

Total area of the figure = 5 + 2 + 4 = 11 sq. units

Figure it Out

1. Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m. 

Ans: Area of the first rectangle:

\[A_1 = 5 \times 10 = 50 \, \text{sq m}\]

Area of the second rectangle:

\[A_2 = 2 \times 7 = 14 \, \text{sq m}\]

Total area:

\[\text{Total area} = A_1 + A_2 = 50 + 14 = 64 \, \text{sq m}\]

Assume the length of the new rectangle is 8 meters. Then, the width will be:

\[\text{Width} = \frac{64}{8} = 8 \, \text{m}\]

The dimensions of the rectangle are 8 m × 8 m.

2. The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden. 

Ans: Given:

Length \( l = 50 \, \text{m} \)

Area \( A = 1000 \, \text{sq m} \)

Formula:

\[A = l \times w\]

Where \( w \) is the width.

Solution:

\[1000 = 50 \times w\]

Divide both sides by 50:

\[w = \frac{1000}{50} = 20 \, \text{m}\]

The width of the garden is 20 meters.

3. The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted. 

Ans: Area of the room:

\[A_{\text{room}} = 5 \times 4 = 20 \, \text{sq m}\]

Area of the carpet:

\[A_{\text{carpet}} = 3 \times 3 = 9 \, \text{sq m}\]

Area not carpeted:

\[\text{Area not carpeted} = A_{\text{room}} - A_{\text{carpet}} = 20 - 9 = 11 \, \text{sq m}\]

The area not carpeted is 11 square meters.

4. Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn? 

Ans: Area of the garden:

\[A_{\text{garden}} = 15 \times 12 = 180 \, \text{sq m}\]

Area of one flower bed:

\[A_{\text{bed}} = 2 \times 1 = 2 \, \text{sq m}\]

Total area of all four flower beds:

\[A_{\text{beds total}} = 4 \times 2 = 8 \, \text{sq m}\]

Area available for lawn:

\[A_{\text{lawn}} = A_{\text{garden}} - A_{\text{beds total}} = 180 - 8 = 172 \, \text{sq m}\]

The area available for the lawn is 172 square meters.

5. Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions. 

Ans:

6. On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border? 

Ans:

7. Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area. 

Ans:

8. A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here? 

a. The area of each rectangle is larger than the area of the square. 

b. The perimeter of the square is greater than the perimeters of both the rectangles added together. 

c. The perimeters of both the rectangles added together is always $1\frac{1}{2}$  times the perimeter of the square. 

d. The area of the square is always three times as large as the areas of both rectangles added together. 

Ans: c.  The perimeters of both the rectangles added together is always $1\frac{1}{2}$  times the perimeter of the square. 

Benefits of NCERT Solutions for Class 6 Maths Chapter 6 Perimeter and Area

• Provides easy-to-follow explanations about perimeter and area, helping students understand how to calculate these measurements for various shapes.

• Offers clear steps and methods for calculating perimeter and area, making it simpler for students to learn and apply these concepts effectively.

• Helps students grasp the basics of perimeter and area, which is essential for understanding more advanced topics in geometry.

• Includes various practice problems that enhance students' ability to recognize and work with different shapes, improving their problem-solving skills.

• The FREE PDF download allows students to study and practice at their own pace, making learning more convenient and adaptable.

Important Study Material Links for Class 6 Maths Chapter 6 - Perimeter and Area

Conclusion

NCERT Solutions for Class 6 Maths Chapter 6, "Perimeter and Area," helps students understand how to calculate perimeter and area for different shapes. The clear explanations make it easier to learn these important concepts and their applications. Practising the exercises strengthens student’s skills in measuring shapes, which is crucial for more advanced geometry topics. With the solutions available as a FREE PDF download, students can study and review at their own pace, making learning both convenient and effective.

Chapter-wise NCERT Solutions Class 6 Maths

The chapter-wise NCERT Solutions for Class 6 Maths are given below. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Related Important Links for Maths Class 6

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