NCERT Solutions for Class 6 Maths Chapter 6 Perimeter And Area Ex 6.3

Exercise 6.3

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

Ans:

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:

$\text{Width} = \dfrac{64}{8} = 8 \, \text{m}$

Answer:

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:

$w = \dfrac{1000}{50} = 20 \, \text{m}$

Answer: 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:

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

Answer: 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:

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

Answer: 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 1 Exercise 6.3 Perimeter and Area

• The solutions explain the concepts of perimeter and area in simple terms, making it easier for students to understand the basics.

• Detailed step-by-step explanations help students learn how to approach and solve different types of perimeter and area problems.

• These solutions provide ample practice, helping students to prepare thoroughly for their exams.

• The solutions offer a quick and effective way to revise key concepts and problem-solving techniques before exams.

• Regular practice using these solutions boosts students' confidence in tackling various perimeter and area questions in their exams.

Class 6 Maths Chapter 1: Exercises Breakdown

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

Conclusion

NCERT Solutions for Class 6 Maths Chapter 1 Exercise 6.3 on Perimeter and Area provides a solid foundation for students to grasp essential mathematical concepts. By offering clear, step-by-step explanations, these solutions simplify complex problems and make learning easier. Regular practice with these solutions not only aids in exam preparation but also enhances problem-solving skills and boosts confidence. Using these solutions ensures that students can effectively tackle questions related to perimeter and area, helping them succeed in their studies.

Chapter-Specific NCERT Solutions for 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

Along with this, students can also download additional study materials provided by Vedantu for Maths Class 6.