What type of questions will you find in Exercise 6.3?

Exercise 6.3 contains word problems linking the concept of area and perimeter to real life.

Why is Exercise 6.3 vital in Chapter 6?

Exercise 6.3 is important because it allows students to practice not only reading and interpreting a word problem, but also to approach the problem in a practical way using the specific mathematical process for each question.

How will the exercise contribute to olympiad preparation?

The exercise builds application based thinking, useful to Olympiad problem solving skills in maths.

How does NCERT solutions help in Exercise 6.3?

The solutions that we provide within each NCERT solutions improve comprehension and assist students with an approach to the word problems.

NCERT Solutions

Class 6

Maths

Chapter 6 Perimeter And Area

Exercise 6.3

NCERT Solutions Class 6 Maths Chapter 6 Perimeter and Area Exercise 6.3

Exercise 6.3 of Class 6 Maths Chapter 6 – Perimeter and Area – will give you practice in solving word problems on perimeter and area. The problems asked here involve real situations, such as a length of fence or the area of a floor. This exercise helps you apply the formulas accurately and know what the question is asking.

This exercise is developed as per the NCERT syllabus and helps you in developing skills in reading, understanding and solving practical problems by applying mathematics. These types of questions are very frequently asked in school examinations and are useful for learning how to prepare your answers properly to get better marks.

You can download the complete NCERT Solutions for Class 6 Maths Chapter 6 Exercise 6.3 for free as a pdf file here. The explanations of the answers are written as simple steps so that the reader can easily follow and learn from. They help for revision, clearing doubts and preparing for exams with confidence.

1.0Download NCERT Solutions Class 6 Maths Chapter 6 Perimeter and Area Exercise 6.3: Free PDF

Exercise 6.3 provides word problems based on the perimeter and area of shapes and helps you learn how to apply the formulas in real-world problems. Get the complete NCERT Solutions for Class 6 Maths Chapter 6 Exercise 6.3 and download the free PDF from below:

NCERT Solutions for Class 6 Maths Chapter 6 - Exercise 6.3

2.0NCERT Solutions Class 6 Chapter 6 Perimeter and Area: All Exercises

NCERT Solutions Class 6 Maths Chapter 4 Exercise 6.1

NCERT Solutions Class 6 Maths Chapter 4 Exercise 6.2

NCERT Solutions Class 6 Maths Chapter 4 Exercise 6.3

3.0NCERT Class 6 Maths Chapter 6 Perimeter and Area Exercise 6.3: Detailed Solutions

6.3 Area of a Triangles

Find the areas of the below by dividing them into rectangles and triangles.
Sol.
(a)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	20	$20 \times 1 = 20$
Half-filled squares	0	0
More than half-filled squares	4	$4 \times 1 = 4$
Less than half-filled squares	4	0

$∴$ Total area of the figure $= 20 + 4 = 24$ sq. units (b)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	25	$25 \times 1 = 25$
Half-filled squares	0	0
More than half-filled squares	4	$4 \times 1 = 4$
Less than half-filled squares	4	0

$∴$ Total area of the figure $= 25 + 4 = 29$ sq. units (c)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	36	$36 \times 1 = 36$
Half-filled squares	2	$2 \times 1/2 = 1$
More than half-filled squares	8	$8 \times 1 = 8$
Less than half-filled squares	6	$6 \times 0 = 0$

$∴$ Total area of the figure $= 36 + 1 + 8 = 45$ sq. units (d)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	13	$13 \times 1 = 13$
Half-filled squares	0	0
More than half-filled squares	3	$3 \times 1 = 3$
Less than half-filled squares	2	$2 \times 0 = 0$

$∴$ Total area of the figure $= 13 + 3 = 16$ sq. units (e)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	5	$5 \times 1 = 5$
Half-filled squares	4	$4 \times 1/2 = 2$
More than half-filled squares	4	$4 \times 1 = 4$
Less than half-filled squares	3	$3 \times 0 = 0$

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

Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: $5 m \times 10 m$ and $2 m \times 7 m$ . Sol. Here, Area of rectangle $1 = 5 \times 10 = 50$ sq. m Area of rectangle $2 = 2 \times 7 = 14$ sq. m The Sum of the areas of these 2 rectangles $= 50 + 14 = 64$ sq. m Now, the total area of the rectangle $= 64$ Let's say the sides of the rectangle are Length $= x$ and Width $= y$ Area of rectangle $= x \times y$ Hence $x \times y = 64$ xy $= 64$ Let's say $x = 1$ , then $y = \frac{64}{1} = 64$ if $x = 2$ , then $y = \frac{64}{2} = 32$ Hence the dimensions of the rectangle are ( $1 \times 64$ ), ( $2 \times 32$ )
The area of a rectangular garden that is 50 m long is 1000 sq m . Find the width of the garden.
Sol. Length of the garden $= 50 m$ Area of the garden $= 1000 m^{2}$ Length $\times$ Width $= 1000$ $50 \times$ Width $= 1000$ Width $= \frac{1000}{50} = 20 m$ Therefore, the width of the garden $= 20 m$
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.
Sol. A square carpet of side 3 m . Area of the floor $=$ length $\times$ breath Area of the floor $= 5 \times 4 = 20 m^{2}$ Area of the square carpet $= 3 \times 3 = 9 m^{2}$ Now, we will subtract the square carpet area from the floor's area to get the area of the floor that is not carpeted. Hence, the area of the floor that is not carpeted $= 20 - 9 = 11 m^{2}$ Thus, the area of the floor that is not carpeted is $11 m^{2}$ .
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?
Sol. Length of garden $= 15 m$ Width of garden $= 12 m$ So, the area of the garden $= 15 \times 12$ sq. $m = 180$ sq. m Now, the length of the flower bed $= 2 m$ Width of flower bed $= 1 m$ Area of the flower bed $= 2 \times 1$ sq. $m = 2$ sq. m Since, the area of four flower beds $= 2 \times 4$ sq. $m = 8$ sq. m Now the area is available for laying down a lawn $= (180 - 8)$ sq. $m = 172$ sq. m
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. Sol.
Shape A has an area of 18 sq. units. $∴$ Possible sides are $18 \times 1, 2 \times 9, 6 \times 3$ Also, shape $B$ has an area of 20 sq. units. $∴$ Possible sides are $20 \times 1, 4 \times 5, 10 \times 2$ Given shape A has a longer perimeter than shape B, hence two such shapes satisfying the given conditions are: Here Perimeter of shape $A = 9 + 2 + 9 + 2 = 22$ units Here Perimeter of shape $B = 5 + 4 + 5 + 4 = 18$ units
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? Sol. Perimeter of the rectangle border $= 2 \times$ (Length $\times$ Width) $= 2 \times (1 + 1.5)$ $= 2 \times 2.5$ $= 5 cm$
Draw a rectangle of size 12 units $\times 8$ units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area.
Sol. Area of given rectangle $= 12 \times 8 = 96$ units $^{2}$ And area of new rectangle $= \frac{1}{2} \times 96 = 48$ sq. units $∴$ Possible sides of new rectangle are $12 \times 4, 16 \times 3, 8 \times 6, 1 \times 48$ $∴$ Hence dimensions of the new rectangle fill in the rectangle of $12 \times 8$ units $^{2}$ $= 8$ units $\times 6$ units
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.
Now in the above square piece side of square $= 1$ unit area of square $= 1 \times 1 = 1$ sq. unit. and perimeter of square $= 1 + 1 + 1 + 1 = 4$ units. Now after folding the above square piece in half becomes 2 rectangles
Perimeter of rectangle $R_{1} = 1 + \frac{1}{2} + 1 + \frac{1}{2} = 3$ units. Area of rectangle $R_{1} = \frac{1}{2} \times 1 = \frac{1}{2}$ sq. unit. Perimeter of rectangle $R_{2} = 1 + \frac{1}{2} + 1 + \frac{1}{2} = 3$ units. Area of rectangle $R_{2} = \frac{1}{2} \times 1 = \frac{1}{2}$ sq. unit. a. Now, area of rectangle $R_{1} =$ area of rectangle $R_{2} = \frac{1}{2} < 1$ . Hence, option (a) is not true. b. Here perimeter of square $= 4$ units and perimeters of both the rectangles $= 3 + 3 = 6$ units. which is greater than 4 units. Hence option (b) is not true c. Here perimeters of both the rectangles $= 6$ units And perimeter of square $= 4$ units $\times 1 \frac{1}{2} = 4 \times \frac{3}{2} = 6$ units The perimeters of both the rectangle added together are $1 \frac{1}{2}$ times the perimeter of the square. Hence, option (c) is true. d. Here, the area of the square $= 4$ units and areas of both the rectangles $= \frac{1}{2} + \frac{1}{2} = 1$ unit. The area of the square is four times the area of both rectangles. Hence, option (d) is not true.

4.0Key Features and Benefits for Class 6 Maths Chapter 6 Exercise 6.3

The problems contain word problems which use the area and perimeter of squares and rectangles.
The questions are based on real-life situations to develop understanding and use of concepts.
Solving NCERT solutions helps foster problem-solving abilities required for various exams.
It helps develop students for maths olympiads as practice, for applied geometry questions.
The step-by-step solutions make understanding and revision of important concepts easier especially by the time the exam period arrives.

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