NCERT Solutions for Class 6 Maths Chapter 6 Perimeter and Area

Perimeter and area class 6 questions and answers are essential for students who want to learn how to measure the world around them. In this chapter, we explore how to measure the outside lines and the flat space inside different shapes. Whether you are building a fence for a backyard or picking out a new carpet for your room, these rules help you find the right answer. These solutions cover every activity and problem found in the Ganita Prakash textbook.

Learning About Perimeter and Area Class 6 Questions and Answers

When you start this chapter, you will see how shapes like squares, rectangles, and triangles work. It is important to know the difference between the two main ideas. Imagine you are playing in a park. If you walk all the way around the edge of the park along the fence, the distance you walked is the Perimeter. If you look at the grass covering the ground inside that fence, the size of that grassy space is the Area.

Students looking for perimeter and area class 6 questions and answers will find that these rules are used in real life every single day. Architects use them to design tall buildings, and farmers use them to know how many seeds to buy for their fields. In the Ganita Prakash book, you are not just memorizing numbers. You are learning to think like a scientist by counting edges and measuring surfaces. Finding the perimeter and area class 6 solutions questions and answers helps you see these steps in action so you can do them yourself.

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Perimeter and Area Class 6 Questions and Answers

Below are the ncert class 6th maths chapter 6 help students understand important concepts related to Perimeter and Area. 

6.1 Perimeter Figure it Out (Page No. 132)

Question 1. Find the missing terms:
(a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?.
(b) Perimeter of a square = 20 cm; length of a side = ?.
(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?.

Solution:

(a) Given, perimeter of rectangle =14 cm, breadth = 2 cm
Perimeter of rectangle = 2 (length + breadth)
⇒ 14 cm = 2 (length + 2 cm)
⇒ 7 cm = length + 2 cm
⇒ length = 5 cm

(b) Given, perimeter of a square = 20 cm
⇒ 4 × side = 20 cm
⇒ side = 5 cm
Therefore, length of a side = 5 cm

(c) Perimeter of a rectangle = 12 m
⇒ 2 (length + breadth) = 12 m
⇒ 3 m + breadth = 6 m
⇒ breadth = 3 m

Question 2. A rectangle having side lengths 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?
Solution:
Given, length of rectangle = 5 cm and breadth = 3 cm
We know that
perimeter of rectangle = 2 × (length × breadth)
= 2 × (5 + 3) = 16 cm

Now, if we bend the wire to form a square, the total length of the wire (16 cm) will be divided equally among the four sides of the square.
So, each side of the square = 

 Perimeter 4= 164  = 4 cm

Question 3. Find the length of the third side of a triangle with a perimeter of 55 cm and two sides of length 20 cm and 14 cm, respectively.
Solution:
Let ABC be the given triangle such that AB = 20 cm, BC = 14 cm
So, perimeter = AB + BC + CA = 55 cm
⇒ 55 = 20 + 14 + CA
⇒ CA = 55 – (20 + 14)
⇒ CA = 21 cm
∴ the length of the third side of the triangle = 21 cm.

Read More - NCERT Solutions for Class 6 Maths Chapter 3 Number Play

Question 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 meter?

Solution:
The length of the fence is the perimeter of the rectangular park.
Given that the length of the rectangular park = 150 m and breadth = 120 m
∴ Perimeter = 2(l + b)
= 2(150 + 120)
= 2(270)
= 540 m
Now cost of fencing per meter = ₹ 40
Cost of fencing the rectangular park = ₹ 40 × 540 = ₹ 21600

Question 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?
Solution:
(a) A square

A square has 4 equal sides.

So,
Length of each side = Total length ÷ 4
= 36 ÷ 4
= 9 cm

(b) An equilateral triangle

A triangle with all equal sides is called an equilateral triangle, which has 3 equal sides.

Length of each side = Total length ÷ 3
= 36 ÷ 3
= 12 cm

(c) A regular hexagon

A regular hexagon has 6 equal sides.

Length of each side = Total length ÷ 6
= 36 ÷ 6
= 6 cm

Question 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?

Solution:
Perimeter of a rectangular field = 2 (length + breadth)
= 2 (230 m + 160 m)
= 780 m
The farmer wants 3 rounds of rope to fence.
Total length of rope needed = 780 m × 3
= 2340 m

6.1 Perimeter Figure it Out (Page No. 133 – 134)

Question 1. Find out the total distance Akshi has covered in 5 rounds.
Solution:
Akshi runs on a rectangular track with a length of 70 metres and a breadth of 40 metres.
∴ Perimeter of track = 2 × (length + breadth)
= 2 × (70 + 40) = 220 m
Since, the distance covered in one round = 220 m
∴ Total distance covered in 5 rounds = 5 × 220 m
= 1100 m

Question 2. Find out the total distance Toshi has covered in 7 rounds.
Who ran a longer distance?
Solution:
Toshi runs on a rectangular track with a length of 60 m and breadth of 30 m.
∴ Perimeter of track = 2 × (length + breadth)
= 2 × (60 + 30) = 180 m
Since, the distance covered in ope round = 180 m
∴ Total distance covered in 7 rounds = 7 × 180 m
= 1260 m
So, Toshi ran a longer distance.

Question 3. Think and mark the positions as directed:
(a) Mark ‘A’ at the point where Akshi will be after she runs 250 m.
(b) Mark ‘B’ at the point where Akshi will be after she runs 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 runs 250 m.
(e) Mark ‘Y’ at the point where Toshi will be after she runs 500 m.
(f) Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘Z’.
Solution:
(a) Mark ‘A’ at the point where Akshi will be after she runs 250 m:

• 1 round = 220 m

• 250 – 220 = 30 m more
  So, ‘A’ will be 30 meters ahead of the starting point.

(b) Mark ‘B’ at the point where Akshi will be after she runs 500 m:

• 2 full rounds = 2 × 220 = 440 m

• 500 – 440 = 60 m more
  So, ‘B’ will be 60 meters ahead of the starting point after 2 rounds.

(c) Akshi ran 1000 m. How many full rounds? Mark her position as ‘C’:

• 1000 ÷ 220 = 4 full rounds and 120 m more
  So, ‘C’ will be 120 meters ahead of the starting point.

(d) Mark ‘X’ at the point where Toshi will be after she runs 250 m:

• 1 round = 180 m

• 250 – 180 = 70 m more
  So, ‘X’ will be 70 meters ahead of the starting point.

(e) Mark ‘Y’ at the point where Toshi will be after she runs 500 m:

• 2 full rounds = 2 × 180 = 360 m

• 500 – 360 = 140 m more
  So, ‘Y’ will be 140 meters ahead of the starting point.

(f) Toshi ran 1000 m. How many full rounds? Mark her position as ‘Z’:

• 1000 ÷ 180 = 5 full rounds and 100 m more
  So, ‘Z’ will be 100 meters ahead of the starting point.

6.2 Area Figure it Out (Page No. 138)

Question 1. The area of a rectangular garden is 300 square meters and its length is 25 meters. What is the width?

Answer:
We know:  Area = Length × Width

So,  300 = 25 × Width
Width = 300 ÷ 25 = 12 meters

Question 2. Find the cost of tiling a rectangular plot 500 m long and 200 m wide. The cost is ₹8 per 100 square meters.

Answer:
First, find the area:

Area = 500 × 200 = 1,00,000 sq m

Now, how many 100 sq m in 1,00,000?
1,00,000 ÷ 100 = 1,000 units

Cost per 100 sq m = ₹8
So, total cost = 1,000 × 8 = ₹8,000

Question 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?

Solution:
First, find the area of the grove:
100 × 50 = 5,000 sq m
Each tree needs 25 sq m of space.  So, number of trees = 5,000 ÷ 25 = 200
Maximum number of trees = 200

6.2 Area Figure it Out (Page No. 139)

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

Question 1. How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D and E?

Solution:
Shape D is twice as big as shape C. This means that if you place two shape C pieces together. Then, they exactly cover shape D.
The relationship between these shapes
Shape D can be completely filled by combining shape C and shape E. So, area of shape D is equal to the sum of the area of shape C and E.
Each of shapes C and E has half the area of shape D.

Question 2. Which shape has more area: Shape D or F? Give reasons for your answer.

Solution:
We can see that two times Shape C forms Shape D. Similarly two times Shape C forms Shape F. Thus, both Shape D and F are equal.

Question 3. Which shape has more area: Shape F or G? Give reasons for your answer.
Solution:
Since the medium triangle and the rhomboid are each made up of two small tangram triangles, they each have an area 2x that of the small triangle. Hence both have the same area.

Question 4.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, and 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.]
Solution:
Shape A has twice the area of shape G.

Question 5. Can you now figure out the area of the big square formed with all seven pieces in terms of the area of Shape C?
Answer:
Let’s say the area of C = x
Area of D = Area of 2C = 2x
Area of E = Area of C = x
Area of F = Area of 2C = 2x
Area of G = Area of 2C = 2x
Area of A = Area of 2F = 2 × 2x = 4x
Area of B = Area of A = 4x
Hence total area of big shape = Area of A + B + C + D + E + F + G
= 4x + 4x + x + 2x + x + 2x + 2x
= 16x
= 16C
That means the area of a big square is 16 times the area of shape C.

Question 6. 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.
Solution:
When arranging the 7 pieces to form a rectangle, the area of the rectangle will be the same as that of area of square.

Area of rectangle = 16 × area of shape C.

Question 7. Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Give an explanation for your answer.
Solution:
For the same area, a square always has the smallest perimeter.
So the perimeter of the square is less than that of the rectangle.

6.3 Area of A Triangle Figure it Out (Page 144)

Question 1. Find the areas of the figures below by dividing them into rectangles and triangles.
Solution:
(a) Figure have 20 full rectangles + 4 more than half rectangles + 4 less than half rectangles
= 20 × 1 + 4 × 1 + 4 × 0
= 20 + 4
= 24 sq. units

(b) Figure have 24 full rectangles, 2 half rectangles, 3 more than half and 3 less than half
∴ Area of figure = 24 + 1 + 2 × 12  + 3 × 1 + 3 × 0
= 24 + 1 + 3 + 0
= 28 sq. units

(c) Figure have 36 full rectangles, 2 half rectangles, 9 more than half, and 10 less than half rectangles
∴ Area of figure = 36 × 1 + 2 × 12 + 9 × 1 + 10 × 0
= 36 + 1 + 9 + 0
= 46 sq. units.

(d) Figure have 13 full rectangles, 1 half, 2 more than half and 2 less than half.
∴ Area of figure = 13 × 1 + 1 × 12 + 2 × 1 + 2 × 0
= 13 + 0.5 + 2
= 15.5 sq. units

(e) Figure have 5 full rectangles, 5 half, 3 more than half and 4 less than half.
∴ Area of figure = 5 × 1 + 5 × 12 + 3 × 1 + 4 × 0
= 5 + 2.5 + 3
= 10.5 sq. units

6.3 Area of A Triangle Figure it Out (Page 149)

Question 1. Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10m and 2m × 7m.
Solution:
Dimensions of rectangle 1: 5 m × 10m
Dimensions of rectangle 2:2m × 7m
Area of rectangle 1 = 50 sq m
Area of rectangle 2 = 14 sq m
Now, area of rectangle = sum of areas of rectangle 1 and 2 = 50 sq m + 14 sq m = 64 sq m
So possible dimensions of a rectangle with area 64 sq m are 1 m × 64 m; 2 m × 32 m; 4 m × 16 m; 8 m × 8 m, etc.

Question 2.  Four flower beds having sides 2 m long and 1 m wide are dug at the four comers of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
Solution:
We have, length of the garden = 15 m
and width of the garden = 12 m
∴ The area of the garden = Length × Width
= 15 × 12
= 180 sq. m

Now, also given that length of a flower bed = 2 m
and width of a flower bed = 1 m
The area of a flower bed = Length × Width
= 2 × 1
= 2 sq. m

Total area of 4 flower beds = 4 × 2
= 8 sq. m

The area available for laying down a lawn
= Area of the garden – Area of 4 flower beds
= 180 – 8
= 172 sq. m

Question 3. 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.
Solution:
For shape A, can be arrange it as 9 units by 2 units, giving a perimeter of 22 units.

For shape B, can be arrange it as 5 units by 4 units, giving a perimeter of 18 units.

Question 4. 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?
Solution:
Perimeter of the border = 2(1 + 1.5) = 5 cm2

Question 5. 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.
Solution:
As given, the Area of the rectangle of size 12 units × 8 units = 96 sq units
And the area of an inner rectangle that occupies exactly half the area = 48 sq units

Intext Questions

(Page No. 133)
Akshi and Toshi start running along the rectangular tracks as shown in the figure. Akshi runs along the outer track and completes 5 rounds. Toshi runs along the inner track and completes 7 rounds. Now, they are wondering who ran more. Find out who ran the longer distance.

Solution:

Here, perimeter of rectangular track PQRS = 2 × (l + b)
= 2 × (70 + 40)
= 2 × 110
= 220 m
and perimeter of rectangular track ABCD = 2 × (60 + 30)
= 2 × 90
= 180 m

Deep Dive: (Page No. 134)
In races, usually, there is a common finish line for all the runners. Here are two square running tracks with an inner track of 100 m on each side and an outer track of 150 m on each side. The common finishing line for both runners is shown by the flags in the figure which are in the center of one of the sides of the tracks. If the total race is 350 m, then we have to find out where the starting positions of the two runners should be on these two tracks so that they both have a common finishing line after they run for 350 m. Mark the starting points of the runner on the inner track as ‘A ’ and the runner on the outer track as ‘B’.

Solution:
Inner Track (100 m per side)
Perimeter Calculation: The perimeter of the inner track is (4 times 100 = 400) meters.
Distance to Run: The runner on the inner track needs to run 350 meters.
Starting Position (A): Since the perimeter is 400 meters, the runner will start 50 meters before the common finish line (400 – 350 = 50 meters).
Outer Track (150 m per side)
Perimeter Calculation: The perimeter of the outer track is (4 times 150 = 600) meters.
Distance to Run: The runner on the outer track also needs to run 350 meters.
Starting Position (B): Since the perimeter is 600 meters, the runner will start 250 meters before the common finish line (600 – 350 = 250 meters).

Perimeter of a Regular Polygon (Page No. 135)
Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalize your understanding of the perimeter of other regular polygons.
Solution:
Some common objects with regular shapes and calculating their perimeters:
Here are a few examples:
1. Square Table:

Shape: Square
Side Length = 1 meter
Perimeter = 4 × 1 = 4 meters

2. Equilateral Triangle Clock:

Shape: Equilateral Triangle
Side Length = 30 cm
Perimeter = 3 × 30 = 90 cm

3. Hexagonal Tile:

Shape: Regular Hexagon
Side Length = 10 cm
Perimeter = 6 × 10 = 60 cm
In general, the Perimeter of a Regular Polygon = (Number of sides) × (Side length of a polygon) units.

Let’s Explore! (Page No. 141)
On a squared grid paper (1 square = 1 square unit), make as many rectangles as you can whose lengths and widths are a whole number of units such that the area of the rectangle is 12 square units.
(a) Which rectangle has the greatest perimeter?
(b) Which rectangle has the least perimeter?
(c) If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer.
Solution:

Perimeter of (a) = 2(1 + 24) = 2 × 25 = 50 units
Perimeter of (b) = 2(2 + 12) = 2 × 14 = 28 units
Perimeter of (c) = 2(4 + 6) = 2 × 10 = 20 units
Perimeter of (d) = 2(3 + 8) = 2 × 11 = 22 units
(a) Clearly rectangle (a) has the greatest perimeter.
(b) Obviously rectangle (c) has the least perimeter.
(c) Yes, it is possible to predict the shape of a rectangle with the greatest and least perimeter for a given area. Here’s how:

Greatest Perimeter: For a given area, the rectangle with the greatest perimeter will have one side as small as possible. This essentially means that the rectangle becomes very elongated.

For example, if the area is 24 square units, a rectangle with dimensions 1 unit by 24 units will have the greatest perimeter.

Example: Area = 24 square units
Dimensions = 1 unit by 24 units
Perimeter = 2(1 + 24) = 50 units

Least Perimeter: The rectangle with the least perimeter for a given area will be as close to a square as possible. This is because a square has the smallest perimeter for a given area among all rectangles.

Example: Area = 24 square units
Dimensions = 4 units by 6 units (since 4 × 6 = 24)
Perimeter = 2(4 + 6) = 20 units

Reasoning
Greatest Perimeter: When one side is minimized, the other side must be maximized to maintain the same area. This increases the sum of the sides, thus increasing the perimeter.
Least Perimeter: A square or a shape close to a square minimizes the sum of the sides for a given area, thus minimizing the perimeter.

Perimeter = 6 + 3 + 1 + 1 + 1 + 1 + 1 + 1 + 3 = 18 units.

Read More - NCERT Solutions for Class 6 Social Science Chapter 1 Locating Places on the Earth

The Official Perimeter and Area Answer Guide Explained

To get the perimeter and area class 6 questions with answers cbse right, we have to look at how formulas help us save time. A formula is like a math secret that lets you find an answer quickly without measuring every single inch with a ruler.

For a square, we know that all four sides are exactly the same length. Instead of adding "side + side + side + side," we can just use the formula: Perimeter = 4 × Side. This is much faster! If you have a shape with six equal sides, called a regular hexagon, the rule is just as easy: Perimeter = 6 × Side.

Think of it like a puzzle. The perimeter and area class 6 solutions questions and answers show that if you know one part of a shape, you can find the others. For example, if you know the total area of a rectangular room and you know how long it is, you can find the width just by dividing. This is how smart math helps us solve mysteries.

The Magic of Tangrams

The Ganita Prakash book introduces something very cool called Tangram pieces. A Tangram is a set of seven flat shapes that you can move around to make new pictures, like animals or people.

The amazing thing about Tangrams is that even when you move the pieces to make a new shape, the total area stays exactly the same. This teaches us that area is about the amount of surface, no matter what shape it takes. Using a good guide for your studies is the secret to doing well in class and understanding these fun puzzles.

Class 6 Perimeter and Area Question and Answer Examples Daily

Let’s look at some simple perimeter and area class 6 questions and answers examples. These will help you get ready for your class tests and make you feel like a math pro.

• Question 1: You have a piece of string that is 36 cm long. You bend it to make a perfect square. How long is each side of your square?

• Answer: Since the whole string is the perimeter, we share that length between the 4 sides. Side = 36 / 4 = 9 cm.

• Question 2: Imagine a wire rectangle that is 5 cm long and 3 cm wide. If you straighten the wire and then bend it into a square, how long will each side of the square be?

• Answer: First, find the total wire length by finding the perimeter of the rectangle: 2 × (5 + 3) = 16 cm. Now, divide that for the square’s 4 sides: 16 / 4 = 4 cm.

• Question 3: A farmer has a big field that is 100 meters long and 50 meters wide. He wants to plant coconut trees. If each tree needs 25 square meters of space to grow, how many trees can he fit?

• Answer: First, find the total space (Area): 100 × 50 = 5000 sq m. Then, see how many tree-sized spots fit: 5000 / 25 = 200 trees.

Whether you are using a perimeter and area class 6 questions and answers pdf or reading your book, the steps stay the same. Noticing these tiny details is the best way to get better at Maths.

Doing Your Best with NCERT Solutions

Working with perimeter and area class 6 questions and answers is the best way to do well in your school work. You will learn about very interesting things, like the "Matha Pachchi" tracks. This is a part of the book where friends like Akshi and Toshi run on different paths.

Don’t let the big numbers in the book scare you. The main point is always to look at what the question is asking and pick the right formula. You can find the answers by looking closely at the units.

• If you see cm or m, it is usually about Perimeter (a line).

• If you see sq cm or sq m, it is always about Area (a surface).

Common mistakes to avoid include using diagonal or slanted lines when you are trying to find the area of a rectangle. Only use the straight length and width. Also, remember that a perimeter is just a length, but area always needs that special word "square" in the answer! Once you see the pattern of the math, your lessons become much easier and more fun.

 The "Running Track" Lesson (Deep Dive)

In the Ganita Prakash textbook, there is a special lesson about race tracks. Have you ever seen a race on TV where the runners start at different spots on the curve? It looks like some people are starting ahead of others, but it is actually to make the race fair!

Because the lane on the outside of the track is a much bigger circle than the lane on the inside, the perimeter of the outside lane is longer. To make sure everyone runs the exact same 350 meters, the person on the outside starts further ahead. This is a perfect example of how perimeter is used in sports.

 The "Cost" Tip for Real Life

A very important part of this chapter is finding out the cost of things. Suppose you want to tile your kitchen floor. The shop might say the price is "₹8 for every 100 square meters." To find the total price:

1. Find the Total Area of your floor.

2. Divide that area by 100 (because the price is for a group of 100).

3. Multiply that answer by the price (₹8).

This helps grow-ups know exactly how much money they need to spend before they even start working. Math helps people plan better!

Solving Complex Shapes

Sometimes you will see a shape that isn't a simple square or rectangle. It might look like the letter 'L' or a set of stairs. These are called composite shapes. Don't panic! You can solve these by "cutting" the shape into smaller rectangles that you already know how to measure.

For example, if you have an L-shaped room, you can draw a dotted line to turn it into two separate rectangles. Find the area of the first one, find the area of the second one, and then add them together. This "Split and Add" method is a lifesaver for perimeter and area class 6  questions and answers.

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