NCERT Solutions for Maths Class 5 Chapter 11 - Areas And Its Boundary
The chapter Areas and Its Boundaries focuses on dealing with the areas of given shapes and their respective boundaries or perimeters. It helps the students to learn how to calculate the areas and perimeters of various shapes through formulae. It covers the topics: philosophy Areas of Shapes philosophy Perimeters of Shapes philosophy Formulas for Areas and Perimeters NCERT Math-Magic questions are answered in a simple and engaging manner. We have also related 'Learning Concepts and interactive worksheets with the solutions. Our 'Learning Beyond' segment caters to all the probable questions that a child might think out of curiosity. Download Chapter 11 Areas and Its Boundaries in PDF format for free here.
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The NCERT Solutions for Maths Class 5 Chapter 11 - Areas And Its Boundary are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Maths Class 5 Chapter 11 - Areas And Its Boundary
Students can access the NCERT Solutions for Maths Class 5 Chapter 11 - Areas And Its Boundary. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Area and its Boundary
Draw a square of 9 square cm. Write A on it. Draw another square with double the side. Write B on it. Answer these:
1) The perimeter of square A is __________ cm.
2) The side of square B is __________ cm.
3) The area of square B is __________ square cm.
4) The area of square B is __________ times the area of square A.
5) The perimeter of square B is __________ cm.
6) The perimeter of square B is __________ times the perimeter of square A.
Draw two squares, one of side 3 cm and another of side 6 cm.
1) Perimeter of square A is:
Perimeter = 4 × side
= 4 × 3 = 12 cm
2) The side of square B is 6 cm.
3) The area of the square B is:
Area = side × side
= 6 × 6 = 36 square cm.
4) The area of the square A is:
Area = side × side
= 3 × 3 = 9 square cm.
Since 36 = 4 × 9, the area of square B is 4 times the area of square A.
5) Perimeter of square B is:
Perimeter = 4 × side
= 4 × 6 = 24 cm
6) Since 24 = 2 × 12, the perimeter of square B is two times the perimeter of square A.
Parth and Gini bought aam paapad (dried mango slice) from a shop. Their pieces looked like these.
Both could not make out whose piece was bigger. Suggest some ways to find out whose piece is bigger. Discuss.
Do it by yourself. Draw squares of side 1 cm on both the pieces, and count the number of squares to find the bigger piece.
A friend of Parth and Gini showed one way, using small squares.
The length of piece A is 6 cm. So, 6 squares of side 1 cm can be arranged along its length. The width of piece A is 5 cm. So, 5 squares can be arranged along its width.
a) Altogether how many squares can be arranged on it? _______
b) So, the area of piece A = _______ square cm
c) In the same way find the area of piece B.
d) Who had the bigger piece? How much bigger?
a) Six squares of side 1 cm can be placed along the length of A, and five squares can be placed along its width. To find the total number of squares, multiply 6 by 5.
6 × 5 = 30
Therefore, 30 squares can be placed on the piece A, as shown below.
b) Since 30 squares of side 1 cm can be placed on piece A, its area is 30 square cm.
c) The length of piece B is 11 cm. Therefore, 11 squares can be placed along its length.
The width of piece B is 3 cm. Therefore, 3 squares can be placed along its width.
The total number of squares that can be placed on piece B is:
11 × 3 = 33
Therefore, its area is 33 square cm.
d) The area of piece A is 30 square cm, and the area of piece B is 33 square cm. Therefore, piece B is 3 square cm bigger than piece A.
Hence, Gini had 3 cm bigger piece.
This stamp has an area of 4 square cm. Guess how many such stamps will cover this big rectangle.
Guess on your own. Answers may vary. About 28 such stamps can cover the big rectangle.
Take a 15 cm long thread. Make different shapes by joining its ends on this sheet.
A) Which shape has the biggest area? How much? _________ What is the perimeter of this shape? _________
B) Which shape has the smallest area? How much? ________ What is the perimeter of this shape? _________ Also make a triangle, a square, a rectangle and a circle. Find which shape has biggest area and which has the smallest.
Do it by yourself. Answers may vary.
a) Measure the yellow rectangle. It is ________ cm long.
b) How many stamps can be placed along its length? ________
c) How wide is the rectangle? ________ cm.
d) How many stamps can be placed along its width? ________
e) How many stamps are needed to cover the rectangle? ________
f) How close was your earlier guess? Discuss.
g) What is the area of the rectangle? ________ square cm.
h) What is the perimeter of the rectangle? ________ cm.
a) Use a ruler to measure the length of the rectangle. It is 14 cm long.
b) The rectangle is 14 cm long, and each side of the stamp is 2 cm. Divide 14 by 2 to find the number of stamps that can be placed along the length of the rectangle.
14 2 = 7
Therefore, 7 stamps can be placed along its length.
c) Use a ruler to measure the width of the rectangle. It is 8 cm wide.
d) The rectangle is 8 cm wide, and each side of the stamp is 2 cm. Divide 8 by 2 to find the number of stamps that can be placed along the width of the rectangle.
8 2 = 4
Therefore, 4 stamps can be placed along its width.
e) 7 stamps can be placed along the length of the rectangle, and 4 stamps can be placed along its width. To find the total number of stamps that can be placed on the rectangle, multiply 7 by 4.
7 × 4 = 28
Therefore, 28 stamps are needed to cover the rectangle.
f) Do it by yourself.
g) The length of the rectangle is 7 cm, and its width is 4 cm.
Area of a rectangle can be found by:
Area = Length × Width
Therefore, area of the rectangle is:
7 × 4 = 28 square cm.
h) Perimeter of a rectangle can be found by:
Perimeter = 2 (Length + Width)
Therefore, the perimeter of the rectangle is:
2(7 + 4) = 2 × 11 = 22 cm.
a) Arbaz plans to tile his kitchen floor with green square tiles. Each side of the tile is 10 cm. His kitchen is 220 cm in length and 180 cm wide. How many tiles will he need?
220 cm is the length of the kitchen, and each side of the tile is 10 cm
Number of tiles that can be placed along its length
= length of kitchen length of each tile.
= 220 10 = 22
Width of the kitchen is 180 cm.
Number of tiles that can be placed along its breadth
= breadth of kitchen breadth of each tile
= 180 10 = 18
Since 22 tiles can be placed along the length of the kitchen, and 18 tiles can be placed along its width, multiply 22 by 16 to get the total number of tiles required.
22 × 18 = 396
Therefore, 396 tiles are required to tile the kitchen floor.
b) The fencing of a square garden is 20 m in length. How long is one side of the garden?
Length of the fencing is 20 m. Length of the boundary of a square is given by:
Perimeter = 4 × Side
⇒ Side = Perimeter ÷ 4
To find the length of the side, divide 20 by 4.
20 4 = 5
Therefore, one side of the garden is 5 m long.
c) A thin wire 20 centimetres long is formed into a rectangle. If the width of this rectangle is 4 centimetres, what is its length?
Perimeter of a rectangle is found by:
Perimeter = 2(Length + Width)
⇒Length + Width = Perimeter ÷ 2
Perimeter of the rectangle is the length of the wire that is 20 cm, and its width is 4 cm. Therefore, we have
Length + 4 = 20 ÷ 2
⇒ Length + 4 = 10
⇒ Length = 10 – 4 = 6 cm
Therefore, the length of the rectangle is 6 cm.
d) A square carrom board has a perimeter of 320 cm. How much is its area?
All sides of a square are of the same length.
AB = BC = CD = AD = Side
Perimeter of a square is given by:
Perimeter = 4 × Side
⇒ Side = Perimeter ÷ 4
Since the perimeter of the carrom board is 320 cm, we have
Side = 320 4 = 80 cm.
Area of a square is given by:
Area = Side × Side
Therefore, we have
Area of the carrom board = 80 × 80 = 6400 square cm.
e) How many tiles like the triangle given here will fit in the white design? Area of design = ___________ square cm.
Observe the given design. There is 1 square and 4 triangles. Since the triangle is half of a cm square, 6 triangular tiles can fit in the design. Area of design = 3 square cm.
f) Sanya, Aarushi, Manav, and Kabir made greeting cards. Complete the table for their cards:
Step 1: Consider Sanya's card:
Length = 10 cm, width = 8 cm. It is a rectangle.
Perimeter = 2(length + width) = 2(10 + 8) = 2 × 18 = 36 cm
Area = length × width = 10 × 8 = 80 square cm
Step 2: Consider Manav’s card:
Length = 11 cm, perimeter = 44 cm
Perimeter = 44 = 4 × 11 = 4 × length.
So, Manav’s card is a square.
Therefore,
Width = Length = 11 cm = Side of the square
Area = Side × Side = 11 × 11 = 121 square cm.
Step 3: Consider Arushi’s card:
Width = 8 cm, area = 80 square cm.
Area = Length × Width
⇒ Length = Area ÷ Width = 80 ÷ 8 = 10 cm. So, it is a rectangle.
Perimeter = 2 (length + width) = 2(10 + 8) = 2 × 18 = 36 cm
Step 4: Consider Kabir’s card:
Perimeter = 40 cm, area = 100 square cm.
Perimeter = 40 = 4 × 10
Area = 10 × 10 = 100 square cm.
So, Kabir’s card is a square with side 10 cm. Therefore,
Length = 10 cm, width = 10 cm.
The correct answer is:
Take a thick paper sheet of length 14 cm and width 9 cm. You can also use an old postcard.
a) What is its area?
b) What is its perimeter?
c) Now cut strips of equal sizes out of it. Using tape join the strips, end to end, to make a belt. How long is your belt? _________
d) What is its perimeter _________
e) Whose belt is the longest in the class? ________
Do it by yourself.
Area = 14 × 9 = 126 square cm.
Perimeter = 14 + 9 + 14 + 9 = 46 cm.
A) Make two squares of one square metre each. Divide your class in two teams. Ready to play! Try these in your teams:
a) How many of you can sit in one square metre? ________
b) How many of you can stand in it? ________
c) Which team could make more children stand in their square? How many? ________
d) Which team could make more children sit in their square? How many?
Do it by yourself with the help of your friends. Answers may vary.
B) Measure the length of the floor of your classroom in metres. Also measure the width.
a) What is the area of the floor of your classroom in square metres? __________
b) How many children are there in your class? _________
c) So how many children can sit in one square metre? __________
d) If you want to move around easily then how many children do you think should be there in one square metre? _________
Do it by yourself as directed. Use a measuring tape to measure the length and width of the floor of your classroom. Answers may vary.
Nasreena is a farmer who wants to divide her land equally among her three children — Chumki, Jhumri and Imran. She wants to divide the land so that each piece of land has one tree. Her land looks like this.
a) Can you divide the land equally? Show how you will divide it. Remember each person has to get a tree. Colour each person’s piece of land differently.
b) If each square on this page is equal to 1 square metre of land, how much land will each of her children get? ___________ square m Chumki, Jhumri and Imran need wire to make a fence.
c) Who will need the longest wire for fencing? __________
d) How much wire in all will the three need? ___________
a) The land is 10 squares long, and 9 squares wide.
Its area = 10 × 9 = 90 squares.
Since she has 3 children, divide 90 by 3 to get the share of each child.
90 ÷ 3 = 30
Therefore, each child will get 30 squares of land.
Now, divide the land into 3 parts in such a way that each part has 30 squares and a tree. There are many ways to divide it. Answers may vary. A sample answer is:
The first part from the left is for Chumki, the middle part is for Jhumri, and the last part is for Imran.
b) Each of her children will get 30 square m of land.
c) Both Chumki and Jhumri will need 28 m of wire for fencing. Imran will need 26 m of wire for fencing his part of the field.
d) Total wire required = 28 + 28 + 26 = 82 m
Look at the table. If you were to write the area of each of these which column would you choose? Make a (
).
The correct answer is:
There are two beautiful lakes near a village. People come for boating and picnics in both the lakes. The village Panchayat is worried that with the noise of the boats the birds will stop coming. The Panchayat wants motor boats in only one lake. The other lake will be saved for the birds to make their nests.
a) How many cm is the length of the boundary of lake A in the drawing? __________ (use thread to find out)
b) What is the length of the boundary of lake B in the drawing?
c) How many kilometres long is the actual boundary of lake A?
d) How many kilometres long is the actual boundary of lake B?
e) A longer boundary around the lake will help more birds to lay their eggs. So which lake should be kept for birds? Which lake should be used for boats?
f) Find the area of lake B on the drawing in square cm. What is its actual area in square km?
a) Use a thread to measure the boundary of lake A. Now, measure the length of the thread using a ruler. It is about 30 cm.
b) Use a thread to measure the boundary of lake B. Now, measure the length of the thread using a ruler. It is about 25 cm.
c) Since 1 cm on the drawing = 1 km on the ground. Therefore, the actual length of the boundary of lake A is 30 km.
d) Since 1 cm on the drawing = 1 km on the ground. Therefore, the actual length of the boundary of lake B is 25 km.
e) The boundary of lake A is larger than the boundary of lake B
Therefore, lake A should be preserved for birds, whereas lake B should be used for boats.
f) In the drawing of lake B, count the complete, half-filled, and more than half filled squares.
Number of complete squares = 15⇒ area = 15 square cm.
Number of half squares = 2 ⇒ area = 1 square cm.
Number of more than half squares = 9 ⇒ area = 9 square cm.
Total area of lake B = 15 + 1 + 9 = 25 square cm
Therefore, actual area of lake B = 25 square km.
The King was very happy with carpenters Cheggu and Anar. They had made a very big and beautiful bed for him. So as gifts the king wanted to give some land to Cheggu, and some gold to Anar. Cheggu was happy. He took 100 metres of wire and tried to make different rectangles. He made a 10 m × 40 m rectangle. Its area was 400 square metres. So he next made a 30 m × 20 m rectangle.
a) What is its area? Is it more than the first rectangle?
b) What other rectangles can he make with 100 metres of wire? Discuss which of these rectangles will have the biggest area.
c) Cheggu’s wife asked him to make a circle with the wire. She knew it had an area of 800 square metres. Why did Cheggu not choose a rectangle? Explain.
d) So Anar also tried many different ways to make a boundary for 800 square metres of land. He made rectangles A, B and C of different sizes. Find out the length of the boundary of each. How much gold wire will he get for these rectangles?
Gold wire for A = ………………… metres
Gold wire for B = ………………… metres
Gold wire for C = ………………… metres
So he will get __________ metres of gold wire!!
a) Area of a rectangle is given by:
Area = Length × Width
Therefore, area of 30 m by 20 m rectangle is:
30 × 20 = 600 square m.
It has a larger area than the first rectangle.
b) With 100 m of wire, he can make many rectangles. Following are some examples:
40 m × 10 m, 48 m × 2 m, 49 m × 1 m, and 45 m × 5 m
The 40 m by 10 m rectangle has the largest area, it is 400 square m.
c) With the same perimeter a circle has a larger area than a rectangle.
d) Perimeter of rectangle A = 2 (40 + 20) = 120 m
Therefore, gold wire for rectangle A is 120 m.
Perimeter of rectangle B = 2 (80 + 10) = 180 m
Therefore, gold wire for rectangle B is 180 m.
Perimeter of rectangle C = 2 (800 + 1) = 1602 m
Therefore, gold wire for rectangle C is 1602 m.
Perimeter of rectangle D = 2 (8000 + 0.1) = 16000.2 m
Therefore, gold wire for rectangle D is 16000.2 m.
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Frequently Asked Questions
The NCERT solution for Class 5 Chapter 11: Areas And Its Boundary is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 5 Chapter 11: Areas And Its Boundary is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 5 Maths Chapter 11 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 5 Maths Chapter 11: Areas And Its Boundary as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 5 Maths Chapter 11 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.