Area and Perimeter Class 6
Mastering Area and Perimeter: Essential Concepts for Sainik School Class 6
For students aiming to join a prestigious institution like Sainik School, having a strong grasp of basic mathematical concepts is crucial. Among these, the topics of Area and Perimeter are fundamental for the Class 6 Maths syllabus. Understanding how to calculate both area and perimeter not only helps in the entrance exam but also lays the groundwork for more advanced geometric concepts in higher grades.
What is Perimeter?
The perimeter is the total distance around the boundary of a twodimensional shape. It is the sum of the lengths of all the sides of the shape. Calculating the perimeter is essential for solving problems related to the outer edge or boundary of geometric figures.
How to Calculate Perimeter
 The perimeter of a Rectangle:
 Formula: Perimeter=2×(Length+Width)
 Example: For a rectangle with a length of 8 cm and a width of 5 cm: Perimeter=2×(8+5)=2×13=26 cm
 The perimeter of a Square:
 Formula: Perimeter=4×Side
 Example: For a square with a side length of 6 cm: Perimeter=4×6=24 cm
 The Perimeter of a Triangle:
 Formula: Perimeter=Side1+Side2+Side3
 Example: For a triangle with sides of 7 cm, 5 cm, and 9 cm: Perimeter=7+5+9=21 cm
What is Area?
The area measures the amount of space enclosed within a shape. It is expressed in square units and is crucial for understanding the size of a surface. Calculating area is fundamental in various practical applications, such as determining the amount of material needed for a project or the space available in a room.
How to Calculate Area
 Area of a Rectangle:
 Formula: Area=Length×Width
 Example: For a rectangle with a length of 8 cm and a width of 5 cm: Area=8×5=40 square cm
 Area of a Square:
 Formula: Area=Side^2
 Example: For a square with a side length of 6 cm: Area=6^2=36 square cm
 Area of a Triangle:
 Formula: Area=1/2×Base×Height
 Example: For a triangle with a base of 10 cm and a height of 4 cm: Area=1/2×10×4=20 square cm
 Area of a Circle:
 Formula: Area=π×Radius ^ 2 (where π≈3.14 or 22/7)
 Example: For a circle with a radius of 7 cm: Area=3.14×7*7 = 3.14×49≈153.86 square cm
Importance of Area and Perimeter in the AISSEE
The concepts of area and perimeter are integral to the Sainik School Class 6 entrance exam. They are commonly tested in various problem types, including word problems and geometric calculations. A thorough understanding of these concepts is essential for solving problems accurately and efficiently.
Types of Problems Involving Area and Perimeter
Simple Calculations:

 Problems where students need to calculate the perimeter or area of basic geometric shapes like rectangles, squares, and triangles.
Example: Calculate the perimeter and area of a rectangle with a length of 12 cm and a width of 7 cm.

 Perimeter: 2×(12+7)=38 cm
 Area: 12×7=84 square cm
Word Problems:


 Reallife scenarios that require students to apply their knowledge of area and perimeter to find solutions.

Example: A rectangular garden has a length of 15 meters and a width of 10 meters. How much fencing is needed to enclose the garden?

 Solution: Perimeter = 2×(15+10)=50 meters
Composite Shapes:


 Problems involving complex shapes made up of simple geometric figures, where students need to find the total area or perimeter.

Example: Find the area of a shape consisting of a rectangle of 8 cm by 5 cm and a square of 4 cm by 4 cm attached.

 Solution: Area of the rectangle = 8×5=40 square cm
 Area of the square = 4×4=16 square cm
 Total Area = 40+16=56 square cm
RealLife Applications:

 Problems that involve practical applications of area and perimeter, such as determining the amount of paint needed for a wall or the fabric required for a garment.
Example: A painter needs to cover a wall that is 10 meters long and 3 meters high. How much paint is needed if one liter covers 5 square meters?

 Solution: Area of the wall = 10×3=30 square meters
 Paint needed = 30/5=6 liters
Tips for Mastering Area and Perimeter
 Memorize Formulas:
 Ensure you have the formulas for area and perimeter memorized for different shapes. Practice using them until they become second nature.
 Draw Diagrams:
 For complex problems, drawing a diagram can help you visualize the shape and understand the problem better.
 Use Units Consistently:
 Always use consistent units when calculating area and perimeter to avoid errors.
 Solve Practice Problems:
 Regularly practice solving area and perimeter problems to build confidence and improve accuracy.
 Check Your Work:
 Doublecheck your calculations and ensure you have used the correct formula for the shape in question.
Common Mistakes to Avoid
 Mixing Up Units:
 Ensure that you are using the same units for all measurements before performing calculations.
 Forgetting to Square Units for Area:
 Remember that area is expressed in square units, while perimeter is expressed in linear units.
 Using Incorrect Formulas:
 Verify that you are using the correct formula for the shape you are working with.
 Ignoring the BODMAS Rule:
 When performing multiple operations, make sure to follow the BODMAS rule to avoid calculation errors.
Conclusion
Understanding area and perimeter is crucial for students preparing for the Sainik School Class 6 entrance exam. These concepts form the basis for solving various mathematical problems and have numerous practical applications. By mastering the formulas, practicing regularly, and applying these concepts to reallife scenarios, students can enhance their problemsolving skills and excel in the AISSEE.
Important Questions Related to Area and Perimeter Class 6
Ques 1: The perimeter of a rectangular room is 200 m. If the breadth of the room is 40 m, find the area of the room.
a) 2500 sq. m.
b) 2400 sq. m.
c) 2600 sq. m.
d) 2225 sq. m.
Answer – b (2400 sq. m. )
Solution:
Perimeter = 200 m
b = 40 m
2 (l + b) = 200
(l + b) = 100 m
l = 100 – 40 = 60 m
Area = l × b
= 60 × 40
= 2400 $m^2$
Ques 2: The perimeter of a rectangle is 30 m and the ratio between length and breadth is 7 : 3. Find the area of the rectangle.
a) 46.55 sq. m.
b) 48.50 sq. m.
c) 45.25 sq. m.
d) 47.25 sq. m.
Answer – d (47.25 sq. m.)
Solution:
2(l + b) = 30
7x + 3x = 15
10 x = 15
x = 3/2
l = 7 *3/2 = 21/2
b = 3 *3/2 = 9/2
l × b = (21/2) * (9/2)
Area = 189/4 = 47.25 sq. m.
Question 3: A rectangular field is 75 meters long and 27 meters wide. The area of a square field is equal to the area of this rectangular field. Find the perimeter of the square field.
a) 160 m
b) 170 m
c) 180 m
d) 165 m
Answer – c (180 m)
Solution:
area of rectangle = 75 × 27
The area of rectangle = area of square= 75 × 27
area of square = 75 × 27= 2025 sq. m.
side of square= (2025)^{1/2}
side of square= 45 m
perimeter of square= 4 × side of square= 4 × 45 = 180 m
For more such questions on Sainik School Area and Perimeter class 6, download Shaurya Bharat app now: https://play.google.com/store/apps/details?id=com.shauryabharat
How to crack the AISSEE 2025 and RMS CET 2024 examinations?
Entrance examinations for Rashtriya Military School and Sainik School will take place in December 2024 and January 2025 respectively for academic session 202526. The best idea to start preparation is to start early so that students get sufficient time to prepare and polish themselves. The Sainik School and RMS Entrance Exams are a tough nut to crack. The standards and competition are way higher than one could think. Getting admission into Sainik School and RMS is the dream of thousands of young children. However, only a few can turn this dream into reality.
To succeed in this competitive environment, candidates need to prepare thoroughly, maintain a disciplined study routine, and focus on developing a strong foundation in the relevant subjects. Regular practice, solving sample papers, and mock tests can help in building confidence and improving performance. Maintaining physical fitness and participating in extracurricular activities can enhance overall candidacy.
It is important to remember that the competition is tough. However, dedication, hard work, and a positive mindset can significantly increase your chances of securing admission to a Sainik School or RMS. Here, Shaurya Bharat can be a perfect guide to students. Shaurya Bharat is the best offline and online platform for students preparing for the Sainik School, RMS, and RIMC entrance examinations.
Offline Classes
We conduct offline coaching for Sainik School and RMS (School + Coaching) at our campus in Jodhpur, Rajasthan.
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Online Classes
Online classes are conducted through the Shaurya Bharat app. The Shaurya Bharat app provides the best content in the form of live classes, video lectures, and ebooks.
Also, it offers topicwise practice tests and patternbased live exams. The live exams are based exactly on the pattern of the main examination. This gives students a wellrequired practice and a look and feel of the main examination. It makes them wellequipped to handle and overcome the examination pressure with minimum ease.
Useful Links
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