Chapter 4: Mensuration - Measuring the World Around Us
🎯 Objective: This chapter aims to build a strong conceptual and practical understanding of measurement. We will explore how to measure the boundaries (perimeter), surfaces (area), and space (volume) of various shapes, and master the units used for length, weight, capacity, time, and money. This knowledge is directly applicable to teaching at both primary and upper-primary levels .
🔲 Section 4.1: Perimeter - The Boundary Line
🚶 4.1.1 Concept of Perimeter
The perimeter is the total distance covered along the boundary of a closed figure. Imagine walking around a park or fencing a garden—the distance you cover is the perimeter. It's a measure of length, so its units are units of length like millimeters (mm), centimeters (cm), meters (m), and kilometers (km).
📐 4.1.2 Perimeter of Rectangle, Square, Triangle
Let's derive the formulas for common shapes.
| Shape | Diagram | Perimeter Formula | Explanation |
|---|---|---|---|
| Rectangle | ▭ | P = 2 × (l + b) where l = length, b = breadth | A rectangle has two lengths and two breadths. So, P = l + b + l + b = 2l + 2b = 2(l + b). |
| Square | ⬛ | P = 4 × s where s = side | All four sides of a square are equal. So, P = s + s + s + s = 4s. |
| Triangle | 🔺 | P = a + b + c where a, b, c are the lengths of the three sides | The perimeter is simply the sum of the lengths of its three sides. |
Solved Examples:
Rectangle: Find the perimeter of a rectangle with length 12 cm and breadth 7 cm.
P = 2 × (l + b) = 2 × (12 + 7) = 2 × 19 = 38 cm.
Square: Find the perimeter of a square with a side of 8.5 m.
P = 4 × s = 4 × 8.5 = 34 m.
Triangle: Find the perimeter of a triangle with sides 4 cm, 5 cm, and 6 cm.
P = 4 + 5 + 6 = 15 cm.
⬢ 4.1.3 Perimeter of Regular Polygons
A regular polygon is a polygon with all sides equal and all angles equal (e.g., equilateral triangle, square, regular pentagon).
Formula: Perimeter of a regular polygon = Number of sides × Length of one side
Examples:
Regular Pentagon (5 sides): If one side is 6 cm, Perimeter = 5 × 6 = 30 cm.
Regular Hexagon (6 sides): If one side is 7 m, Perimeter = 6 × 7 = 42 m.
📝 4.1.4 Word Problems
Problem 1 (Fencing): A farmer has a rectangular field of length 50 m and breadth 30 m. He wants to put a fence around it. If the cost of fencing is ₹15 per meter, what is the total cost?
Step 1: Find the perimeter (length of fence needed).
P = 2 × (l + b) = 2 × (50 + 30) = 2 × 80 = 160 m.
Step 2: Find the total cost.
Cost = Perimeter × Rate = 160 × 15 = ₹2400.
Problem 2 (Border): A square photo of side 25 cm has to be framed. What is the length of the frame required?
The length of the frame is the perimeter of the square photo.
P = 4 × s = 4 × 25 = 100 cm.
🟦 Section 4.2: Area - The Surface Inside
🟩 4.2.1 Concept of Area
The area is the amount of surface enclosed within a closed figure. It's measured in square units, like square millimeters (mm²), square centimeters (cm²), square meters (m²), etc. Think of it as the number of unit squares that can fit inside a shape.
🔲 4.2.2 Area of Rectangle and Square
| Shape | Diagram | Area Formula | Explanation |
|---|---|---|---|
| Rectangle | ▭ | A = l × b where l = length, b = breadth | If you divide a rectangle into rows and columns of unit squares, the total number of squares is length times breadth. |
| Square | ⬛ | A = s × s = s² where s = side | Since all sides of a square are equal, the length and breadth are both 's'. |
Solved Examples:
Rectangle: Find the area of a rectangle with length 15 cm and breadth 8 cm.
A = l × b = 15 × 8 = 120 cm².
Square: Find the area of a square with a side of 11 m.
A = s² = 11 × 11 = 121 m².
🧩 4.2.3 Area of Irregular Shapes Using Grid Method
To find the area of an irregular shape (like a leaf or an odd-shaped polygon), we can use a grid (graph paper).
Place the shape on a grid.
Count the number of full squares completely inside the shape. ➡️ Let this be
F.Count the number of half or more than half squares that are covered by the shape. ➡️ Let this be
H.Ignore squares that are less than half.
The approximate area is the sum of these: Area ≈ (F + H) × Area of one square.
Example: If a leaf covers 40 full squares and 28 half-or-more squares on a 1 cm × 1 cm grid, its approximate area is 40 + 28 = 68 cm².
🤔 4.2.4 Relationship Between Perimeter and Area
It's crucial to understand that there is no fixed relationship between perimeter and area.
Figures with the same perimeter can have different areas.
Figures with the same area can have different perimeters.
Example:
Consider two rectangles with a perimeter of 20 cm.
Rectangle A: l = 9 cm, b = 1 cm. Area = 9 × 1 = 9 cm².
Rectangle B: l = 6 cm, b = 4 cm. Area = 6 × 4 = 24 cm².
Rectangle C: l = 5 cm, b = 5 cm (a square). Area = 5 × 5 = 25 cm².
All have the same perimeter (20 cm), but their areas are vastly different. The square encloses the maximum area for a given perimeter.
🧊 Section 4.3: Volume - The Space Inside
📦 4.3.1 Concept of Volume
The volume of a solid object is the amount of space it occupies. It's measured in cubic units, like cubic centimeters (cm³), cubic meters (m³), etc. Think of it as the number of unit cubes that can fit inside a 3D shape.
🧱 4.3.2 Volume of Cube and Cuboid
| Shape | Diagram | Volume Formula | Explanation |
|---|---|---|---|
| Cuboid | 🧱 | V = l × b × h where l = length, b = breadth, h = height | A cuboid can be filled with layers of unit cubes. The number of cubes is length × breadth × height. |
| Cube | 🧊 | V = s × s × s = s³ where s = side | A cube is a special cuboid where l = b = h = s. |
Solved Examples:
Cuboid: Find the volume of a cuboid with length 10 cm, breadth 5 cm, and height 4 cm.
V = l × b × h = 10 × 5 × 4 = 200 cm³.
Cube: Find the volume of a cube with a side of 6 cm.
V = s³ = 6 × 6 × 6 = 216 cm³.
🔢 4.3.3 Measuring Volume Using Unit Cubes
The volume of a 3D shape can be found by filling it with unit cubes (cubes of side 1 unit). This is a hands-on way to understand the formula.
A cuboid of length 'l', breadth 'b', and height 'h' can hold 'l' number of cubes along its length, 'b' number along its breadth, and 'h' number of layers. The total number of cubes is l × b × h, which is its volume in cubic units.
💧 4.3.4 Relationship Between Capacity and Volume
Volume is the total space an object occupies.
Capacity is the maximum amount of liquid (or any substance) that a container can hold.
For a solid object, its internal volume is its capacity.
| Unit of Volume | Relationship | Unit of Capacity |
|---|---|---|
| 1 cm³ (cubic centimeter) | = 1 mL (milliliter) | 💧 |
| 1000 cm³ | = 1 L (liter) | 🥛 |
| 1 m³ (cubic meter) | = 1000 L | 🛢️ |
⚖️ Section 4.4: Measurement
This section deals with the practical aspects of measuring different attributes of objects.
📏 4.4.1 Length – Units and Conversion
Length is the measurement of distance between two points.
| Unit | Symbol | Relation to Meter |
|---|---|---|
| Millimeter | mm | 1 m = 1000 mm |
| Centimeter | cm | 1 m = 100 cm |
| Meter | m | Base Unit |
| Kilometer | km | 1 km = 1000 m |
Conversion Rules:
To convert a larger unit to a smaller unit (e.g., m to cm), multiply.
Example: 5 m = 5 × 100 = 500 cm.
To convert a smaller unit to a larger unit (e.g., cm to m), divide.
Example: 350 cm = 350 ÷ 100 = 3.5 m.
⚖️ 4.4.2 Weight – Units and Conversion
Weight (or mass) is the measure of how heavy an object is.
| Unit | Symbol | Relation to Gram |
|---|---|---|
| Gram | g | Base Unit |
| Kilogram | kg | 1 kg = 1000 g |
Conversion Rules:
kg → g: Multiply by 1000 (e.g., 2.5 kg = 2.5 × 1000 = 2500 g).
g → kg: Divide by 1000 (e.g., 7500 g = 7500 ÷ 1000 = 7.5 kg).
💧 4.4.3 Capacity – Units and Conversion
Capacity measures the volume of liquid a container can hold.
| Unit | Symbol | Relation to Liter |
|---|---|---|
| Milliliter | mL | 1 L = 1000 mL |
| Liter | L | Base Unit |
Conversion Rules:
L → mL: Multiply by 1000 (e.g., 3.75 L = 3.75 × 1000 = 3750 mL).
mL → L: Divide by 1000 (e.g., 4250 mL = 4250 ÷ 1000 = 4.25 L).
⏰ 4.4.4 Time – Reading Clock, Calendar, Time Intervals, Conversion
Time is a fundamental measure of duration.
Reading a Clock:
The short hand is the hour hand. 🕐
The long hand is the minute hand. 🕐
The seconds hand (in analog clocks) is the thinnest and moves quickly.
Units of Time:
| Unit | Relation |
|---|---|
| Minute | 1 minute = 60 seconds |
| Hour | 1 hour = 60 minutes |
| Day | 1 day = 24 hours |
| Week | 1 week = 7 days |
| Month | 1 month ≈ 4 weeks or 30/31 days |
| Year | 1 year = 12 months = 365 days (366 in a leap year) |
| Leap Year | A year divisible by 4 (e.g., 2024, 2028) |
Time Intervals: The duration between two given times.
Example: Find the time between 9:25 AM and 11:10 AM.
From 9:25 to 10:00 is 35 minutes.
From 10:00 to 11:10 is 1 hour and 10 minutes.
Total interval = 1 hour + (35 + 10) minutes = 1 hour 45 minutes.
💰 4.4.5 Money – Operations Involving Money, Bills
This is the most common real-life application of mathematics.
Operations: Addition, subtraction, multiplication, and division of money in rupees and paise.
Key Fact: ₹1 = 100 paise.
Writing Bills: A bill is a document that itemizes the items purchased, their quantities, their prices, and the total amount to be paid.
Example Bill:
📝 The Stationery Shop
Date: 2024-05-20
| S.No. | Item Description | Quantity | Rate (per item) | Total Cost |
|---|---|---|---|---|
| 1. | Notebooks | 5 | ₹25.00 | ₹125.00 |
| 2. | Pens | 10 | ₹5.00 | ₹50.00 |
| 3. | Geometry Box | 1 | ₹85.00 | ₹85.00 |
| Subtotal | ₹260.00 | |||
| Total Amount | ₹260.00 |
Word Problem:
Rohan bought 2 kg of apples at ₹80 per kg and 3 kg of oranges at ₹50 per kg. He gave a ₹500 note to the shopkeeper. How much money will he get back?
Cost of apples = 2 × 80 = ₹160
Cost of oranges = 3 × 50 = ₹150
Total cost = 160 + 150 = ₹310
Money returned = 500 - 310 = ₹190.
📝 Chapter Summary: Quick Revision Table for PSTET
| Section | Key Concepts | PSTET Focus |
|---|---|---|
| 4.1 Perimeter | Concept, formulas for rectangle, square, triangle, regular polygons, word problems (fencing, borders). | Applying correct formulas in word problems, especially those involving cost. |
| 4.2 Area | Concept, formulas for rectangle and square, grid method for irregular shapes, relationship with perimeter. | Calculating area, finding missing sides, understanding that shapes with same perimeter can have different areas. |
| 4.3 Volume | Concept, formulas for cube and cuboid, unit cube method, relationship between volume and capacity (1 cm³ = 1 mL). | Solving problems involving volume of tanks/boxes and converting between volume and capacity units. |
| 4.4 Measurement | Units and conversions for length, weight, capacity, time, and money; reading clocks; preparing bills. | Performing accurate conversions, calculating time intervals, and solving practical problems involving money and bills. |
🧠 PSTET Preparation Strategy for Mensuration
| Topic | Priority | Preparation Tips |
|---|---|---|
| Perimeter & Area | High | Memorize formulas. Practice many word problems. Focus on distinguishing between these two concepts. |
| Volume & Capacity | High | Master the formula V = l × b × h. Practice problems involving filling/emptying tanks and the relationship between cm³ and mL. |
| Measurement Conversion | Medium | Create a flashcard for conversion rules (e.g., multiply to go to smaller unit, divide to go to larger unit). Practice with all four measures: length, weight, capacity, time. |
| Money & Bills | Medium | Practice addition and subtraction of money. Create sample bills to ensure you understand the process. |
I hope this comprehensive chapter serves as a valuable resource in your PSTET preparation journey. A strong grasp of mensuration and measurement is not just for exams but is a life skill you'll pass on to your students. Happy teaching and best of luck! 👍
