Perimeter and Area

Perimeter — Part 1

Perimeter — The Journey Around a Shape

Imagine a farmer, Mr. Sharma, who has a beautiful rectangular vegetable garden. To protect his crops from animals, he decides to build a fence all around it. He needs to know the exact length of wire required for the fence. How does he calculate this?

He measures the length of all four sides of his garden and adds them up. This total length of the boundary is what we call the Perimeter. The word perimeter comes from two Greek words: peri meaning 'around' and metron meaning 'measure'. So, perimeter literally means "to measure around". It's the total distance you would walk if you traced the edge of a shape exactly once.

{{FORMULA: expr=P = 2 × (l + b) | symbols=P:Perimeter of a rectangle, l:length, b:breadth}}

Definitions & Formulas

Before we start calculating, let's get our key terms and formulas straight. These are the building blocks for everything we'll learn about perimeter.

How Are These Formulas Derived?

The formulas for the perimeter of a rectangle and a square aren't magic! They are just smart, quick ways of adding the side lengths. Let's see how they come about.

1. Deriving the Perimeter of a Rectangle

A rectangle has two pairs of equal sides: two lengths and two breadths.

{{VISUAL: diagram: A labeled rectangle ABCD with length 'l' (top and bottom sides AB and CD) and breadth 'b' (left and right sides BC and DA). Arrows trace the boundary from A → B → C → D → A to illustrate the concept of perimeter.}}

1. The definition of perimeter is the sum of all sides. For a rectangle ABCD, this is:

   Perimeter = AB + BC + CD + DA
   

2. We know that in a rectangle, opposite sides are equal. So, AB = CD = length (l) and BC = DA = breadth (b). Let's substitute these into our equation.

   Perimeter = l + b + l + b
   

3. Now, we can group the like terms together.

   Perimeter = (l + l) + (b + b)
   

4. Adding them gives us:

   Perimeter = 2 × l + 2 × b
   

5. Finally, we can take '2' as a common factor, which gives us the familiar formula.

   Perimeter = 2 × (l + b)
   

2. Deriving the Perimeter of a Square

A square is even simpler because all four of its sides are equal.

{{VISUAL: diagram: A labeled square PQRS with all four sides marked with the same variable 's' to indicate they are equal. The formula P = 4 × s is shown just below the square.}}

1. Again, we start with the basic definition: the perimeter is the sum of all sides.

   Perimeter = side + side + side + side
   

2. Let's call the length of one side 's'. Since all sides are equal, the equation becomes:

   Perimeter = s + s + s + s
   

3. This is simply 's' added to itself four times, which is the same as multiplication.

   Perimeter = 4 × s
   

{{KEY: type=concept | title=The Core Idea of Perimeter | text=Perimeter is always about addition and length. It is a one-dimensional measurement (units are cm, m, km, etc.), not a two-dimensional measurement like area (which uses cm², m², etc.).}}

Solved Numericals

Let's apply these formulas to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Finding the Perimeter of a Square Photo Frame (Easy)

Given: A square photo frame has a side of length 25 cm.

To Find: The perimeter of the photo frame.

Solution:

1. Identify the shape and the formula. The shape is a square, so we use the formula for the perimeter of a square.

   Perimeter of a square = 4 × length of a side
   

2. Substitute the given side length into the formula.

   Perimeter = 4 × 25 cm
   

3. Calculate the final result.

   Perimeter = 100 cm
   

Final Answer: The perimeter of the square photo frame is 100 cm.

Example 2: Cost of Fencing a Rectangular Park (Medium)

{{VISUAL: diagram: A rectangular park for the 'Medium' solved example, with its length labeled as 150 m and breadth as 120 m. A simple line representing a fence is drawn around its boundary.}}

Given: A rectangular park is 150 m long and 120 m wide. The cost of fencing is ₹40 per metre.

To Find: The total cost of fencing the park.

Solution:

1. First, we need to find the total length of the fence required. This is equal to the perimeter of the park.

   Perimeter = 2 × (length + breadth)
   

2. Substitute the given values for length and breadth.

   Perimeter = 2 × (150 m + 120 m)
   

3. Solve the expression inside the brackets first.

   Perimeter = 2 × (270 m)
   

4. Calculate the total perimeter.

   Perimeter = 540 m
   

5. Now, calculate the total cost. The cost is ₹40 for every metre of the fence.

   Total Cost = Perimeter × Cost per metre
   

6. Substitute the values.

   Total Cost = 540 × ₹40
   

7. Calculate the final cost.

   Total Cost = ₹21,600
   

Final Answer: The total cost of fencing the park is ₹21,600.

Example 3: Finding the Missing Side of a Rectangle (Hard)

Given: The perimeter of a rectangular board is 18 m. Its length is 5 m.

To Find: The breadth of the rectangular board.

Solution:

1. Write down the formula for the perimeter of a rectangle.

   Perimeter = 2 × (length + breadth)
   

2. Substitute the known values into the formula. Let 'b' be the unknown breadth.

   18 m = 2 × (5 m + b)
   

3. To find 'b', we need to isolate it. First, divide both sides of the equation by 2.

   18 ÷ 2 = 5 + b
   

4. Simplify the left side.

   9 = 5 + b
   

5. Now, subtract 5 from both sides to find 'b'.

   b = 9 - 5
   

6. Calculate the final value for the breadth.

   b = 4 m
   

Final Answer: The breadth of the rectangular board is 4 m.

Example 4: Multiple Rounds Around a Square Park (Tricky)

Given: Usha takes three full rounds of a square park whose side is 75 m.

To Find: The total distance travelled by Usha.

Solution:

1. First, find the distance Usha covers in one round. This is the perimeter of the square park.

   Perimeter = 4 × side
   

2. Substitute the side length of the park.

   Perimeter = 4 × 75 m
   

   Perimeter = 300 m
   

3. So, the distance covered in one round is 300 m. Usha takes three rounds.

   Total Distance = Distance in one round × Number of rounds
   

4. Substitute the values to find the total distance.

   Total Distance = 300 m × 3
   

5. Calculate the final answer.

   Total Distance = 900 m
   

Final Answer: The total distance travelled by Usha is 900 m.

Try It Yourself

1. The perimeter of a square is 20 cm. What is the length of its side?
2. A rectangle has a perimeter of 14 cm and a breadth of 2 cm. What is its length?
3. A piece of string is 36 cm long. What will be the length of each side if it is used to form a square?

Answer Key: 1. 5 cm, 2. 5 cm, 3. 9 cm

Tips & Tricks

Here are a few shortcuts to help you solve perimeter problems faster and more accurately.

Common Mistakes to Avoid

Many students make small errors that lead to wrong answers. Here’s what to watch out for.

Brain-Teaser Questions

Think you've mastered the basics? Try these tricky questions!

1. A piece of wire is bent into the shape of a rectangle with a length of 10 cm and a breadth of 6 cm. If the wire is straightened and then re-bent to form a square, what will be the length of the square's side? {{VISUAL: diagram: A wire bent into a rectangle with dimensions 10cm and 6cm, then shown as a single straight line, and finally bent into a square with all sides labeled 's'.}}

   💡 Answer: The length of the wire is equal to the perimeter of the rectangle. P = 2 × (10 + 6) = 32 cm. When this 32 cm wire forms a square, its perimeter is 32 cm. Side of the square = Perimeter ÷ 4 = 32 ÷ 4 = 8 cm.

2. The perimeter of square A is 24 cm. The perimeter of square B is double the perimeter of square A. What is the length of the side of square B?

   💡 Answer: Perimeter of square B = 2 × (Perimeter of square A) = 2 × 24 cm = 48 cm. The side of square B = Its perimeter ÷ 4 = 48 ÷ 4 = 12 cm.

3. A rectangular field is 50 m long. The cost of fencing it at ₹20 per metre is ₹3000. What is the breadth of the field?

   💡 Answer: First, find the perimeter. Total Cost = Perimeter × Cost per metre. So, Perimeter = Total Cost ÷ Cost per metre = ₹3000 ÷ ₹20 = 150 m. We know P = 2 × (l + b). So, 150 = 2 × (50 + b). This means 75 = 50 + b. Therefore, breadth (b) = 75 - 50 = 25 m.

Mini Cheatsheet

Here's a quick summary of all the key formulas from this page. Screenshot this for your last-minute revision!

Perimeter — Part 2

Page 2: Perimeter — Part 2: Triangles and Regular Polygons

Welcome back! In our last lesson, we explored the concept of perimeter and learned how to calculate it for squares and rectangles. Now, let's expand our understanding to other common shapes like triangles and a special family of shapes called regular polygons.

Imagine you're designing a unique triangular kite or building a hexagonal sandbox for a playground. To know how much material you need for the border or fence, you'll need to calculate the perimeter. This lesson will give you the tools to measure the boundary of any polygon, no matter how many sides it has!

{{FORMULA: expr=P = n × s | symbols=P:Perimeter of a regular polygon, n:number of equal sides, s:length of one side}}

Definitions & Formulas

Let's organize the key formulas for the shapes we'll be discussing today. Understanding these is the first step to mastering perimeter calculations.

The Logic Behind Regular Polygons

Why is there a special, simple formula for regular polygons? The logic comes directly from the definition of "regular" and "perimeter". Let's break it down.

1. A regular polygon is a closed figure where all sides are equal in length and all angles are equal. Examples include an equilateral triangle (3 sides), a square (4 sides), a regular pentagon (5 sides), and a regular hexagon (6 sides).

2. The perimeter is the sum of the lengths of all its sides.

3. Let's consider a regular polygon with n sides. Let the length of one side be s. Since it's a regular polygon, every single one of its n sides has the same length, s.

{{VISUAL: diagram: A labeled regular hexagon with vertices A, B, C, D, E, F. Each side (AB, BC, CD, etc.) is marked with the same label 's' to show they are equal in length.}}

4. To find the perimeter, we need to add the lengths of all n sides together.

Perimeter = s + s + s + ... (added n times)

5. Repeated addition is simply multiplication. If we add s to itself n times, we get n multiplied by s.

Perimeter = n × s

This simple, powerful formula works for any regular polygon, whether it has 3 sides or 100 sides!

Solved Examples

Let's apply these formulas to solve some problems, starting from easy ones and moving to more challenging ones.

Example 1: The Triangular Park (Easy)

Given: A triangular park has sides measuring 120 m, 90 m, and 75 m.

To Find: The perimeter of the park.

Solution:

1. The perimeter of a triangle is the sum of the lengths of its three sides. The sides are a = 120 m, b = 90 m, and c = 75 m.

P = a + b + c

2. Substitute the given values into the formula.

P = 120 m + 90 m + 75 m

3. Add the numbers to find the total length.

P = 285 m

Final Answer: The perimeter of the triangular park is 285 m.

Example 2: The Missing Side (Medium)

Given: The perimeter of a triangle is 55 cm. Two of its sides are 20 cm and 14 cm.

To Find: The length of the third side.

Solution:

1. Let the three sides be a, b, and c. We are given P = 55 cm, a = 20 cm, and b = 14 cm. We need to find c.

2. The formula for the perimeter is the sum of all sides.

P = a + b + c

3. Substitute the known values into the equation.

55 cm = 20 cm + 14 cm + c

4. First, add the lengths of the two known sides.

55 cm = 34 cm + c

5. To find c, subtract the sum of the known sides from the total perimeter.

c = 55 cm - 34 cm

6. Perform the subtraction.

c = 21 cm

Final Answer: The length of the third side is 21 cm.

Example 3: Fencing a Hexagonal Garden (Hard)

Given: A garden is in the shape of a regular hexagon with each side measuring 18 m. The cost of fencing is ₹40 per metre.

To Find: The total cost of fencing the garden.

{{VISUAL: diagram: A regular hexagonal garden. One side is labeled "18 m". A small icon of a fence post is shown at each vertex.}}

Solution:

1. First, we need to find the total length of the fence required, which is the perimeter of the hexagonal garden. A hexagon has 6 sides (n = 6). Since it's a regular hexagon, all sides are equal (s = 18 m).

2. Use the formula for the perimeter of a regular polygon.

P = n × s

3. Substitute the values for the hexagon.

P = 6 × 18 m

4. Calculate the perimeter.

P = 108 m

5. Now, calculate the total cost. The cost is ₹40 for every metre of fence.

Total Cost = Perimeter × Cost per metre

6. Substitute the values to find the final cost.

Total Cost = 108 × 40

Total Cost = ₹ 4320

Final Answer: The total cost of fencing the garden is ₹4320.

Example 4: The Shape-Shifting Wire (Tricky)

Given: A piece of wire is 36 cm long. It is bent to form an equilateral triangle.

To Find: The length of each side of the equilateral triangle.

Solution:

1. The key idea here is that the total length of the wire does not change, no matter what shape it is bent into. The length of the wire is the perimeter of the shape it forms.

2. Therefore, the perimeter of the equilateral triangle is equal to the length of the wire.

Perimeter (P) = 36 cm

3. An equilateral triangle is a regular polygon with 3 equal sides (n = 3). We need to find the length of one side (s).

{{VISUAL: diagram: A visual flow showing three stages. Stage 1: A straight line labeled "Wire, 36 cm". Stage 2: An arrow pointing to a square, with a question mark on its side. Stage 3: Another arrow pointing to an equilateral triangle, with a question mark on its side.}}

4. Use the perimeter formula for a regular polygon, P = n × s, adapted for an equilateral triangle (n=3).

P = 3 × s

5. Substitute the known perimeter.

36 cm = 3 × s

6. To find the side length s, divide the perimeter by the number of sides.

s = 36 cm ÷ 3

s = 12 cm

Final Answer: The length of each side of the equilateral triangle is 12 cm.

{{KEY: type=concept | title=The Constant Perimeter Principle | text=When a piece of string or wire is bent to form different closed shapes, the total length of the wire remains the same. This means the perimeter of each shape formed will be equal.}}

Tips & Tricks

Here are a few shortcuts and mental checks to help you solve perimeter problems faster and more accurately.

Common Mistakes to Avoid

Many students make small, avoidable errors. Watch out for these common traps!

Brain-Teaser Questions

Ready for a challenge? These questions require you to think a little differently.

1. The L-Shaped Lawn: An L-shaped lawn has the following outer dimensions: the total length is 12 m and the total width is 10 m. What is its perimeter? (You don't need the inner side lengths to solve this!)

   {{VISUAL: diagram: An L-shaped polygon. The total vertical height is labeled 12 m, and the total horizontal width is labeled 10 m. The inner corner's sides are left unlabeled.}}

   💡 Answer: The perimeter is 44 m. Imagine pushing the two inner sides outwards. They would complete a full rectangle of 12 m by 10 m. The perimeter of this rectangle is 2 × (12 + 10) = 2 × 22 = 44 m. The perimeter of the L-shape is the same as the perimeter of the rectangle it fits into!

2. Scaling Up: An equilateral triangle has a side of 5 cm. If you double the length of each side to 10 cm, what happens to its perimeter?

   💡 Answer: The perimeter also doubles. The original perimeter is 3 × 5 = 15 cm. The new perimeter is 3 × 10 = 30 cm. The new perimeter is exactly twice the old one. This is true for any polygon!

3. Two Runners: Usha runs one lap around a square park of side 75 m. Akshi runs one lap around a regular hexagonal park of side 50 m. Who ran a longer distance?

   💡 Answer: They both ran the exact same distance. Usha's distance = Perimeter of square = 4 × 75 m = 300 m. Akshi's distance = Perimeter of hexagon = 6 × 50 m = 300 m.

Mini Cheatsheet

Here's a handy summary of all the perimeter formulas we've covered. Screenshot this for your last-minute revision!

Area — Part 1

Area — Part 1

Welcome back! In our last lesson, we walked along the boundaries of shapes and learned how to measure their perimeter. Now, it's time to step inside those boundaries and explore the space they enclose.

Imagine you want to paint a wall, lay a carpet on a floor, or figure out how much wrapping paper you need for a gift. In all these cases, you're not interested in the length of the border; you need to know the size of the entire surface. This measure of the surface or region enclosed by a closed figure is called its area. While perimeter is a measure of length (1-dimensional), area is a measure of space (2-dimensional), and it tells us "how much stuff fits inside".

{{FORMULA: expr=A = l × w | symbols=A:Area of the rectangle, l:length, w:width}}

Definitions & Formulas

Before we start calculating, let's be clear about the terms and formulas we will use. These are the foundational building blocks for understanding area.

The Logic Behind the Formula

Have you ever wondered why the area of a rectangle is length × width? It’s not just a rule to memorize; it comes from a very simple idea of counting squares. Let's see how.

1. Imagine we have a rectangular floor that is 5 metres long and 4 metres wide.

2. Now, let's cover this floor with square tiles, where each tile is exactly 1 metre long and 1 metre wide. The area of one such tile is 1 square metre (1 sq m).

3. If we lay these tiles along the length (5 m), we can fit exactly 5 tiles in one row.

{{VISUAL: diagram: a 5 by 4 grid of squares. The length along the top is labeled 'length = 5 m' with 5 squares shown. The width along the side is labeled 'width = 4 m' with 4 squares shown. Text inside says 'Total squares = 5 × 4 = 20'.}}

4. Since the width of the floor is 4 m, we can fit 4 such rows of tiles, one below the other.

5. So, we have 4 rows, and each row has 5 tiles. To find the total number of tiles, we multiply the number of rows by the number of tiles in each row.

   Total Tiles = 5 tiles per row × 4 rows = 20 tiles
   

6. Because each tile represents 1 sq m of area, the total area of the floor is 20 sq m. This shows us that Area = length × width. The same logic applies to a square, where length and width are just the same (side × side).

{{KEY: type=concept | title=The Importance of Square Units | text=Area is always measured in square units (like m², cm², sq km). This is because we are measuring how many 1x1 squares can fit into a shape. Never write the unit for area as 'm' or 'cm'; it must be 'm²' or 'sq m'.}}

Solved Examples

Let's apply these formulas to solve some real-world problems. We'll start easy and gradually increase the difficulty.

Example 1: Finding a Missing Dimension (Easy)

Given: The area of a rectangular garden is 300 sq m and its length is 25 m.

To Find: The width of the garden.

Solution:

1. Start with the formula for the area of a rectangle.

   Area = length × width
   

2. Substitute the known values into the formula.

   300 sq m = 25 m × width
   

3. To find the width, we need to rearrange the equation. We can do this by dividing the area by the length.

   width = 300 sq m ÷ 25 m
   

4. Perform the division to get the final answer.

   width = 12 m
   

Final Answer: The width of the garden is 12 m.

Example 2: Calculating Cost Based on Area (Medium)

Given: A rectangular plot of land is 500 m long and 200 m wide. The cost of tiling is ₹8 per hundred sq m.

To Find: The total cost of tiling the entire plot.

Solution:

1. First, calculate the total area of the rectangular plot.

   Area = length × width = 500 m × 200 m
   

   Area = 100,000 sq m
   

2. The cost is given per hundred sq m. So, we need to find out how many 'hundred sq m' blocks are in our total area.

   Number of blocks = Total Area ÷ 100 sq m
   

   Number of blocks = 100,000 ÷ 100 = 1000
   

3. Now, multiply the number of blocks by the cost per block to find the total cost.

   Total Cost = Number of blocks × Cost per block
   

   Total Cost = 1000 × ₹8 = ₹8000
   

Final Answer: The total cost of tiling the plot is ₹8,000.

{{VISUAL: diagram: a large rectangle representing the plot labeled '500 m' and '200 m'. A small 10x10 square is shown in the corner labeled '100 sq m block'. An arrow points to it with the text 'Cost = ₹8'.}}

Example 3: Area of Remaining Portion (Hard)

Given: A floor is 5 m long and 4 m wide. A square carpet with sides of 3 m is laid on it.

To Find: The area of the floor that is not carpeted.

Solution:

1. This problem requires a two-step approach. First, find the total area of the floor.

   Area of floor = length × width = 5 m × 4 m
   

   Area of floor = 20 sq m
   

2. Next, find the area of the square carpet.

   Area of carpet = side × side = 3 m × 3 m
   

   Area of carpet = 9 sq m
   

3. To find the area of the floor not covered by the carpet, subtract the carpet's area from the floor's total area.

   Uncarpeted Area = Area of floor - Area of carpet
   

   Uncarpeted Area = 20 sq m - 9 sq m = 11 sq m
   

Final Answer: The area of the floor that is not carpeted is 11 sq m.

Example 4: Area of a Complex Shape (Tricky)

Given: The L-shaped figure below, with all measurements in metres.

{{VISUAL: diagram: An L-shaped polygon. The top horizontal side is 3m. The right-most vertical side is 2m. The bottom horizontal side is 5m. The left-most vertical side is 4m. The inner corner has sides of 2m (horizontal) and 2m (vertical). A dotted line is shown splitting the L-shape into two rectangles, one 3x2 and the other 5x2 (by extending the line from the inner corner).}}

To Find: The total area of the figure.

Solution:

1. This shape is not a simple rectangle. The trick is to split it into two smaller, familiar rectangles. We can split it vertically or horizontally. Let's split it vertically. This creates two rectangles: Rectangle A on the left and Rectangle B on the right.

2. Calculate the dimensions of Rectangle A. The left side is 4 m. The top side is 3 m.

   Area of Rectangle A = 4 m × 3 m = 12 sq m
   

3. Calculate the dimensions of Rectangle B. The right side is 2 m. The bottom side is the total bottom length (5 m) minus the bottom part of Rectangle A (which is 3 m). So, the width of B is 5 m - 3 m = 2 m.

   Dimensions of Rectangle B = 2 m × 2 m
   

   Area of Rectangle B = 4 sq m
   

4. Add the areas of the two smaller rectangles to get the total area of the L-shaped figure.

   Total Area = Area of A + Area of B
   

   Total Area = 12 sq m + 4 sq m = 16 sq m
   

   Alternative Method: You could also split it horizontally to get a top 3x2 rectangle and a bottom 5x2 rectangle, giving 6 + 10 = 16 sq m. The result is the same!

Final Answer: The total area of the figure is 16 sq m.

Tips & Tricks

Mastering area calculations is easier with a few shortcuts in your toolkit.

Common Mistakes

Many students make small errors that lead to wrong answers. Watch out for these common pitfalls!

Brain-Teaser Questions

Ready to challenge your understanding? Try these problems that require a little extra thought.

1. The length of a rectangle is 10 cm and its area is 60 sq cm. If the length is doubled to 20 cm while the width stays the same, what is the new area?

   💡 Answer: The new area will be 120 sq cm. Since Area = length × width, if you double the length, you double the area.

2. A square tile has a side of 20 cm. How many such tiles are required to cover a rectangular floor of length 5 m and width 4 m?

   💡 Answer: 500 tiles. Floor area = 5 m × 4 m = 20 m². Tile area = 20 cm × 20 cm = 400 cm². Convert floor area to cm²: 20 m² = 20 × 10,000 = 200,000 cm². Number of tiles = 200,000 ÷ 400 = 500.

3. A rectangular park is 40 m long and 30 m wide. A path 2 m wide is built all around the inside of the park. What is the area of the path?

   💡 Answer: 256 sq m. Area of the whole park (outer rectangle) = 40 m × 30 m = 1200 sq m. The inner lawn's length is 40 - 2 - 2 = 36 m. The inner lawn's width is 30 - 2 - 2 = 26 m. Area of inner lawn = 36 m × 26 m = 944 sq m. Area of path = Outer Area - Inner Area = 1200 - 944 = 256 sq m.

Mini Cheatsheet

Here’s a quick summary of everything we covered. Screenshot this for your last-minute revision!

Area — Part 2

Area — Part 2: Estimating with Grids

Concept Introduction

Imagine you are a geographer studying a beautiful, irregularly shaped lake from a satellite image. How would you measure its surface area? You can't use a simple length × width formula. This is where estimation becomes a powerful tool. By overlaying a grid of known-sized squares (like graph paper) onto the image of the lake, you can count the squares it covers. This method allows us to find an approximate area for any shape, no matter how complex.

This technique is used in many real-world fields, from cartography (map making) to biology (measuring the area of a leaf to study photosynthesis) and even medicine (assessing the size of a skin rash). It's a practical skill that turns a complex problem into a simple counting exercise.

{{FORMULA: expr=Area of Rectangle = Length × Width | symbols=Length: the longer side of the rectangle, Width: the shorter side of the rectangle}}

The Method of Estimation

When we place an irregular shape on a squared grid paper, it rarely covers the squares perfectly. To get a consistent and reasonable estimate of the area, we follow a set of simple rules.

The Logic: Why This Method Works

Estimating area with a grid is a process of approximation. The rules are designed to balance out the errors. The parts we ignore (less than half) are assumed to be compensated for by the parts we round up (more than half).

1. Trace the Figure: First, trace the outline of the irregular shape onto a sheet of squared paper (or graph paper). Each small square on the paper represents 1 square unit.

2. Count the Full Squares: Look inside the outline and count every single square that is fully covered by the shape. Let's call this count F.

3. Count the Partial Squares: Now, examine the squares along the boundary.

   • Count the squares that are more than half-covered. Let's call this M.
   • Count the squares that are exactly half-covered. Let's call this H.
   • Ignore all squares that are less than half-covered.

4. Calculate the Total Area: The estimated area is the sum of the values from the squares you've counted.

   • Total Area ≈ (Number of full squares) + (Number of more-than-half squares) + (Number of half squares × ½)
   • In short: Area ≈ F + M + (H × ½)

This method provides a remarkably good estimate, and the more squares you use (a finer grid), the more accurate your estimate becomes!

{{VISUAL: diagram: An irregular leaf shape drawn over a grid paper. Squares are color-coded: green for 'full', yellow for 'more than half', blue for 'exactly half', and red for 'less than half' (which would be ignored).}}

Solved Examples & Practice

Here we apply the concepts of area calculation for regular shapes and estimation for irregular ones.

Example 1: Area of a Tiled Floor (Easy)

Given: A rectangular room is 8 m long and 6 m wide. It needs to be tiled with square tiles of side 2 m.

To Find: How many square tiles are needed to cover the entire floor?

Solution:

1. First, calculate the total area of the rectangular floor. The formula is Area = Length × Width.

   Area of floor = 8 m × 6 m = 48 sq m
   

2. Next, calculate the area of a single square tile. The formula is Area = Side × Side.

   Area of one tile = 2 m × 2 m = 4 sq m
   

3. To find the number of tiles needed, divide the total area of the floor by the area of one tile.

   Number of tiles = (Area of floor) ÷ (Area of one tile) = 48 ÷ 4 = 12
   

Final Answer: 12 tiles are needed to cover the floor.

Example 2: Area of a Composite Shape (Medium)

Given: The L-shaped figure below, with all measurements in metres.

{{VISUAL: diagram: An L-shaped polygon with vertices labeled. The vertical arm is 10m high and 3m wide. The horizontal arm is 7m long and 4m wide from the base. Dimensions are clearly marked.}}

To Find: The total area of the figure.

Solution:

1. This shape is a composite figure. We can find its area by splitting it into two simpler rectangles. Let's split it horizontally. This creates a top rectangle (A) and a bottom rectangle (B).

2. Calculate the area of Rectangle A (the top part).

   • Its length is given as 3 m.
   • Its width is the total height minus the height of the bottom part: 10 m - 4 m = 6 m.

   Area of A = 3 m × 6 m = 18 sq m
   

3. Calculate the area of Rectangle B (the bottom part).

   • Its length is given as 7 m.
   • Its width is given as 4 m.

   Area of B = 7 m × 4 m = 28 sq m
   

4. Add the areas of the two rectangles to get the total area of the L-shaped figure.

   Total Area = Area of A + Area of B = 18 sq m + 28 sq m = 46 sq m
   

Final Answer: The total area of the figure is 46 sq m.

Example 3: Estimating Area on a Grid (Hard)

Given: A drawing of a pond on a 1 cm × 1 cm grid paper. The drawing covers:

• 35 full squares.
• 18 squares that are more than half-covered.
• 8 squares that are exactly half-covered.
• 12 squares that are less than half-covered.

To Find: The estimated area of the pond in square centimetres.

Solution:

1. Apply the rules of grid estimation. We will assign values to each category of squares.

2. Value from full squares: Each of the 35 full squares is counted as 1 sq cm.

   Area from full squares = 35 × 1 = 35 sq cm
   

3. Value from 'more than half' squares: Each of the 18 squares is counted as 1 sq cm.

   Area from 'more than half' squares = 18 × 1 = 18 sq cm
   

4. Value from 'exactly half' squares: Each of the 8 squares is counted as ½ sq cm.

   Area from 'exactly half' squares = 8 × ½ = 4 sq cm
   

5. Value from 'less than half' squares: These 12 squares are ignored and counted as 0.

   Area from 'less than half' squares = 12 × 0 = 0 sq cm
   

6. Sum up the areas from all categories to get the total estimated area.

   Total Estimated Area = 35 + 18 + 4 + 0 = 57 sq cm
   

Final Answer: The estimated area of the pond is 57 sq cm.

Example 4: Area Logic with Tangrams (Tricky)

Given: A standard 7-piece tangram set. Let the area of the smallest triangle piece be 1 unit.

{{VISUAL: diagram: A classic 7-piece tangram set, with each piece labeled: two large triangles (A, B), one medium triangle (C), two small triangles (D, E), one square (F), and one parallelogram (G).}}

To Find: What is the area of the entire square formed by all 7 pieces, in terms of the smallest triangle's area?

Solution:

1. This is a puzzle of relative areas. We don't need a formula, just observation. Let the area of one small triangle (like D or E) be 1 unit.

2. Observe the relationships between the pieces:

   • The square piece (F) can be formed by two small triangles. So, Area of F = 2 units.
   • The medium triangle (C) can also be formed by two small triangles. So, Area of C = 2 units.
   • The parallelogram (G) can also be formed by two small triangles. So, Area of G = 2 units.

3. Now look at the large triangles (A and B). Each large triangle can be formed by combining the medium triangle and two small triangles, or by four small triangles.

   Area of one large triangle (A or B) = 4 units
   

4. To find the total area of the complete tangram square, add up the areas of all 7 pieces.

   • 2 Large Triangles (A, B): 2 × 4 = 8 units
   • 1 Medium Triangle (C): 1 × 2 = 2 units
   • 2 Small Triangles (D, E): 2 × 1 = 2 units
   • 1 Square (F): 1 × 2 = 2 units
   • 1 Parallelogram (G): 1 × 2 = 2 units

5. Sum all these values.

   Total Area = 8 + 2 + 2 + 2 + 2 = 16 units
   

Final Answer: The area of the entire square is 16 times the area of the smallest triangle.

Try It Yourself

1. By splitting the following figure into rectangles, find its area (all measures are in cm). The shape is a "plus" sign, with the central square being 3 cm × 3 cm, and four 2 cm × 3 cm rectangles attached to each side.
2. A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
3. Estimate the area of a circle of radius 7 cm drawn on a 1 cm grid paper. (Hint: Trace it and count squares. Your answer will be an approximation).

Answer Key: 1. 33 sq cm, 2. 400 sq m, 3. Approx. 154 sq cm (The actual area is πr² ≈ (22/7) × 7² = 154 sq cm, so a good estimation should be close to this value).

{{KEY: type=concept | title=Area vs. Perimeter | text=Two shapes can have the same area but very different perimeters. For a fixed area, a shape that is more "compact" and square-like will generally have a smaller perimeter than a long, thin shape. For example, a 6×6 square (Area=36, Perimeter=24) has a smaller perimeter than a 1×36 rectangle (Area=36, Perimeter=74).}}

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. You have 36 meters of fencing. What is the largest rectangular area you can enclose? What are the dimensions of this rectangle?

   💡 Answer: The largest area is enclosed by a 9 m × 9 m square, which gives an area of 81 sq m. For a fixed perimeter, a square always encloses the maximum possible rectangular area.

2. A square paper of area 100 sq cm is cut into two identical, non-overlapping rectangles. What is the perimeter of one of these rectangles?

   💡 Answer: The original square was 10 cm × 10 cm. Cutting it in half creates two rectangles of size 10 cm × 5 cm. The perimeter of one rectangle is 2 × (10 + 5) = 2 × 15 = 30 cm.

3. You take a rectangular sheet of paper and cut out one of its four corners as a small square. Has the perimeter of the sheet increased, decreased, or stayed the same?

   💡 Answer: The perimeter stays the same. When you cut out a corner square, you remove two sides (let's say length x and width x). But, you also create two new sides of the same length (x and x) in the cut-out area. So, the total length of the boundary remains unchanged!

Mini Cheatsheet

Area of a Triangle

Area of a Triangle

Concept Introduction

Have you ever cut a sandwich diagonally? You start with a square or rectangular slice of bread and, with one simple cut, you get two identical triangles. Did you notice that each triangular half is exactly the same size? This simple, everyday action holds the key to understanding a very important concept in geometry: the area of a triangle.

Just like we can find the space covered by a rectangular floor, we can also measure the space inside a triangular shape, like a pizza slice, a sail on a boat, or a triangular garden patch. The formula we use is directly connected to the area of a rectangle. By understanding this connection, you'll see that finding the area of a triangle is as easy as slicing a sandwich in half!

{{FORMULA: expr=Area = ½ × b × h | symbols=A:Area of the triangle, b:length of the base, h:height of the triangle}}

Definitions & Formulas

To calculate the area of a triangle, we need two key measurements: the base and the height. It's crucial to understand what these terms mean.

The fundamental formula that connects these is:

Area = ½ × base × height

Derivation: From Rectangle to Triangle

Why is there a ½ in the formula? The answer lies in the triangle's relationship with a rectangle. Let's see how.

1. Start with a Rectangle: Imagine a rectangle with length l and breadth b. We know its area is l × b.

   {{VISUAL: diagram: A rectangle ABCD with length AB labeled 'l' and breadth BC labeled 'b'.}}

2. Draw the Diagonal: Now, draw a diagonal line connecting two opposite corners, say from A to C. This cut divides the rectangle into two triangles: ΔABC and ΔADC.

3. Two Equal Halves: If you were to cut the rectangle along this diagonal, you would find that the two triangles are perfectly identical. They are congruent, which means they have the same size and shape. Therefore, they must have the same area.

4. Area of One Triangle: Since the two triangles together make up the whole rectangle, the area of one triangle must be exactly half the area of the rectangle.

   Area of ΔABC = ½ × Area of Rectangle ABCD
   

5. Relate to Triangle's Formula: For triangle ΔABC, the side AB becomes its base and the side BC becomes its height (since it's a right-angled triangle). So, we can replace l with base (b) and b with height (h).

   Area of ΔABC = ½ × (length × breadth)
   

   Area of ΔABC = ½ × (base × height)
   

This simple derivation shows that any triangle's area is half the area of a rectangle that can be drawn around it with the same base and height.

{{KEY: type=concept | title=The Rectangle Connection | text=The area of a triangle is always half the area of a rectangle with the same base and height. This is the fundamental reason for the '½' in the formula.}}

Solved Examples

Let's apply this formula to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Direct Calculation (Easy)

Given: A triangle with a base of 10 cm and a height of 6 cm.

To Find: The area of the triangle.

Solution:

1. Write down the formula for the area of a triangle.

   Area = ½ × base × height
   

2. Substitute the given values for base (b = 10 cm) and height (h = 6 cm).

   Area = ½ × 10 cm × 6 cm
   

3. Perform the multiplication. You can multiply 10 × 6 first, or take half of 10 first. Let's do the latter.

   Area = 5 cm × 6 cm
   

4. Calculate the final area. Remember to use square units.

   Area = 30 cm²
   

Final Answer: The area of the triangle is 30 cm².

Example 2: Finding a Missing Dimension (Medium)

Given: The area of a triangular field is 200 m². Its base is 40 m long.

To Find: The height (altitude) of the field.

Solution:

1. Start with the area formula.

   Area = ½ × base × height
   

2. Substitute the known values: Area = 200 m² and base = 40 m. The height h is what we need to find.

   200 m² = ½ × 40 m × h
   

3. Simplify the right side of the equation.

   200 m² = 20 m × h
   

4. To find h, we need to isolate it. Divide both sides of the equation by 20 m.

   h = 200 m² ÷ 20 m
   

5. Calculate the result.

   h = 10 m
   

Final Answer: The height of the triangular field is 10 m.

Example 3: Area of a Composite Figure (Hard)

Given: A shape that looks like a house, formed by a square with a side of 8 m and an isosceles triangle sitting on top of it. The total height of the figure from the bottom of the square to the tip of the triangle is 11 m.

To Find: The total area of the figure.

{{VISUAL: diagram: A composite figure shaped like a house. The bottom part is a square labeled with side 8m. A triangle sits on top. The total height from the bottom base to the top vertex is labeled 11m.}}

Solution:

This problem requires us to break the figure into two parts: a square and a triangle.

1. Calculate the area of the square. The formula for the area of a square is side × side.

   Area_square = 8 m × 8 m = 64 m²
   

2. Determine the dimensions of the triangle.

   • The base of the triangle is the same as the top side of the square, so base = 8 m.
   • The height of the triangle is the total height minus the height of the square.
   • Height_triangle = Total height - Side of square = 11 m - 8 m.

   Height_triangle = 3 m
   

3. Calculate the area of the triangle. Use the triangle area formula with the dimensions we just found.

   Area_triangle = ½ × base × height
   

   Area_triangle = ½ × 8 m × 3 m
   

   Area_triangle = 4 m × 3 m = 12 m²
   

4. Find the total area. Add the area of the square and the area of the triangle.

   Total Area = Area_square + Area_triangle
   

   Total Area = 64 m² + 12 m² = 76 m²
   

Final Answer: The total area of the figure is 76 m².

Example 4: The Area Maze Puzzle (Tricky)

Given: A large rectangle is divided into four smaller regions. The areas of three regions are given: 15 sq cm, 13 sq cm, and 26 sq cm. A small rectangle inside has a side length of 2 cm.

To Find: The area of the fourth, unknown region (marked with '?').

{{VISUAL: diagram: An "Area Maze". A large rectangle divided into four smaller rectangular regions. Top-left area=15 sq cm, top-right area=13 sq cm, bottom-right area=26 sq cm, bottom-left area is unknown '?'. A vertical line separates left/right, a horizontal line separates top/bottom. The horizontal line has a small segment within the bottom-right rectangle marked as 2 cm in height.}}

Solution:

This puzzle seems hard, but it relies on the basic area formula (Area = length × width). Let's use logic.

1. Focus on the right side. We have two rectangles stacked vertically, with areas 13 cm² and 26 cm². Notice they share the same width. The ratio of their areas must be equal to the ratio of their heights.

   Area_top_right / Area_bottom_right = Height_top_right / Height_bottom_right
   

   13 / 26 = ½
   

   This means the height of the top-right rectangle is half the height of the bottom-right rectangle.

2. Use the given dimension. The problem states a small part of the bottom-right rectangle's height is 2 cm. Ah, looking closely at the diagram, we can deduce the heights. The area of the bottom-right is 26 cm² and the top-right is 13 cm². They share a width. This means the height of the bottom rectangle is double the height of the top one. Let the width be w. Let the height of the top one be h1 and bottom one be h2. h1 × w = 13 and h2 × w = 26. Dividing these gives h1/h2 = 13/26 = 1/2.

   Let's re-examine the puzzle logic, as these can be tricky. Often, it's about shared lengths and widths.

3. A simpler approach: Proportionality. In a rectangle divided into four smaller rectangles like this, the product of the areas of diagonally opposite rectangles is equal. Area(top-left) × Area(bottom-right) = Area(top-right) × Area(bottom-left)

4. Let's verify this logic. Let the widths be w1, w2 and heights be h1, h2.

   • Top-left Area = w1 × h1
   • Bottom-right Area = w2 × h2
   • Top-right Area = w2 × h1
   • Bottom-left Area = w1 × h2

   Product of diagonals:

   • (w1 × h1) × (w2 × h2)
   • (w2 × h1) × (w1 × h2) These are indeed equal! So the property holds.

5. Apply the property to solve. 15 × 26 = 13 × ?

6. Calculate the product on the left. 390 = 13 × ?

7. Solve for the unknown area. ? = 390 ÷ 13 ? = 30 cm²

Final Answer: The area of the unknown region is 30 cm².

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. If you double the base of a triangle but keep its height the same, what happens to its area? What if you double the base and halve the height?

   💡 Answer: Doubling the base doubles the area (Area = ½ × (2b) × h = 2 × (½ × b × h)). If you double the base and halve the height, the area stays the same! (Area = ½ × (2b) × (h/2) = ½ × b × h).

2. A large triangle has a base of 20 m and a height of 12 m. Inside it, a smaller triangle is removed. The smaller triangle has the same base but its height is ⅓ of the larger triangle's height. What is the area of the remaining part?

   💡 Answer: Area of large triangle = ½ × 20 × 12 = 120 m². Height of small triangle = ⅓ × 12 = 4 m. Area of small triangle = ½ × 20 × 4 = 40 m². The remaining area is 120 m² - 40 m² = 80 m².

3. The area of a square is 64 cm². A triangle has the same area as the square. If the height of the triangle is 8 cm, what is its base?

   💡 Answer: The area of the triangle is 64 cm². We know Area = ½ × base × height. So, 64 = ½ × base × 8. This simplifies to 64 = 4 × base. Therefore, the base is 64 ÷ 4 = 16 cm.

Mini Cheatsheet

Summary & Quick Revision

Chapter 6: Perimeter and Area - Summary & Quick Revision

Concept Introduction

Imagine you're helping your parents redecorate your room. You want to put a new decorative border along all the walls. To know how much border to buy, you need to measure the total length around the room. This total length is the perimeter. Now, you also want to buy a new carpet that covers the entire floor. To get the right size, you need to measure the total surface of the floor. This surface measurement is the area. Perimeter and area are two fundamental concepts in geometry that help us measure and understand the space around us, from planning gardens and building houses to creating art and designing maps.

{{FORMULA: expr=A = l × b | symbols=A:Area of a rectangle, l:length, b:breadth}}

Definitions & Formulas

Here are the key terms and formulas you'll need for this chapter. Remember that perimeter is a measure of length (in cm, m, km) while area is a measure of a two-dimensional surface (in cm², m², km²).

Derivation: Area of a Triangle

Have you ever wondered why the formula for the area of a triangle has that ½ in it? The answer comes from a simple relationship between a triangle and a rectangle. Let's see how.

1. Start with any rectangle, let's call it ABCD. Its area is simply length × breadth. Let's consider side AB as the length and AD as the breadth.

{{VISUAL: diagram: A rectangle ABCD with vertices labeled. A diagonal AC is drawn across it, splitting it into two triangles, ABC and ADC.}}

2. Now, draw a diagonal line connecting two opposite corners, for example, from A to C. This diagonal cuts the rectangle perfectly into two triangles: triangle ABC and triangle ADC.

3. If you were to cut these two triangles out with scissors, you would find that they are identical. They can be placed on top of each other and they would match up perfectly. This means they must have the exact same area.

4. Since the two triangles together make up the whole rectangle and they have equal areas, the area of one triangle must be exactly half the area of the rectangle.

Area of Triangle ABC = ½ × Area of Rectangle ABCD

5. We know the area of the rectangle is length × breadth. In the context of the triangle ABC, the length of the rectangle (AB) acts as the base of the triangle, and the breadth of the rectangle (BC) acts as its height.

6. Therefore, by substituting the terms for a triangle, we get the famous formula:

Area of a Triangle = ½ × base × height

{{KEY: type=concept | title=Rectangle-Triangle Connection | text=Every triangle can be seen as exactly half of a rectangle that shares the same base and height. This is the fundamental reason for the '½' in its area formula.}}

Solved Examples

Let's work through some problems to master these concepts. We'll start easy and move to more challenging questions.

Example 1: Basic Calculations (Easy)

Given: A rectangular park is 40 m long and 25 m wide.

To Find: The length of the fence required to surround it and the total area of the park.

Solution:

1. The length of the fence is the perimeter of the park. We use the formula for the perimeter of a rectangle.

P = 2 × (length + breadth)

2. Substitute the given values for length (l = 40 m) and breadth (b = 25 m).

P = 2 × (40 m + 25 m) = 2 × 65 m = 130 m

3. The total area of the park is found using the area formula for a rectangle.

A = length × breadth

4. Substitute the given values.

A = 40 m × 25 m = 1000 m²

Final Answer:

The length of the fence required is 130 m, and the area of the park is 1000 m².

Example 2: Finding a Missing Side (Medium)

Given: The area of a rectangular cardboard sheet is 150 cm². Its breadth is 10 cm.

To Find: The length of the cardboard sheet and its perimeter.

Solution:

1. We are given the area and the breadth. We start with the area formula.

Area = length × breadth

2. Substitute the known values into the formula. Here, A = 150 cm² and b = 10 cm.

150 cm² = length × 10 cm

3. To find the length, we rearrange the formula by dividing the area by the breadth.

length = 150 cm² ÷ 10 cm = 15 cm

4. Now that we know both length (15 cm) and breadth (10 cm), we can find the perimeter.

Perimeter = 2 × (length + breadth)

5. Substitute the values we found.

Perimeter = 2 × (15 cm + 10 cm) = 2 × 25 cm = 50 cm

Final Answer:

The length of the cardboard sheet is 15 cm and its perimeter is 50 cm.

Example 3: Area of a Composite Figure (Hard)

Given: A square carpet with a side of 4 m is laid on a rectangular floor of length 7 m and breadth 5 m.

To Find: The area of the floor that is not carpeted.

Solution:

1. This problem requires us to find the difference between two areas: the total area of the floor and the area covered by the carpet. First, let's calculate the area of the rectangular floor.

{{VISUAL: diagram: A large rectangle representing the floor (labeled 7m by 5m). Inside it, a smaller square representing the carpet (labeled 4m by 4m) is shown. The area between the rectangle and the square is shaded and labeled 'Area not carpeted'.}}

Area_floor = length × breadth = 7 m × 5 m = 35 m²

2. Next, we calculate the area of the square carpet.

Area_carpet = side × side = 4 m × 4 m = 16 m²

3. To find the area that is not carpeted, we subtract the carpet's area from the floor's total area.

Area_uncovered = Area_floor - Area_carpet

4. Substitute the values we calculated.

Area_uncovered = 35 m² - 16 m² = 19 m²

Final Answer:

The area of the floor that is not carpeted is 19 m².

Example 4: Cost Calculation (Tricky)

Given: A rectangular field is 100 m long and 80 m wide. The cost of fencing is ₹15 per meter. The cost of leveling the field is ₹10 per square meter.

To Find: The total cost of fencing and leveling the field.

Solution:

1. This is a two-part problem. Fencing relates to the perimeter, and leveling relates to the area. First, we calculate the perimeter of the field.

Perimeter = 2 × (100 m + 80 m) = 2 × 180 m = 360 m

2. Now, calculate the total cost of fencing. We multiply the perimeter by the cost per meter.

Cost_fencing = Perimeter × Rate = 360 m × ₹15/m = ₹5400

3. Next, we calculate the area of the field that needs to be leveled.

Area = 100 m × 80 m = 8000 m²

4. Now, calculate the total cost of leveling. We multiply the area by the cost per square meter.

{{VISUAL: diagram: A rectangular field labeled 100m by 80m. Arrows along the boundary indicate 'Fencing (Perimeter)'. Shading inside the rectangle indicates 'Leveling (Area)'.}}

Cost_leveling = Area × Rate = 8000 m² × ₹10/m² = ₹80000

5. Finally, we find the total cost by adding the cost of fencing and the cost of leveling.

Total Cost = Cost_fencing + Cost_leveling = ₹5400 + ₹80000

Total Cost = ₹85400

Final Answer:

The total cost of fencing and leveling the field is ₹85,400.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A piece of wire is 48 cm long. It is first bent to form a square. Then, it is unbent and re-bent to form a rectangle whose length is 14 cm. Which shape encloses more area and by how much?

   💡 Answer: Square: Side = 48/4 = 12 cm. Area = 12×12 = 144 cm². Rectangle: Perimeter = 48 cm. 2(14 + b) = 48 → 14 + b = 24 → b = 10 cm. Area = 14×10 = 140 cm². The square encloses more area by 4 cm².

2. A rectangular park is 60 m long and 40 m wide. A path 2 m wide is constructed all around inside the park. Find the area of the path.

   💡 Answer: Area of Park (Outer) = 60 × 40 = 2400 m². The inner rectangle (excluding path) has dimensions: length = 60-2-2=56m, breadth = 40-2-2=36m. Area of Inner Rectangle = 56 × 36 = 2016 m². Area of Path = Outer Area - Inner Area = 2400 - 2016 = 384 m².

3. How many square tiles of side 25 cm will be required to pave a rectangular hall which is 10 m long and 8 m wide?

   💡 Answer: First, make units consistent. Hall length = 10 m = 1000 cm. Hall breadth = 8 m = 800 cm. Area of Hall = 1000 cm × 800 cm = 800,000 cm². Area of one tile = 25 cm × 25 cm = 625 cm². Number of tiles = Area of Hall / Area of one tile = 800,000 ÷ 625 = 1280 tiles.

Mini Cheatsheet

Here is a quick summary of all the formulas from this chapter. Screenshot this for your last-minute revision!