The Baudhayana-Pythagoras Theorem

Doubling and Halving a Square

Page 1: Doubling and Halving a Square

Concept Introduction

Imagine you're doing origami, and you have a perfect square piece of paper. Your instructions say you need a new square with exactly double the area. Your first instinct might be to simply double the length of the sides. But if you do that, you'll find the new square is actually four times bigger! This puzzle isn't new; it dates back to ancient India, where the mathematician Baudhāyana tackled this exact problem in his work, the Śulba-Sūtras, around 800 BCE.

He asked: How can one construct a square having double the area of a given square? The answer he found was elegant and surprising. It didn't involve complex calculations but a simple geometric trick. The secret, he discovered, lies not in the sides of the square, but in its diagonal. This single line, cutting across the square, holds the key to perfectly doubling its area, a foundational concept that leads us to one of the most famous theorems in all of mathematics.

{{FORMULA: expr=Area_new = 2 × Area_old | symbols=Area_new:Area of the larger square, Area_old:Area of the original square}}

Definitions & Formulas

Before we dive into the logic, let's clarify the key terms and formulas we will be using.

The Logic: Why the Diagonal Works

Baudhāyana's claim that "The diagonal of a square produces a square of double the area" is a powerful statement. But why is it true? We can understand this with a simple visual proof.

1. Start with any square. Let's call it the "original square." Its area is made up of the space it occupies.

2. Draw a single diagonal line connecting two opposite corners. This action immediately divides the square into two identical (or congruent) isosceles right triangles. So, the area of the original square is equal to the area of these two triangles combined.

{{VISUAL: diagram: An original square ABCD is shown. The diagonal AC is drawn, dividing the square into two shaded triangles, ΔABC and ΔADC.}}

3. Now, let's use this diagonal as the side for a new square. This new square will be tilted compared to the first one.

4. To see the relationship between their areas, we can draw some helpful "east-west" (horizontal) and "north-south" (vertical) lines that extend from the sides of our original square.

5. These lines perfectly complete a larger bounding box around our new, tilted square. Look closely at the picture that forms. The original square is made of 2 small triangles.

6. The new square, built on the diagonal, is clearly made up of 4 of these exact same small triangles.

{{VISUAL: diagram: A square ABCD with diagonal AC. A new, larger square ACGF is constructed on the diagonal. Dashed lines from the corners of the original square extend to show that the original square consists of 2 small right triangles, while the new square on the diagonal consists of 4 identical small right triangles.}}

Since the new square contains four triangles and the original square contains two, the new square's area must be exactly double the area of the original one. It's a simple, elegant proof by sight!

{{KEY: type=concept | title=The Triangle Counting Method | text=The easiest way to remember this proof is to visualize the triangles. The original square is 2 triangles. The square built on its diagonal is 4 identical triangles. Therefore, the area is doubled (4 triangles is 2 × 2 triangles).}}

Solved Examples

Example 1: Doubling a Basic Square (Easy)

Given: A square with a side length of 6 cm.

To Find: The area of a new square constructed on its diagonal.

Solution:

1. First, calculate the area of the original square. The side s is 6 cm.

   Area_original = s² = 6²
   

2. Perform the calculation.

   Area_original = 36 cm²
   

3. According to Baudhāyana's principle, the area of the square on the diagonal is double the area of the original square.

   Area_new = 2 × Area_original
   

4. Substitute the value of the original area.

   Area_new = 2 × 36 = 72 cm²
   

Final Answer: The area of the new square is 72 cm².

Example 2: Finding the Side of a Doubled Square (Medium)

Given: An original square with an area of 50 m².

To Find: The side length of the new square that has double this area.

Solution:

1. First, determine the area of the new, doubled square.

   Area_new = 2 × Area_original = 2 × 50
   

   Area_new = 100 m²
   

2. The area of any square is given by A = s². We know the area of the new square is 100 m². Let its side be s_new.

   s_new² = 100
   

3. To find the side length s_new, we need to take the square root of the area.

   s_new = √100
   

4. Calculate the square root.

   s_new = 10 m
   

Final Answer: The side length of the new square is 10 m.

Example 3: Halving a Square (Hard)

Given: A large square placemat has a side length of 14 cm. A smaller square is created by joining the midpoints of its sides.

To Find: The area of the smaller inner square.

Solution:

1. This problem is the reverse of doubling. The smaller, inner square is formed by folding the corners of the larger square. This construction creates a square whose area is half that of the larger one. First, find the area of the large square.

   Area_large = s² = 14²
   

   Area_large = 196 cm²
   

{{VISUAL: diagram: A large square of side 14cm. A smaller, tilted square PQRS is inscribed by joining the midpoints of the sides of the large square. The area of the large square is marked as 196cm², and the area of the small square is marked as 'A_small'.}}

2. The principle of halving a square states that this inner square (formed by joining midpoints) has half the area of the outer square.

   Area_small = ½ × Area_large
   

3. Substitute the area of the large square to find the area of the small one.

   Area_small = ½ × 196
   

4. Calculate the final area.

   Area_small = 98 cm²
   

Final Answer: The area of the smaller inner square is 98 cm².

Example 4: The Diagonal as a Side (Tricky)

Given: The diagonal of Square A is 8 cm long. This diagonal is used as the side of a new Square B.

To Find: The area of the original Square A.

Solution:

1. We are given the side of Square B, which is 8 cm. This side is the same as the diagonal of Square A. First, let's find the area of Square B.

   Area_B = (side_B)² = 8²
   

   Area_B = 64 cm²
   

2. We know from Baudhāyana's principle that the area of the square on the diagonal (Square B) is double the area of the original square (Square A).

   Area_B = 2 × Area_A
   

3. We can rearrange this formula to solve for the area of Square A.

   Area_A = Area_B ÷ 2
   

4. Substitute the value for the area of Square B.

   Area_A = 64 ÷ 2 = 32 cm²
   

Final Answer: The area of the original Square A is 32 cm².

Tips & Tricks

Mastering this concept is easier with a few shortcuts in mind.

Common Mistakes

It's easy to fall into a few common traps when doubling or halving areas. Here's how to stay on track.

Brain-Teaser Questions

1. A square garden has an area of 40 m². If you build a new square garden whose side length is equal to the diagonal of the old garden, what is the area of the new garden?

   💡 Answer: The new garden's side is the old garden's diagonal. Therefore, the new garden's area is double the old one's. Area = 2 × 40 = 80 m².

2. You have a square piece of cloth with an area of 98 cm². You fold its four corners in so their tips meet exactly at the center. This creates a new, smaller square. What is the side length of this new square?

   💡 Answer: Folding the corners to the center is the same as creating the "halved" square. The new area is 98 / 2 = 49 cm². The side length is the square root of the area, so s = √49 = 7 cm.

3. A sequence of squares is made. Square 2 has double the area of Square 1. Square 3 has double the area of Square 2. If the side of Square 1 is 5 cm, what is the side of Square 3?

   💡 Answer: Area of Square 1 = 5² = 25 cm². Area of Square 2 = 2 × 25 = 50 cm². Area of Square 3 = 2 × 50 = 100 cm². The side of Square 3 is √100 = 10 cm. Notice the side length doubled (10 cm) while the area quadrupled (100 cm²).

Mini Cheatsheet

Hypotenuse of an Isosceles Right Triangle

Page 2: Hypotenuse of an Isosceles Right Triangle

Welcome back! In our last session, we explored how a simple paper-folding activity revealed a deep geometric truth. Now, we'll formalize that discovery and dive into the fascinating relationship between the sides of a special kind of triangle.

{{FORMULA: expr=c² = 2a² | symbols=c:length of the hypotenuse, a:length of the two equal sides}}

Concept Introduction

Have you ever looked at a perfectly square tile on the floor and imagined cutting it exactly in half, corner to corner? You've just created two identical triangles! These are not just any triangles; they are isosceles right triangles.

Imagine a carpenter cutting a square piece of plywood with sides of 1 meter each. If they make a single straight cut from one corner to the opposite corner, how long is that cut? It's clearly longer than 1 meter, but how much longer? This cut length is the hypotenuse, and finding its exact value leads us to a very special and famous number in mathematics. This section will give you the tools to calculate that length precisely.

Definitions & Formulas

Before we begin, let's be clear on the terms we'll be using. An isosceles right triangle is a triangle with one right angle (90°) and two equal sides.

{{VISUAL: diagram: an isosceles right triangle labeled with vertices P, Q, R. Angle at Q is 90°. Sides PQ and QR are labeled 'a'. The hypotenuse PR is labeled 'c'.}}

The Logic: Deriving the Formula

How can we find a relationship between the equal sides (a) and the hypotenuse (c)? We can use a powerful method based on areas, just like the ancient Indian mathematicians did.

1. Start with the sides. Consider an isosceles right triangle with equal sides of length a. Let's build a square on one of these sides. The area of this square is a × a.

   Area of square on side 'a' = a²
   

2. Construct a square on the hypotenuse. Now, let's build a larger square using the hypotenuse, c, as its side. The area of this new, larger square is c × c.

   Area of square on hypotenuse 'c' = c²
   

3. Relate the two areas. As we saw in the paper-folding activity, the square built on the hypotenuse is made up of exactly four of our original isosceles right triangles. The smaller square (with side a) is made up of only two such triangles.

   This means the square on the hypotenuse has exactly double the area of the square on one of the equal sides.

   {{VISUAL: diagram: a square of side 'a' (area a²) next to a larger square of side 'c' (area c²). The larger square is shown to be composed of two squares of side 'a', demonstrating that Area(c²) = 2 × Area(a²).}}

4. Form the equation. We can now write this relationship as an equation: Area of square on hypotenuse = 2 × Area of square on a side

   c² = 2 × a²
   

This simple but profound formula, c² = 2a², is the key to solving all problems involving isosceles right triangles.

Solved Examples

Let's apply this formula to solve some problems, starting with the simplest case.

Example 1: The Simplest Case (The Origin of √2)

Given: An isosceles right triangle with equal sides of length 1 unit (a = 1).

To Find: The length of the hypotenuse (c).

Solution:

1. Start with our derived formula that connects the hypotenuse and the equal sides.

   c² = 2a²
   

2. Substitute the given value a = 1 into the formula.

   c² = 2 × (1)²
   

3. Calculate the value of c².

   c² = 2 × 1 = 2
   

4. To find c, we must find the number which, when multiplied by itself, gives 2. This number is called the square root of 2.

   c = √2
   

Final Answer: The hypotenuse has a length of √2 units.

A Deeper Look at √2: What exactly is √2? As the NCERT text shows, it's not a simple fraction or a decimal that ends.

• 1.4² = 1.96 (too small)
• 1.5² = 2.25 (too big)
• 1.41² = 1.9881 (closer, but still too small)
• 1.414² = 1.999396 (even closer!)

We can get closer and closer, but we'll never land exactly on 2. √2 is a non-terminating, non-repeating decimal (an irrational number) that starts as 1.41421356… For most calculations, we use an approximation like 1.414.

Example 2: Finding the Hypotenuse with Larger Sides

Given: An isosceles right triangle with equal sides of length 12 cm (a = 12).

To Find: The length of the hypotenuse (c).

Solution:

1. Use the standard formula for an isosceles right triangle.

   c² = 2a²
   

2. Substitute the value a = 12 into the equation.

   c² = 2 × (12)²
   

3. First, calculate the square of 12.

   c² = 2 × 144
   

4. Now, perform the multiplication.

   c² = 288
   

5. The length of the hypotenuse is the square root of 288. We can find bounds for this value. We know 16² = 256 and 17² = 289. Since 288 is between 256 and 289, √288 must be between 16 and 17.

   c = √288
   

Final Answer: The hypotenuse is √288 cm long, which is a value between 16 cm and 17 cm.

Example 3: Working Backwards to Find the Sides

Given: The hypotenuse of an isosceles right triangle is 10 meters long (c = 10).

To Find: The length of the two equal sides (a).

Solution:

1. Begin with the same master formula.

   c² = 2a²
   

2. This time, we substitute the known value of c.

   (10)² = 2a²
   

3. Calculate the square of 10.

   100 = 2a²
   

4. To isolate a², we need to divide both sides of the equation by 2.

   a² = 100 ÷ 2
   

   a² = 50
   

5. To find a, we take the square root of 50.

   a = √50
   

{{VISUAL: diagram: an isosceles right triangle. The hypotenuse is labeled '10 m'. The two equal legs are labeled 'a'. The formula 10² = 2a² is shown next to the diagram.}}

Final Answer: The length of each equal side is √50 meters. (We know 7²=49 and 8²=64, so the answer is slightly more than 7 meters).

Example 4: The 'Tricky' Square Root Case

Given: The hypotenuse of an isosceles right triangle is √72 units (c = √72).

To Find: The length of the other two sides (a).

Solution:

1. Write down the governing formula.

   c² = 2a²
   

2. Substitute c = √72. Be careful with the parentheses.

   (√72)² = 2a²
   

3. Remember that squaring a square root cancels it out. (√x)² = x.

   72 = 2a²
   

4. Now it's a simple equation. Divide both sides by 2 to find a².

   a² = 72 ÷ 2
   

   a² = 36
   

5. Find the square root of 36 to get the value of a.

   a = √36 = 6
   

Final Answer: Each of the two equal sides has a length of 6 units.

{{KEY: type=concept | title=The Core Relationship | text=In any isosceles right triangle, the square of the hypotenuse is ALWAYS double the square of one of its equal sides. This single idea, c² = 2a², allows you to find any missing side.}}

Tips & Tricks

Use these shortcuts to solve problems faster once you understand the main concept.

Common Mistakes to Avoid

Many students make small errors that lead to the wrong answer. Be careful to avoid these!

Brain-Teaser Questions

Ready for a challenge? These questions require you to combine what you've learned.

1. The area of an isosceles right triangle is 32 square units. What is the length of its hypotenuse?

   💡 Answer: The formula for the area of a triangle is ½ × base × height. In our case, this is Area = ½ × a × a = a²/2. So, a²/2 = 32, which means a² = 64, and a = 8. Now find the hypotenuse: c² = 2a² = 2 × 64 = 128. Therefore, c = √128 units.

2. A square park has a straight path that runs diagonally from one corner to the opposite. The path is √200 meters long. What is the area of the park?

   💡 Answer: The diagonal path is the hypotenuse (c) of two isosceles right triangles that make up the square. The sides of the park are the equal sides (a). We are given c = √200. Using the formula: c² = 2a² → (√200)² = 2a² → 200 = 2a². Dividing by 2, we get a² = 100. The area of the square park is a × a = a². So, the area is 100 square meters.

3. You have two identical square photos, each with an area of 50 cm². You cut both photos in half diagonally. Can you arrange these four triangular pieces to form one large square? What would be the length of the side of this new large square?

   💡 Answer: Yes, you can. This is the reverse of our derivation. Arranging the four triangles with their right-angle corners meeting at the center forms a new square. The area of this new square is the sum of the areas of the two original squares: 50 cm² + 50 cm² = 100 cm². The side length of a square with an area of 100 cm² is √100 = 10 cm.

Mini Cheatsheet

Screenshot this table for quick revision before a test!

Combining Two Different Squares — Part 1

Combining Two Different Squares — Part 1

Concept Introduction

In our last lesson, we discovered a neat trick: if you take two identical squares, you can combine them to form a single larger square. The side of this new square is simply the diagonal of the original squares. But what happens if the squares aren't the same size? Imagine you are a designer with one small square tile and one medium square tile. You need to create a single, larger square tile that has the exact same total area as your two different tiles combined. How would you determine the side length of this new, larger square?

This is a much more interesting puzzle! The brilliant Indian mathematician Baudhāyana, in his ancient text called the Śulba-Sūtra, provided a truly remarkable and elegant solution. He showed that this problem is deeply connected to a special kind of triangle: the right-angled triangle. His method allows us to perfectly combine the areas of any two different squares into one.

{{FORMULA: expr=a² + b² = c² | symbols=a:length of one short side, b:length of the other short side, c:length of the hypotenuse}}

Definitions & Formulas

The relationship discovered by Baudhāyana connects the sides of a right-angled triangle. Before we dive in, let's define the key terms.

The core relationship, known as Baudhāyana's Theorem, is that the area of the square on the hypotenuse is the sum of the areas of the squares on the other two sides.

The Logic Behind Baudhāyana's Method

Why does this method work? Baudhāyana didn't just state the rule; he provided a visual, constructive proof. Let's walk through his brilliant geometric argument. The goal is to show that a square with side c (the hypotenuse) has an area equal to a² + b².

1. Start with the Squares. Imagine you have two squares. One has a side length of a, so its area is a². The second has a side length of b, so its area is b².

2. Form a Right-Angled Triangle. Baudhāyana's first instruction is to create a right-angled triangle where the two perpendicular sides are the side lengths of our squares, a and b. The third side, the hypotenuse, we will call c.

{{VISUAL: diagram: Baudhāyana's geometric proof setup. A large square of side 'b' is shown next to a smaller square of side 'a'. Inside the larger square, a rectangle is marked with sides 'a' and 'b'. The diagonal of this rectangle is drawn and labeled 'c'.}}

3. Construct a New Square. Now, construct a square whose side is the hypotenuse, c. Baudhāyana's claim is that the area of this new square, c², is exactly a² + b². To prove this, he uses a clever arrangement of triangles. He arranges four identical copies of our a-b-c right-angled triangle.

4. Visual Rearrangement (The Proof). The four triangles are placed such that their hypotenuses form the sides of a new, tilted square. The total area of this arrangement can be viewed in two ways.

5. Equating the Areas. The diagram in the NCERT text shows a beautiful dissection. The area of the square on the hypotenuse can be broken into pieces. These exact same pieces can be rearranged to perfectly form the original two squares of sides a and b. This visual proof confirms that the area is conserved.

6. The Final Statement. This geometric demonstration proves that the area of the square built on the hypotenuse (c²) is indeed the sum of the areas of the squares built on the other two sides (a² and b²). This gives us the famous equation:

a² + b² = c²

{{KEY: type=concept | title=Baudhāyana's Theorem | text=In any right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.}}

Solved Examples

Let's apply this powerful theorem to solve some problems.

Example 1: Finding the Hypotenuse (Easy)

Given: A right-angled triangle has shorter sides of lengths 3 cm and 4 cm.

To Find: The length of the hypotenuse.

Solution:

1. Let the lengths of the shorter sides be a = 3 cm and b = 4 cm. Let the hypotenuse be c.

2. According to Baudhāyana's Theorem, a² + b² = c². Substitute the given values.

3² + 4² = c²

3. Calculate the squares of the numbers.

9 + 16 = c²

4. Add the values to find c².

25 = c²

5. To find c, take the square root of 25.

c = √25

c = 5

Final Answer: The length of the hypotenuse is 5 cm.

Example 2: Finding a Shorter Side (Medium)

Given: A right-angled triangle has a hypotenuse of length 17 cm and one short side of length 8 cm.

To Find: The length of the third side.

Solution:

1. Here, the hypotenuse c = 17 cm and one short side a = 8 cm. We need to find the other short side, b.

{{VISUAL: diagram: A right-angled triangle with the side opposite the 90-degree angle labeled 'c = 17 cm'. One of the other sides is labeled 'a = 8 cm' and the remaining side is labeled 'b = ?'.}}

2. Start with Baudhāyana's Theorem.

a² + b² = c²

3. Substitute the known values into the equation.

8² + b² = 17²

4. Calculate the squares.

64 + b² = 289

5. To find b², we need to isolate it. Subtract 64 from both sides of the equation.

b² = 289 - 64

b² = 225

6. Now, find b by taking the square root of 225.

b = √225

b = 15

Final Answer: The length of the third side is 15 cm.

Example 3: Real-World Application (Hard)

Given: A ladder 10 m long is placed against a vertical wall. The foot of the ladder is 6 m away from the base of the wall.

To Find: How high up the wall the ladder reaches.

Solution:

1. This scenario forms a right-angled triangle. The ladder is the hypotenuse (c = 10 m). The distance from the wall is one of the shorter sides (b = 6 m). The height on the wall is the other shorter side (a).

{{VISUAL: diagram: A ladder leaning against a vertical wall, forming a right-angled triangle. The ladder is labeled as the hypotenuse 'c = 10 m'. The ground distance from the wall to the ladder's base is labeled 'b = 6 m'. The height the ladder reaches on the wall is labeled 'a = ?'.}}

2. We use the theorem a² + b² = c².

3. Substitute the given values.

a² + 6² = 10²

4. Calculate the squares.

a² + 36 = 100

5. Isolate a² by subtracting 36 from both sides.

a² = 100 - 36

a² = 64

6. Find a by taking the square root of 64.

a = √64

a = 8

Final Answer: The ladder reaches 8 m high up the wall.

Example 4: Diagonal of a Rectangle (Tricky)

Given: A rectangular park has a length of 12 m and a width of 5 m.

To Find: The length of the diagonal path that cuts across the park.

Solution:

1. A rectangle's corners are all right angles (90°). The diagonal of a rectangle divides it into two identical right-angled triangles.

2. The length and width of the rectangle become the two shorter sides of the right-angled triangle (a = 5 m, b = 12 m). The diagonal is the hypotenuse (c).

3. Apply Baudhāyana's Theorem.

a² + b² = c²

4. Substitute the values of the sides.

5² + 12² = c²

5. Calculate the squares and add them.

25 + 144 = c²

169 = c²

6. Find c by taking the square root of 169.

c = √169

c = 13

Final Answer: The length of the diagonal path is 13 m.

Tips & Tricks

Use these shortcuts to solve problems faster and with more confidence.

Common Mistakes to Avoid

Many students make small errors when first using this theorem. Here’s what to watch out for.

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. A square has a side length of 7 cm. What is the exact length of its diagonal?

   💡 Answer: The diagonal splits the square into two right-angled triangles. The sides are a=7 and b=7. So, c² = 7² + 7² = 49 + 49 = 98. The diagonal's length is c = √98 cm. This can also be written as 7√2 cm.

2. Two vertical poles are 12 m apart. One pole is 6 m high and the other is 11 m high. What is the distance between the tops of the two poles?

   💡 Answer: Imagine a horizontal line from the top of the shorter pole to the taller pole. This creates a right-angled triangle. The base b is the distance between the poles (12 m). The height a is the difference in the pole heights (11 m - 6 m = 5 m). The distance between the tops is the hypotenuse c. So, c² = 5² + 12² = 25 + 144 = 169. The distance is c = √169 = 13 m.

3. A carpenter has a triangular piece of wood with side lengths 8 cm, 15 cm, and 17 cm. Can this piece be used as a perfect corner support (a 90° angle)?

   💡 Answer: Yes. To check for a right angle, see if the sides satisfy Baudhāyana's theorem. The longest side, 17 cm, must be the hypotenuse. Let's check: Is 8² + 15² = 17²? 64 + 225 = 289. Since 17² is also 289, the equation holds true. This means the triangle is a right-angled triangle, and the angle between the 8 cm and 15 cm sides is exactly 90°.

Mini Cheatsheet

Screenshot this summary for quick revision!

Combining Two Different Squares — Part 2 and Right-Triangles Having Integer Sidelengths

Combining Two Different Squares & The Baudhāyana-Pythagoras Theorem

Welcome back! In our last session, we explored how the ancient Indian mathematician Baudhāyana provided a method to combine two identical squares. Now, we'll delve deeper into his genius by understanding how to combine two different squares. This exploration will lead us directly to one of the most fundamental and famous theorems in all of mathematics.

Imagine an architect designing a park. She wants to create a perfectly square plaza whose area is exactly equal to the sum of a 30m × 30m square garden and a 40m × 40m square fountain area. How long should each side of the new plaza be? It's not as simple as adding the side lengths. This is the exact problem Baudhāyana solved over 2800 years ago, and his solution is a cornerstone of geometry, construction, and navigation.

{{FORMULA: expr=a² + b² = c² | symbols=a:length of one shorter side, b:length of the other shorter side, c:length of the hypotenuse}}

The Baudhāyana-Pythagoras Theorem

This theorem provides the mathematical relationship between the sides of any right-angled triangle. It's often called the Pythagorean Theorem in many parts of the world, but Baudhāyana stated it centuries earlier in his Śulba-sūtras. To honour this heritage, we refer to it as the Baudhāyana-Pythagoras Theorem.

Visual Proof: Paper Cutting & Rearrangement

Baudhāyana's genius was not just in stating the theorem, but also in providing intuitive ways to see why it works. Let's follow a hands-on proof he inspired.

1. Start with two different squares. Let their side lengths be a and b. Their areas are a² and b². Our goal is to create a single new square with an area of a² + b².

2. Make two cuts. Place the squares side-by-side. On the larger square (side b), mark a point at length a from one corner. Make a cut from this point to the opposite corner. Make a second cut from the same point to the adjacent corner. This divides the shape into three distinct pieces.

{{VISUAL: diagram: Two squares of side 'a' and 'b' placed adjacent. The larger square 'b²' is shown with two cuts, creating three pieces that will be rearranged.}}

3. Rearrange the pieces. Now, take these three pieces and rearrange them. You will find that they fit together perfectly to form a single, larger square.

4. Analyze the new square. The side length of this new square is equal to the length of the first cut you made. That cut was the hypotenuse of a right-angled triangle with sides a and b. Let's call this length c.

5. The Result. The area of this new square is c × c, or c². Since it was formed by combining the areas of the first two squares, its area must also be a² + b².

Therefore, we have visually demonstrated that for a right-angled triangle with sides a, b, and hypotenuse c:

a² + b² = c²

{{KEY: type=concept | title=The Area Connection | text=The Baudhāyana-Pythagoras theorem is fundamentally about areas. It states that the area of the square built on the hypotenuse of a right-angled triangle is exactly equal to the sum of the areas of the squares built on the other two sides.}}

Solved Examples

Let's put the theorem into practice with some examples, from simple to more challenging.

Example 1: Finding the Hypotenuse (Easy)

Given: A right-angled triangle has shorter sides of lengths 5 cm and 12 cm.

To Find: The length of its hypotenuse.

Solution:

1. Let the shorter sides be a = 5 cm and b = 12 cm. We need to find the hypotenuse c.

2. Apply the Baudhāyana-Pythagoras theorem: a² + b² = c².

3. Substitute the given values into the formula.

   5² + 12² = c²
   

4. Calculate the squares.

   25 + 144 = c²
   

5. Add the values.

   169 = c²
   

6. To find c, take the square root of 169.

   c = √169 = 13
   

Final Answer: The length of the hypotenuse is 13 cm.

Example 2: Finding a Shorter Side (Medium)

Given: A right-angled triangle has a hypotenuse of length 17 cm and one short side of length 8 cm.

To Find: The length of the third side.

Solution:

1. Let the hypotenuse be c = 17 cm and one short side be a = 8 cm. We need to find the other short side, b.

2. Start with the Baudhāyana-Pythagoras theorem: a² + b² = c².

3. Rearrange the formula to solve for b².

   b² = c² - a²
   

4. Substitute the given values.

   b² = 17² - 8²
   

5. Calculate the squares.

   b² = 289 - 64
   

6. Perform the subtraction.

   b² = 225
   

7. Take the square root to find b.

   b = √225 = 15
   

Final Answer: The length of the third side is 15 cm.

Example 3: Real-World Application (Hard)

Given: A 10-meter ladder leans against a vertical wall. The base of the ladder is 6 meters away from the base of the wall.

To Find: How high up the wall does the ladder reach?

Solution:

1. Visualize the situation. The ladder, the wall, and the ground form a right-angled triangle.
   • The ladder is the hypotenuse, c = 10 m.
   • The distance from the wall is one short side, b = 6 m.
   • The height on the wall is the other short side, a.

{{VISUAL: diagram: A ladder leaning against a wall. The ladder is labeled 'c = 10m', the ground distance is 'b = 6m', and the height on the wall is 'a = ?'. The right angle at the base of the wall is marked.}}

2. We use the formula rearranged to find a: a² = c² - b².

3. Substitute the values.

   a² = 10² - 6²
   

4. Calculate the squares.

   a² = 100 - 36
   

5. Perform the subtraction.

   a² = 64
   

6. Take the square root to find a.

   a = √64 = 8
   

Final Answer: The ladder reaches 8 meters high up the wall.

Example 4: The Converse of the Theorem (Tricky)

Given: A triangle has side lengths of 7 cm, 10 cm, and 12 cm.

To Find: Is this a right-angled triangle?

Solution:

1. For a triangle to be right-angled, it must satisfy the Baudhāyana-Pythagoras theorem a² + b² = c².

2. First, identify the longest side. This must be the hypotenuse (c) if it is a right-angled triangle. Here, c = 12. The other two sides are a = 7 and b = 10.

3. Now, we check if the sum of the squares of the shorter sides equals the square of the longest side.

   • Calculate a² + b²:

     7² + 10² = 49 + 100 = 149
     

   • Calculate c²:

     12² = 144
     

4. Compare the two results.

   149 ≠ 144
   

5. Since a² + b² is not equal to c², the theorem is not satisfied.

Final Answer: No, this is not a right-angled triangle.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A perfect square room has a floor area of 18 square meters. What is the exact length of the diagonal of this room?

   💡 Answer: The area is s² = 18, so the side length s = √18. The diagonal d is the hypotenuse of a right triangle with sides s and s. So, d² = s² + s² = 18 + 18 = 36. Therefore, the diagonal d = √36 = 6 meters.

2. You are at corner A of a rectangular field that is 30 meters wide and 40 meters long. You want to walk to the diagonally opposite corner C. How much shorter is it to walk directly across the diagonal than to walk along the two sides (A to B, then B to C)?

   💡 Answer: Walking along the sides is 30 + 40 = 70 meters. The diagonal d is the hypotenuse, so d² = 30² + 40² = 900 + 1600 = 2500. So, d = √2500 = 50 meters. The shortcut is 70 - 50 = 20 meters shorter.

{{VISUAL: diagram: A rectangle ABCD with sides AB=40m and BC=30m. The diagonal AC is drawn and labeled 'd'. Arrows show the path along the sides versus the diagonal path.}}

3. Can a right-angled triangle have all its sides as odd integers? Why or why not?

   💡 Answer: No. Let the sides be a and b. If a and b are both odd, then a² is odd and b² is odd. The sum of two odd numbers (a² + b²) is always even. So, c² must be even, which means c must also be even. Therefore, you cannot have all three sides (a, b, c) be odd integers.

Mini Cheatsheet

A Long-Standing Open Problem, Further Applications & Summary

Page 5 of 5: A Long-Standing Problem, Further Applications & Summary

{{FORMULA: expr=a² + b² = c² | symbols=a:length of one leg, b:length of the other leg, c:length of the hypotenuse}}

A Bridge from Ancient Geometry to Modern Puzzles

The Baudhāyana-Pythagoras theorem is more than just a formula about triangles; it's a fundamental principle of how space works. We've seen how it helps find distances and check for right angles. But its influence goes much deeper. The simple equation a² + b² = c² has a famous cousin: aⁿ + bⁿ = cⁿ. For centuries, mathematicians wondered if this equation had any whole number solutions for n greater than 2.

This question, known as Fermat's Last Theorem, became one of the most famous unsolved problems in mathematics. It was only in 1994 that it was finally proven by Andrew Wiles, confirming that no such solutions exist. This shows how a simple idea from geometry can lead to centuries of deep mathematical exploration.

In this final section, we move from pure theory to practical problem-solving. We will see how to apply the theorem to real-world shapes and scenarios, just as engineers, architects, and designers do every day.

Definitions & Formulas

Here are the key terms and the central formula for this section. Remember, correctly identifying each part of the triangle is the most important first step.

Logic: From Words to a Right Triangle

Many real-world problems don't give you a triangle directly. They describe a situation. The key skill is to translate the story into a diagram and find the hidden right-angled triangle. Let's use the logic from Bhāskarāchārya’s famous lotus problem as a guide.

1. Read and Visualize: Read the problem carefully. Try to draw a simple sketch of the situation described. This is the most crucial step. Don't worry about making it perfect; a rough diagram is enough.

2. Identify the Right Angle: Look for perpendicular lines in your drawing. A wall meeting the ground, a pole standing upright, or the cardinal directions (North and East) all form right angles. This corner is the key to finding your triangle.

3. Label the Sides: Once you have the right-angled triangle, label the sides with the known lengths given in the problem. Assign a variable, like x, to the length you need to find.

4. Express All Sides: Sometimes, a side might be an expression like x + 1 (as in the lotus problem). Make sure all three sides of the triangle are represented by a number or an expression.

5. Apply the Theorem: Substitute the lengths of the three sides into the Baudhāyana-Pythagoras formula: a² + b² = c². Be extra careful to identify the hypotenuse (c) correctly—it's always the side opposite the right angle.

6. Solve the Equation: Use algebra to solve the equation for the unknown variable x. This will give you the answer to the problem.

Solved Examples

Let's walk through some problems, increasing in difficulty, to see this logic in action.

Example 1: The Diagonal of a TV Screen (Easy)

Given: A square television screen has a side length of 40 cm.

To Find: The length of its diagonal.

Solution:

1. The diagonal of a square divides it into two identical right-angled triangles. The two sides of the square become the legs (a and b), and the diagonal is the hypotenuse (c).

   {{VISUAL: diagram: A square with side 's' and a diagonal 'd' drawn, forming a right-angled triangle with sides s, s, and d.}}

2. Here, a = 40 cm and b = 40 cm. We need to find c (the diagonal).

3. Apply the Baudhāyana-Pythagoras theorem: a² + b² = c².

   40² + 40² = c²
   

4. Calculate the squares.

   1600 + 1600 = c²
   

5. Add the values.

   3200 = c²
   

6. Find the square root to get c. We can write 3200 as 1600 × 2.

   c = √3200 = √(1600 × 2) = 40√2
   

Final Answer: The length of the diagonal is 40√2 cm.

Example 2: The Kite-Shaped Window (Medium)

Given: A rhombus-shaped window has diagonals of length 24 inches and 70 inches.

To Find: The length of one side of the window frame (the perimeter). First, let's find the side length.

Solution:

1. A key property of a rhombus is that its diagonals bisect each other at a right angle (90°). This divides the rhombus into four identical right-angled triangles.

2. The legs of each triangle are half the length of the diagonals.

   • Leg a = ½ × 70 inches = 35 inches.
   • Leg b = ½ × 24 inches = 12 inches.

3. The hypotenuse of each small triangle is the side of the rhombus itself. Let's call it c.

4. Apply the theorem: a² + b² = c².

   35² + 12² = c²
   

5. Calculate the squares.

   1225 + 144 = c²
   

6. Add them up.

   1369 = c²
   

7. Find the square root of 1369. (You might recognize this or use estimation. 30²=900, 40²=1600. It ends in 9, so the root must end in 3 or 7. Let's try 37). 37 × 37 = 1369.

   c = 37
   

Final Answer: The length of one side of the window frame is 37 inches.

Example 3: The Traffic Sign (Hard)

Given: An equilateral triangle-shaped traffic sign has a side length of 60 cm.

To Find: The area of the traffic sign. (Hint: Area = ½ × base × height).

Solution:

1. First, we need to find the height of the triangle. Let's draw an altitude (the height) from the top vertex down to the base. In an equilateral triangle, this altitude bisects the base and forms a right angle.

2. This creates a right-angled triangle with:

   • Hypotenuse c = side of the equilateral triangle = 60 cm.
   • One leg a = half the base = ½ × 60 cm = 30 cm.
   • The other leg b = the height (h) of the triangle, which is what we need to find.

3. Apply the theorem to this new triangle: a² + b² = c².

   30² + h² = 60²
   

4. Calculate the squares.

   900 + h² = 3600
   

5. Solve for h² by subtracting 900 from both sides.

   h² = 3600 - 900 = 2700
   

6. Find the height h. We can simplify √2700 as √(900 × 3).

   h = √2700 = 30√3 cm
   

7. Now that we have the height, we can find the area of the original equilateral triangle. Area = ½ × base × height.

   Area = ½ × 60 × 30√3
   

   Area = 30 × 30√3 = 900√3
   

Final Answer: The area of the traffic sign is 900√3 cm².

Example 4: The Sliding Ladder (Tricky)

Given: A 13-meter ladder is placed against a vertical wall. Its base is 5 meters away from the wall. The base of the ladder is then pulled 7 meters further away from the wall.

To Find: How far down the wall does the top of the ladder slide?

Solution:

This is a two-part problem. We need to find the initial height and the final height.

Part 1: Initial Position

1. The ladder, wall, and ground form a right-angled triangle.

   • Hypotenuse c = length of the ladder = 13 m.
   • Base a = distance from the wall = 5 m.
   • Height b = initial height on the wall (h₁).

2. Apply the theorem: a² + b² = c².

   5² + h₁² = 13²
   

   25 + h₁² = 169
   

3. Solve for h₁².

   h₁² = 169 - 25 = 144
   

   h₁ = √144 = 12 m
   

   So, the ladder initially reaches 12 m up the wall.

Part 2: Final Position

1. The base is pulled 7 meters further away. The new base distance is 5 m + 7 m = 12 m.

   • Hypotenuse c is still the ladder's length = 13 m.
   • New base a = 12 m.
   • New height b = final height on the wall (h₂).

2. Apply the theorem again.

   12² + h₂² = 13²
   

   144 + h₂² = 169
   

3. Solve for h₂².

   h₂² = 169 - 144 = 25
   

   h₂ = √25 = 5 m
   

   The ladder's final height is 5 m.

Part 3: Find the Difference

1. The distance the ladder slid down is the difference between the initial and final heights.

   Distance slid = h₁ - h₂ = 12 m - 5 m = 7 m
   

Final Answer: The top of the ladder slides 7 meters down the wall.

{{KEY: type=concept | title=Always Identify the Hypotenuse | text=In any word problem, the first and most critical step is to correctly identify the hypotenuse ('c'). It is always the longest side and is directly opposite the 90° angle. A mistake here will make the entire calculation incorrect.}}

Tips & Tricks

Use these shortcuts to solve problems faster and with more confidence.

Common Mistakes

Here are some frequent errors students make. Study them to avoid falling into the same traps!

Brain-Teaser Questions

Ready for a challenge? These questions require you to apply the theorem in more complex ways.

1. The 3D Challenge: What is the length of the longest straight pencil that can fit inside a rectangular box with dimensions 12 cm in length, 9 cm in width, and 8 cm in height?

   💡 Answer: This requires two applications of the theorem. First, find the diagonal of the base (d): d² = 12² + 9² = 144 + 81 = 225, so d = 15 cm. Now, this base diagonal, the height of the box, and the space diagonal (D) form a new right triangle. D² = d² + height² = 15² + 8² = 225 + 64 = 289. So, the longest pencil length is D = √289 = 17 cm.

2. The Lookout Towers: Two towers stand on level ground. Tower A is 20 meters tall and Tower B is 28 meters tall. The distance between the bases of the towers is 15 meters. What is the distance between the tops of the two towers?

   💡 Answer: Imagine a horizontal line from the top of Tower A to a point on Tower B. This forms a rectangle and a right-angled triangle on top. The base of this triangle is the distance between the towers (15 m). The height of the triangle is the difference in tower heights (28 m - 20 m = 8 m). The distance between the tops is the hypotenuse (c). So, c² = 15² + 8² = 225 + 64 = 289. The distance is c = √289 = 17 meters.

3. The Tilted Square: On a coordinate grid, a square has an area of 10 square units. If one of its vertices is at the origin (0,0), what are the possible integer coordinates of an adjacent vertex?

   💡 Answer: If the area is 10, the side length (s) is √10. Let the adjacent vertex be at coordinates (x, y). The distance from (0,0) to (x, y) is the side length. Using the theorem (as the distance formula), we have x² + y² = s². So we need x² + y² = 10. The integer pairs (x, y) that work are (1, 3), (3, 1), (-1, 3), (3, -1), etc. One possible coordinate is (1, 3), since 1² + 3² = 1 + 9 = 10.

Mini Cheatsheet

Screenshot this table for your last-minute revision. It covers all the key formulas and concepts from this chapter.