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The Baudhayana-Pythagoras Theorem - A Long-Standing Open Problem

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Baudhayana-Pythagoras theorem states that in a right-angled triangle, the area of the square on the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides. Ancient Indian mathematician Baudhayana expressed this as: 'The diagonal of an oblong produces by itself both the areas which the two sides produce separately.' This is mathematically represented as a2+b2=c2a^2 + b^2 = c^2.

A right-angled triangle showing sides a, b and hypotenuse c.
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A 'Pythagorean Triplet' consists of three positive integers a,b,a, b, and cc, such that a2+b2=c2a^2 + b^2 = c^2. For any natural number m>1m > 1, a general form of a triplet can be generated using (2m,m2−1,m2+1)(2m, m^2 - 1, m^2 + 1). This provides a systematic way to find sets of numbers that satisfy the theorem without trial and error.

Boxes representing the three components of a Pythagorean triplet.
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The theorem is widely used to find the length of the diagonal of a rectangle. If a rectangle has length ll and breadth ww, the diagonal dd forms a right-angled triangle with the sides. Thus, d=l2+w2d = \sqrt{l^2 + w^2}.

A rectangle with a diagonal line dividing it into two right-angled triangles.
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The theorem also applies to real-world distance problems, such as finding the shortest distance between two points on a coordinate plane or determining the height reached by a ladder leaning against a wall.

📐Formulae

a2+b2=c2a^2 + b^2 = c^2

c=a2+b2c = \sqrt{a^2 + b^2}

Pythagorean Triplet: (2m,m2−1,m2+1) for m>1\text{Pythagorean Triplet: } (2m, m^2 - 1, m^2 + 1) \text{ for } m > 1

💡Examples

Problem 1:

Verify if the numbers 6,8, and 106, 8, \text{ and } 10 form a Pythagorean triplet.

Solution:

Let a=6a = 6, b=8b = 8, and c=10c = 10. Calculating squares: a2=62=36a^2 = 6^2 = 36 b2=82=64b^2 = 8^2 = 64 c2=102=100c^2 = 10^2 = 100 Check the sum: 36+64100\begin{array}{r} 36 \\ + 64 \\ \hline 100 \end{array} Since 36+64=10036 + 64 = 100, we have a2+b2=c2a^2 + b^2 = c^2.

Explanation:

To check for a triplet, we square the two smaller numbers and see if their sum equals the square of the largest number.

Problem 2:

Find the length of the diagonal of a rectangle whose length is 12 cm12\text{ cm} and breadth is 5 cm5\text{ cm}.

Solution:

In a rectangle, the diagonal forms a right-angled triangle with the length and breadth. Let a=12a = 12 and b=5b = 5. Using the Baudhayana-Pythagoras theorem: c2=122+52c^2 = 12^2 + 5^2 c2=144+25c^2 = 144 + 25 c2=169c^2 = 169 c=169=13c = \sqrt{169} = 13 The diagonal is 13 cm13\text{ cm}.

Explanation:

The diagonal of a rectangle acts as the hypotenuse. We calculate the square of the sides, add them, and find the square root of the result.

Problem 3:

Write a Pythagorean triplet whose smallest member is 88.

Solution:

Let 2m=82m = 8, which gives m=4m = 4. Now find the other two members: m2−1=42−1=16−1=15m^2 - 1 = 4^2 - 1 = 16 - 1 = 15 m2+1=42+1=16+1=17m^2 + 1 = 4^2 + 1 = 16 + 1 = 17 The triplet is (8,15,17)(8, 15, 17). Check: 82+152=64+225=289=1728^2 + 15^2 = 64 + 225 = 289 = 17^2.

Explanation:

We use the general form (2m,m2−1,m2+1)(2m, m^2 - 1, m^2 + 1) to derive the triplet starting from the even number given.

Problem 4:

A 13 m13\text{ m} long ladder is placed against a wall such that it reaches a window 12 m12\text{ m} high. Find the distance of the foot of the ladder from the base of the wall.

Diagram showing a ladder leaning against a wall forming a right triangle.

Solution:

Let the distance of the foot of the ladder from the wall be xx. In the right-angled triangle formed: Hypotenuse (ladder length) c=13 mc = 13\text{ m} Height (wall) a=12 ma = 12\text{ m} Base (distance) b=xb = x By Baudhayana-Pythagoras theorem: a2+b2=c2a^2 + b^2 = c^2 122+x2=13212^2 + x^2 = 13^2 144+x2=169144 + x^2 = 169 x2=169−144x^2 = 169 - 144 x2=25x^2 = 25 x=25=5 mx = \sqrt{25} = 5\text{ m} The foot of the ladder is 5 m5\text{ m} away from the wall.

Explanation:

We model the wall, ground, and ladder as a right-angled triangle. The ladder acts as the hypotenuse. We substitute the known values into the formula and solve for the unknown base.

Problem 5:

Calculate the perimeter of a rhombus whose diagonals measure 16 cm16\text{ cm} and 30 cm30\text{ cm}.

Rhombus with diagonals showing a right-angled triangle inside.

Solution:

In a rhombus, diagonals bisect each other at right angles (90∘90^\circ). Let the diagonals be d1=16 cmd_1 = 16\text{ cm} and d2=30 cmd_2 = 30\text{ cm}. Half of the diagonals are 8 cm8\text{ cm} and 15 cm15\text{ cm}. These half-diagonals form the legs of a right-angled triangle where the side of the rhombus is the hypotenuse (ss). By Baudhayana-Pythagoras theorem: s2=82+152s^2 = 8^2 + 15^2 s2=64+225s^2 = 64 + 225 s2=289s^2 = 289 s=289=17 cms = \sqrt{289} = 17\text{ cm} Perimeter of rhombus =4×s= 4 \times s Perimeter =4×17=68 cm= 4 \times 17 = 68\text{ cm}.

Explanation:

Using the property that rhombus diagonals bisect perpendicularly, we find the side length using the theorem on one of the four internal triangles, then multiply by 4 for the total perimeter.