krit.club logo

The Baudhayana-Pythagoras Theorem - Right-Triangles Having Integer Sidelengths

Grade 8CBSE

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

🔑Concepts

•

The Baudhayana-Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Ancient Indian mathematician Baudhayana described this property long before Pythagoras.

A right-angled triangle labeled with base a, perpendicular b, and hypotenuse c.
•

A Pythagorean Triplet consists of three positive integers (a,b,c)(a, b, c) such that a2+b2=c2a^2 + b^2 = c^2. For any natural number m>1m > 1, a general form for these triplets is (2m,m2−1,m2+1)(2m, m^2 - 1, m^2 + 1). Note that m2+1m^2 + 1 is always the largest side (the hypotenuse).

•

Integer Sidelengths: While many right triangles have sides with square roots (irrational numbers), the Baudhayana-Pythagoras theorem is most commonly applied in Grade 8 to triangles where all three side lengths are whole numbers (integers), such as (3,4,5)(3, 4, 5) or (5,12,13)(5, 12, 13).

•

Visualizing Squares: The theorem can be seen as the relationship between the areas of squares built on each side of the triangle. The area of the square on the hypotenuse equals the sum of the areas of the squares on the other two sides.

Diagram showing squares attached to the sides of a right triangle to represent a squared, b squared, and c squared.

📐Formulae

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

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

Hypotenuse=Base2+Perpendicular2\text{Hypotenuse} = \sqrt{\text{Base}^2 + \text{Perpendicular}^2}

💡Examples

Problem 1:

Check if (8,15,17)(8, 15, 17) is a Pythagorean triplet.

Solution:

Identify the sides: a=8,b=15,c=17a = 8, b = 15, c = 17. Calculate a2+b2a^2 + b^2: 82+152=64+2258^2 + 15^2 = 64 + 225. 64+225289\begin{array}{r} 64 \\ + 225 \\ \hline 289 \end{array} Calculate c2c^2: 172=28917^2 = 289. Since a2+b2=c2a^2 + b^2 = c^2, the numbers (8,15,17)(8, 15, 17) form a Pythagorean triplet.

Explanation:

We square the two smaller numbers, find their sum using vertical addition, and verify if it equals the square of the largest number.

Problem 2:

Write a Pythagorean triplet whose smallest member is 1212.

Solution:

We use the general form 2m,m2−1,m2+12m, m^2 - 1, m^2 + 1. Let 2m=12  ⟹  m=62m = 12 \implies m = 6. Now find the other two members: m2−1=62−1=36−1=35m^2 - 1 = 6^2 - 1 = 36 - 1 = 35. m2+1=62+1=36+1=37m^2 + 1 = 6^2 + 1 = 36 + 1 = 37. The triplet is (12,35,37)(12, 35, 37). Check: 122+352=144+1225=1369=37212^2 + 35^2 = 144 + 1225 = 1369 = 37^2.

Explanation:

By setting the given even number to 2m2m, we find the value of mm and then use it to calculate the remaining two integers of the triplet.

Problem 3:

In a right triangle, the two sides containing the right angle are 9 cm9\text{ cm} and 12 cm12\text{ cm}. Find the hypotenuse.

Solution:

Let a=9a = 9 and b=12b = 12. We need to find cc. c2=a2+b2c^2 = a^2 + b^2 c2=92+122=81+144c^2 = 9^2 + 12^2 = 81 + 144. 81+144225\begin{array}{r} 81 \\ + 144 \\ \hline 225 \end{array} c=225=15c = \sqrt{225} = 15. The hypotenuse is 15 cm15\text{ cm}.

Explanation:

According to the Baudhayana-Pythagoras theorem, we sum the squares of the base and perpendicular to find the square of the hypotenuse.

Problem 4:

Find a Pythagorean triplet whose one member is 1414.

A right triangle with sides labeled 14, 48 and hypotenuse 50.

Solution:

Let 2m=142m = 14. Then m=7m = 7. The other two members are: m2−1=72−1=49−1=48m^2 - 1 = 7^2 - 1 = 49 - 1 = 48 m2+1=72+1=49+1=50m^2 + 1 = 7^2 + 1 = 49 + 1 = 50 So, the triplet is (14,48,50)(14, 48, 50). Verification: 142+482=196+2304=2500=50214^2 + 48^2 = 196 + 2304 = 2500 = 50^2.

Explanation:

We use the general form (2m,m2−1,m2+1)(2m, m^2 - 1, m^2 + 1). Since 1414 is even, we set 2m=142m = 14 to find mm.

Problem 5:

A ladder 25 m25\text{ m} long reaches a window of a building 20 m20\text{ m} above the ground. Determine the distance of the foot of the ladder from the building.

A diagram of a ladder leaning against a wall forming a right triangle.

Solution:

Let the distance be xx. By the Baudhayana-Pythagoras theorem: x2+202=252x^2 + 20^2 = 25^2 x2+400=625x^2 + 400 = 625 x2=625−400x^2 = 625 - 400 x2=225x^2 = 225 x=225=15x = \sqrt{225} = 15 Therefore, the distance is 15 m15\text{ m}.

Explanation:

The ladder, the wall, and the ground form a right-angled triangle where the ladder is the hypotenuse.