krit.club logo

The Baudhayana-Pythagoras Theorem - Combining Two Different Squares

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. This is expressed as a2+b2=c2a^2 + b^2 = c^2.

Right-angled triangle showing sides a, b, and hypotenuse c.
•

Combining two squares: If you have two separate squares with side lengths xx and yy, the area of a new square formed by the sum of their areas will have a side length zz such that z=x2+y2z = \sqrt{x^2 + y^2}.

Visual addition of two smaller squares to form a larger square.
•

Pythagorean triplets are sets of three positive integers (a,b,c)(a, b, c) that satisfy the relation a2+b2=c2a^2 + b^2 = c^2. Common examples include (3,4,5)(3, 4, 5) and (8,15,17)(8, 15, 17).

•

The converse of the theorem is also true: if the sum of the squares of two sides of a triangle equals the square of the third side, then the triangle must be a right-angled triangle.

📐Formulae

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

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

Area of Square on Hypotenuse=Area of Square on Base+Area of Square on Altitude\text{Area of Square on Hypotenuse} = \text{Area of Square on Base} + \text{Area of Square on Altitude}

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

💡Examples

Problem 1:

A square has a side of 15 cm15 \text{ cm} and another square has a side of 20 cm20 \text{ cm}. If these two squares are combined to form a new larger square, what will be the length of the side of the new square?

Solution:

Let the side of the first square be a=15a = 15 and the second square be b=20b = 20. The area of the new square c2c^2 is the sum of the areas of the two squares: c2=a2+b2c^2 = a^2 + b^2 c2=152+202c^2 = 15^2 + 20^2 c2=225+400c^2 = 225 + 400 To find the total area: 225+400625\begin{array}{r} 225 \\ + 400 \\ \hline 625 \end{array} c2=625c^2 = 625 c=625=25 cmc = \sqrt{625} = 25 \text{ cm}

Explanation:

According to the Baudhayana-Pythagoras theorem, the side of a square whose area is the sum of two other squares is the hypotenuse of a right triangle with the given sides.

Problem 2:

Find a Pythagorean triplet whose smallest member is 1212.

Solution:

We use the general form of a triplet (2m,m2−1,m2+1)(2m, m^2 - 1, m^2 + 1). Let 2m=122m = 12, which gives m=6m = 6. Now, find the other two members:

  1. m2−1=62−1=36−1=35m^2 - 1 = 6^2 - 1 = 36 - 1 = 35
  2. 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=136912^2 + 35^2 = 144 + 1225 = 1369 372=136937^2 = 1369

Explanation:

By setting the even number 1212 to the 2m2m part of the formula, we can solve for mm and then find the remaining sides of the right-angled triangle.

Problem 3:

In a right-angled triangle, the hypotenuse is 13 cm13 \text{ cm} and one side is 5 cm5 \text{ cm}. Find the third side.

Solution:

Let the hypotenuse be c=13c = 13 and one side be a=5a = 5. We need to find bb. Using a2+b2=c2a^2 + b^2 = c^2: 52+b2=1325^2 + b^2 = 13^2 25+b2=16925 + b^2 = 169 b2=169−25b^2 = 169 - 25 To find the difference: 169−25144\begin{array}{r} 169 \\ - 25 \\ \hline 144 \end{array} b2=144b^2 = 144 b=144=12 cmb = \sqrt{144} = 12 \text{ cm}

Explanation:

We rearrange the formula to solve for a leg: b2=c2−a2b^2 = c^2 - a^2, and then take the square root of the result.

Problem 4:

Two squares have sides 9 cm9 \text{ cm} and 12 cm12 \text{ cm} respectively. Find the side length of a third square whose area is equal to the sum of the areas of these two squares.

Right triangle with legs 9 and 12, hypotenuse marked as side of the new square.

Solution:

Area of first square=92=81 cm2\text{Area of first square} = 9^2 = 81 \text{ cm}^2 Area of second square=122=144 cm2\text{Area of second square} = 12^2 = 144 \text{ cm}^2 Sum of areas=81+144=225 cm2\text{Sum of areas} = 81 + 144 = 225 \text{ cm}^2 Side of new square=225=15 cm\text{Side of new square} = \sqrt{225} = 15 \text{ cm}

Explanation:

The side of the new square is the hypotenuse of a right triangle where the other two sides are 9 cm9 \text{ cm} and 12 cm12 \text{ cm}.

Problem 5:

Verify if (7,24,25)(7, 24, 25) is a Pythagorean triplet.

Right triangle with sides 7, 24 and hypotenuse 25.

Solution:

a2+b2=72+242a^2 + b^2 = 7^2 + 24^2 =49+576=625= 49 + 576 = 625 c2=252=625c^2 = 25^2 = 625 Since 72+242=252, it is a Pythagorean triplet.\text{Since } 7^2 + 24^2 = 25^2 \text{, it is a Pythagorean triplet.}

Explanation:

We calculate the squares of the two smaller numbers and check if their sum equals the square of the largest number.