The Baudhayana-Pythagoras Theorem — Formula Sheet

Area of REST = 2 × Area of PEAR = 2 × 1 = 2 sq. units

Doubling the side length of a square makes its area 4 times the original

PQRS has half the area of the original square

Hypotenuse of unit isosceles right triangle: c^2 = 2, so c = √2

1 < √2 < 2

√2 ≈ 1.41421356...

General isosceles right triangle relation: c^2 = 2a^2

Baudhāyana's Theorem: a^2 + b^2 = c^2

The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides

3^2 + 4^2 = 5^2

(3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2 = (5k)^2

(n - 1)^2 + (2n - 1) = n^2

For the unit isosceles right triangle, c^2 = 2 and c = √2.

1 < √2 < 2.

√2 lies between 1 and 2, and more closely between 1.414 and 1.415.

If √2 = m/n, then 2n² = m².

For equal sides of length a and hypotenuse c, the relation is c² = 2a².