The Baudhāyana-pythagoras Theorem: questions with answers

Square the candidates: $1.414^2 = 1.999396$, which is less than 2, and $1.415^2 = 2.002225$, which is more than 2. So $\sqrt{2}$ lies between 1.414 and 1.415. It is not exactly 1.41, so it does not sit between 1.410 and 1.411.

The outer square holds all eight congruent triangles and the tilted square holds four of them. So the outer square has double the area of the tilted square.

Doubling the side gives a square of 4 times the area, not 2 times. The middle square has area 16 square units, and the four pieces add one more square of 16, so the new square has 32 square units.

The square whose corners are the midpoints of the sides takes half of each quarter of the given square, so it has half the area. A square on half a side has only a quarter of the area, and a square on the diagonal has double.

The hypotenuse is the longest side, so its square is the sum of the other two squares. 17² − 8² = 289 − 64 = 225, and √225 = 15 cm.

The longest side is opposite the right angle. Its square is 3² + 4² = 9 + 16 = 25, so the side is 5 cm.

The square on the hypotenuse has twice the area of the square on an equal side. That square has area 3² = 9, so the answer is twice 9, which is 18 sq. units.

The square drawn on the hypotenuse has twice the area of the unit square, so its area is 2 sq. units. A square of area 2 has side √2, so the hypotenuse is √2 units.

Scaling multiplies every term by 10, so (3, 4, 5) becomes (30, 40, 50). It is still a triple: 900 + 1600 = 2500 = 50².

The square on the hypotenuse has area 16 + 32 = 48 sq units, three times the area 16. Its side is $\sqrt{48}$ units, which is not a whole number.

Every Baudhāyana triple, such as (3, 4, 5) and its multiples, solves $x^2 + y^2 = z^2$, so there are infinitely many. For every power above 2, Fermat's Last Theorem says there is none.

The new square's area is 36 + 64 = 100 square units, so its side is √100 = 10 units. The side is the hypotenuse of the right triangle with perpendicular sides 6 and 8.

A side going $a$ across and $b$ up gives area $a^2 + b^2$, and one of $a$, $b$ may be 0. So the areas are 1, 2, 4, 5, 8, 9, 10, 13 and so on, and areas such as 3, 6, 7 and 11 never occur.

The side 17 comes from adding 5 and 12. The theorem adds the squares: 25 + 144 = 169, so the square on the hypotenuse has side $\sqrt{169} = 13$.

The swayed stem is the hypotenuse, so $3^2 + x^2 = (x + 1)^2$. Moving $x^2$ across, $(x + 1)^2 - x^2 = 3^2 = 9$, which is also $2x + 1$.