Squares and Square Roots | FIO

Question 15

True or False—Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.

Solution

Understand the Question

A Baudhāyana triple (or Pythagorean triple) consists of three positive integers $(a, b, c)$ satisfying $a^2 + b^2 = c^2$ .
A triple is called primitive if the numbers have no common factor other than $1$ , meaning $\text{HCF}(a, b, c) = 1$ .
If $\text{HCF}(a, b, c) = k > 1$ , dividing each term by $k$ yields a primitive triple $(x, y, z)$ , making $(a, b, c) = (kx, ky, kz)$ a scaled version of that primitive triple.
Therefore, every Baudhāyana triple is either primitive (when $k = 1$ ) or a scaled version of a primitive triple (when $k > 1$ ).

Step 1 · Examine Primitive Triples

A Baudhāyana triple $(a, b, c)$ satisfies: $a^2 + b^2 = c^2$

For example, consider $(3, 4, 5)$ :

\begin{aligned} 3^2 + 4^2 &= 9 + 16 = 25 \\ 5^2 &= 25 \\ 3^2 + 4^2 &= 5^2 \end{aligned}

Since $\text{HCF}(3, 4, 5) = 1$ , the triple $(3, 4, 5)$ is a primitive Baudhāyana triple.

Step 2 · Examine Scaled Triples

Consider $(6, 8, 10)$ :

\begin{aligned} 6^2 + 8^2 &= 36 + 64 = 100 \\ 10^2 &= 100 \\ 6^2 + 8^2 &= 10^2 \end{aligned}

Here, $\text{HCF}(6, 8, 10) = 2$ .

Dividing each term by $2$ gives: $(6 \div 2, 8 \div 2, 10 \div 2) = (3, 4, 5)$

Thus, $(6, 8, 10)$ is a scaled version of the primitive triple $(3, 4, 5)$ scaled by a factor of $2$ .

Step 3 · Prove the General Case

Let $(a, b, c)$ be any Baudhāyana triple such that $a^2 + b^2 = c^2$ .

Let $k = \text{HCF}(a, b, c)$ . We can express: $a = kx, \quad b = ky, \quad c = kz$ where $x, y, z$ are positive integers with $\text{HCF}(x, y, z) = 1$ .

Substitute these into $a^2 + b^2 = c^2$ :

\begin{aligned} (kx)^2 + (ky)^2 &= (kz)^2 \\ k^2x^2 + k^2y^2 &= k^2z^2 \end{aligned}

Dividing the entire equation by $k^2$ ( $k \neq 0$ ):

\begin{aligned} \dfrac{k^2x^2}{k^2} + \dfrac{k^2y^2}{k^2} &= \dfrac{k^2z^2}{k^2} \\[0.6em] x^2 + y^2 &= z^2 \end{aligned}

Hence, $(x, y, z)$ is a primitive Baudhāyana triple, and $(a, b, c) = (kx, ky, kz)$ is:

A primitive triple if $k = 1$ .
A scaled version of the primitive triple $(x, y, z)$ if $k > 1$ .

Answer

True

Common Mistakes

Assuming non-primitive triples cannot satisfy $a^2 + b^2 = c^2$ : Multiplying all three numbers of any valid triple by an integer $k$ always results in another valid triple because $k^2$ factors out from both sides.
Confusing pairwise coprime with overall HCF: For a triple to be primitive, $\text{HCF}(a, b, c) = 1$ , which means there is no single common factor greater than $1$ shared across all three terms.

More questions in FIO

Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut in the following way.

Can you arrange these pieces to create a square with double the area of either square?

The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.

(i) 3

(ii) 4

(iii) 6

(iv) 8

(v) 9

The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: Find the area of the square composed of two such right triangles.]

If a right-angled triangle has shorter sides of lengths $5\text{ cm}$ and $12\text{ cm}$ , then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhāyana's Theorem.

If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhāyana's Theorem.

Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudhāyana's Śulba-Sūtra, Verse 1.10)

Let $a$ , $b$ and $c$ denote the length of the sides of a right triangle, with $c$ being the length of the hypotenuse. Find the missing sidelength in each of the following cases:

(i) $a = 5$ , $b = 7$

(ii) $a = 8$ , $b = 12$

(iii) $a = 9$ , $c = 15$

(iv) $a = 7$ , $b = 12$

(v) $a = 1.5$ , $b = 3.5$

Find 5 more Baudhāyana triples using this idea.

Does this method yield non-primitive Baudhāyana triples?

[Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

Q10

Are there primitive triples that cannot be obtained through this method? If yes, give examples.

Q11

Find the diagonal of a square with sidelength $5 \text{ cm}$ .

Q12

Find the missing sidelengths in the following right triangles:

Q13

Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.

Q14

Is the hypotenuse the longest side of a right triangle? Justify your answer.

Q15

True or False—Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.

Q16

Give 5 examples of rectangles whose sidelengths and diagonals are all integers.

Q17

Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.

Q18

(i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. units?

(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Q19

Find the area of an equilateral triangle with sidelength $6\text{ units}$ . [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]