Squares and Square Roots | FIO

Question 9

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.]

Solution

Understand the Question

A Baudhāyana (Pythagorean) triple is a set of three positive integers $(a, b, c)$ satisfying $a^2 + b^2 = c^2$ , where $c$ is the hypotenuse.
A triple is primitive if $\text{HCF}(a, b, c) = 1$ , and non-primitive if $\text{HCF}(a, b, c) > 1$ .
In this method, one of the smaller sides is always one less than the hypotenuse ( $c - b = 1$ ).
Because consecutive integers are coprime (their difference is $1$ ), any common divisor of $b$ and $c$ must divide $1$ , meaning the $\text{HCF}$ of the triple is always $1$ .

Step 1 · Derive General Form of the Triples

Let the sides of the triangle be $a$ , $b$ , and hypotenuse $c$ . Given that one leg is one less than the hypotenuse, let $b = c - 1$ .

Using $a^2 + b^2 = c^2$ $a^2 + (c - 1)^2 = c^2$

$a^2 + c^2 - 2c + 1 = c^2$

$a^2 - 2c + 1 = 0$

\begin{aligned} 2c &= a^2 + 1 \\[0.6em] c &= \dfrac{a^2 + 1}{2} \end{aligned}

Now find $b$

\begin{aligned} b &= \dfrac{a^2 + 1}{2} - 1 \\[0.6em] &= \dfrac{a^2 + 1 - 2}{2} \\[0.6em] &= \dfrac{a^2 - 1}{2} \end{aligned}

For $b$ and $c$ to be integers, $a$ must be an odd integer. The generated triple is $\left(a, \dfrac{a^2 - 1}{2}, \dfrac{a^2 + 1}{2}\right)$ .

Step 2 · Find the HCF of the Generated Triples

Find the difference between the two largest sides, $c$ and $b$

\begin{aligned} c - b &= \dfrac{a^2 + 1}{2} - \dfrac{a^2 - 1}{2} \\[0.6em] &= \dfrac{(a^2 + 1) - (a^2 - 1)}{2} \\[0.6em] &= \dfrac{2}{2} \\[0.6em] &= 1 \end{aligned}

Since $c - b = 1$ , any common factor dividing $b$ and $c$ must also divide their difference, $1$ . $\text{HCF}(a, b, c) = \text{HCF}(a, b, c - b) = \text{HCF}(a, b, 1) = 1$

Since $\text{HCF}(a, b, c) = 1$ , the triple is always primitive.

Step 3 · Verify with Examples

For odd integer $a = 7$

\begin{aligned} b &= \dfrac{7^2 - 1}{2} = \dfrac{48}{2} = 24 \\[0.6em] c &= \dfrac{7^2 + 1}{2} = \dfrac{50}{2} = 25 \end{aligned}

Triple: $(7, 24, 25)$ with $\text{HCF}(7, 24, 25) = 1$ (primitive).

For odd integer $a = 9$

\begin{aligned} b &= \dfrac{9^2 - 1}{2} = \dfrac{80}{2} = 40 \\[0.6em] c &= \dfrac{9^2 + 1}{2} = \dfrac{82}{2} = 41 \end{aligned}

Triple: $(9, 40, 41)$ with $\text{HCF}(9, 40, 41) = 1$ (primitive).

Answer

No, this method does not yield non-primitive Baudhāyana triples; it only yields primitive triples.

Common Mistakes

Overlooking Coprime Consecutive Integers: Forgetting that two consecutive integers $b$ and $c = b + 1$ are always coprime ( $\text{HCF}(b, b+1) = 1$ ), which ensures $\text{HCF}(a, b, c)$ is always $1$ .
Even Values for $a$ : Trying to substitute an even number for $a$ , which gives fractional values for $b$ and $c$ rather than integer triples.

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.]