Squares and Square Roots | FIO

Question 1

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?

Solution

Understand the Question

Each original square has a side length $s$ and area $s^2$ .
Cutting two identical squares diagonally produces $4$ congruent right-angled isosceles triangles with legs of length $s$ .
The total area of all $4$ pieces combined is $2 \times s^2 = 2s^2$ , which is double the area of one original square.
By arranging the $4$ pieces so that their right-angled vertices meet at the center, the hypotenuses form the four outer edges of a new, larger square.

Step 1 · Find Dimensions of the Triangular Pieces

Let the side length of each original square be $s$ .

$\text{Area of one square} = s^2$

Cutting two identical squares along their diagonals yields $4$ congruent right-angled isosceles triangles with legs of length $s$ . Diagram 1

By Pythagoras theorem, the hypotenuse $h$ of each triangle is

\begin{aligned} h^2 &= s^2 + s^2 \\ &= 2s^2 \\ h &= \sqrt{2s^2} = s\sqrt{2} \end{aligned}

Step 2 · Arrange Pieces to Form a New Square

Place the $4$ triangles such that their right-angled vertices meet at the center. Diagram 2

The hypotenuses of the $4$ triangles form the outer sides of the new square.

$\text{Side length of new square} = s\sqrt{2}$

\begin{aligned} \text{Area of new square} &= (\text{side length})^2 \\ &= (s\sqrt{2})^2 \\ &= s^2 \times (\sqrt{2})^2 \\ &= s^2 \times 2 \\ &= 2s^2 \end{aligned}

The area of this new square ( $2s^2$ ) is double the area of an original square ( $s^2$ ).

Answer

Yes. Arrange the four triangular pieces with their right-angled vertices meeting at the center to form a square of side length $s\sqrt{2}$ and area $2s^2$ .

Common Mistakes

Mismatched Orientation: Arranging the pieces with their hypotenuses touching instead of their right-angled vertices, which creates a rectangle or parallelogram rather than a single square.
Doubling the Side Instead of Area: Assuming a square with double the area has side length $2s$ (which yields an area of $4s^2$ ). The correct side length is $s\sqrt{2}$ .

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.]