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

We will arrange the four triangular pieces to form a larger square.

Step 1 — Understanding the pieces

Let us consider one of the original square papers. Let its side length be $s$ . The area of this square is $s \times s = s^2$ . When this square is cut diagonally, it forms two identical right-angled triangles. The two shorter sides (legs) of each triangle are equal to the side length of the square, which is $s$ . The longest side (hypotenuse) of each triangle can be found using the Pythagorean theorem. Let $h$ be the hypotenuse.

$h^2 = s^2 + s^2$

$h^2 = 2s^2$

$h = \sqrt{2s^2}$

$\boxed{h = s\sqrt{2}}$

We have two such squares, so we have a total of four identical right-angled triangles. The total area of these four triangles is $2 \times s^2$ . We need to form a new square with this total area.

Diagram 1

Step 2 — Arranging the pieces

To form a new square, we can place the four triangles such that their right-angle vertices meet at the center. The hypotenuses of the triangles will then form the outer boundary of the new shape. When arranged this way, the four hypotenuses form the four sides of a larger square. The side length of this new square will be equal to the hypotenuse of one of the triangles. So, the side length of the new square is $s\sqrt{2}$ . Let us calculate the area of this new square.

$\text{Area of new square} = (\text{side length})^2$

$\text{Area of new square} = (s\sqrt{2})^2$

$\text{Area of new square} = s^2 \times (\sqrt{2})^2$

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

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

This new square has an area that is double the area of one original square ( $s^2$ ).

Diagram 2

Answer

The four triangular pieces can be arranged by placing their right-angle vertices together at the center. This forms a larger square whose sides are made up of the hypotenuses of the triangles. The area of this new square is double the area of one original square.

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 cm and 12 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 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. unit?

(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 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]