2. The Baudhāyana-Pythagoras Theorem

The diagonal of the original square is a side of the dotted square. Each side of the original square makes 45° with this diagonal.

So the sides of the original square bisect the 90° angles of the dotted square. A line that bisects an angle of a square passes through the opposite vertex. So the extended sides pass through the vertices of the dotted square.

Each small triangle is half of a square equal to the original square, cut along a diagonal. All of them have two equal sides (the side of the original square) with a right angle between them. So they are congruent by SAS.

No. Its area is 12 × 12 = 14 of the original area.

So 4 such squares fill the original square.

PR and QS join the midpoints of opposite sides, so each equals the side of the paper. They are equal, bisect each other and are perpendicular. So PQRS is a square.

PR and QS divide the paper into 4 small squares. Each small square is cut in half by a side of PQRS. PQRS has 4 of these 8 halves. So its area is half.

Yes. We get 4 congruent isosceles right triangles. Place them so that their right angles meet at one point and their long sides (hypotenuses) are outside.

They form a square whose side is the diagonal of the original square. Its area is double.

c² = 2a².

(i) c = √18. 4.2² = 17.64, 4.3² = 18.49. So 4.2 < c < 4.3

(ii) c = √32. 5.6² = 31.36, 5.7² = 32.49. So 5.6 < c < 5.7

(iii) c = √72. 8.4² = 70.56, 8.5² = 72.25. So 8.4 < c < 8.5

(iv) c = √128. 11.3² = 127.69, 11.4² = 129.96. So 11.3 < c < 11.4

(v) c = √162. 12.7² = 161.29, 12.8² = 163.84. So 12.7 < c < 12.8

The square on the hypotenuse has area 100. It is double the square on a side. So a² = 50.

Each side = √50 ≈ 7.07 (between 7.0 and 7.1).

If a = b, the right triangle is isosceles. The square on the hypotenuse is then the square on the diagonal, which has double the area. This agrees with our earlier method.

In each right triangle, the two acute angles are x and 90° − x.

At each corner of the new figure, the angles x, the corner angle and 90° − x lie on a straight line. So the corner angle = 180° − x − (90° − x) = 90°.

All four sides are equal and all angles are 90°, so it is a square.

c² = 25 + 144 = 169. c = 13 cm

b² = 289 − 64 = 225. b = 15 cm

Triple: Let the side be a. Draw the diagonal (length √2a). Make a right triangle with legs a and the diagonal. Its hypotenuse² = a² + 2a² = 3a². The square on this hypotenuse has triple area.

Five times: Make a right triangle with legs a and 2a. Hypotenuse² = a² + 4a² = 5a². The square on this hypotenuse has five times the area.

(i) c² = 25 + 49 = 74 → c = √74 ≈ 8.6

(ii) c² = 64 + 144 = 208 → c = √208 ≈ 14.4

(iii) b² = 225 − 81 = 144 → b = 12

(iv) c² = 49 + 144 = 193 → c = √193 ≈ 13.9

(v) c² = 2.25 + 12.25 = 14.5 → c = √14.5 ≈ 3.8

(3, 4, 5), (6, 8, 10), (5, 12, 13), (9, 12, 15), (8, 15, 17), (12, 16, 20)

900 + 1600 = 2500 = 50². 90000 + 160000 = 250000 = 500². So both are triples.

Yes, 5, 12 and 13 have no common factor greater than 1.

Primitive triples up to 20: (3, 4, 5), (5, 12, 13), (8, 15, 17)

Scaled versions, e.g. (6, 8, 10), (9, 12, 15), (10, 24, 26), (16, 30, 34), are not primitive, because they have the common factor k.

(9, 12, 15) with f = 3 gives (3, 4, 5), which is a triple.

In general, a² + b² = c². Dividing both sides by f²: (af)² + (bf)² = (cf)². So yes.

Use (n − 1)² + (2n − 1) = n² where 2n − 1 is an odd square.

• 49 = 2 × 25 − 1 → (7, 24, 25)
• 81 = 2 × 41 − 1 → (9, 40, 41)
• 121 = 2 × 61 − 1 → (11, 60, 61)
• 169 = 2 × 85 − 1 → (13, 84, 85)
• 225 = 2 × 113 − 1 → (15, 112, 113)

No. Here the hypotenuse n and the side n − 1 are consecutive numbers. Consecutive numbers have no common factor other than 1. So every triple obtained is primitive.

Yes. In this method, the hypotenuse is always 1 more than a side. Triples like (8, 15, 17) and (20, 21, 29) are primitive but do not have this property.

d² = 25 + 25 = 50. d = 5√2 ≈ 7.07 cm

(a) Legs 7 and 9: hypotenuse = √130 ≈ 11.4

(b) Legs 4 and 10: hypotenuse = √116 ≈ 10.8

(c) Leg 40, hypotenuse 41: other leg = √81 = 9

(d) Leg 10, hypotenuse √200: other leg = √200 − 100 = 10

(e) Legs 10 and √150: hypotenuse = √250 ≈ 15.8

(f) Leg 27, hypotenuse 45: other leg = √2025 − 729 = √1296 = 36

The diagonals bisect each other at right angles. Half-diagonals are 12 and 35.

Side² = 144 + 1225 = 1369. Side = 37 units

Yes. c² = a² + b², so c² is greater than a² and b². Hence c is greater than a and b.

True. If a triple has a common factor, divide by the HCF. We get a primitive triple, and the original is its scaled version.

3 × 4 (diagonal 5), 6 × 8 (10), 5 × 12 (13), 8 × 15 (17), 7 × 24 (25).

Required area = 49 − 25 = 24.

Draw a right triangle with hypotenuse 7 and one leg 5: draw a 5-unit line, make a right angle at one end, and cut the perpendicular with an arc of 7 from the other end.

The other leg is √24. The square on this leg has area 24 sq. units.

(a) Yes: a tilted square whose side goes 1 across and 1 up (side √2).

(b) No. 3 cannot be written as a sum of two squares.

(c) Yes: a 2 × 2 square.

(d) Yes: a tilted square whose side goes 2 across and 1 up (side √5).

(ii) The area is a² + b² for whole numbers a, b. Possible areas: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, …

The altitude bisects the base (the two halves are congruent by RHS). So it makes a right triangle with hypotenuse 6 and base 3.

Height = √36 − 9 = √27 ≈ 5.2.

Area = 12 × 6 × √27 = 3√27 ≈ 15.6 sq. units

Open the box labelled RED. It cannot have red balls, so it has blue or green.

If it has blue balls: the box labelled GREEN cannot be green, so it has red, and the box labelled BLUE has green.

If it has green balls: the box labelled BLUE cannot be blue, so it has red, and the box labelled GREEN has blue.