Exercise practice 2: The Baudhayana-Pythagoras Theorem

4.1 Figure it Out: generating Baudhayana triples (page 50)

The chapter builds triples from odd squares. The identity (n - 1)² + (2n - 1) = n² becomes a triple when the odd number 2n - 1 is itself a perfect square.

2.5 Question 1: Five more triples from odd squares

Statement. Using the odd-square idea, find five more Baudhayana triples.

Answer. (24, 7, 25), (40, 9, 41), (60, 11, 61), (84, 13, 85) and (112, 15, 113).

Steps. For an odd square m² = 2n - 1, we get n = (m² + 1)/2, and the triple is (n - 1, m, n).

1. m = 7, so m² = 49 and n = 25. The triple is (24, 7, 25). Check: 576 + 49 = 625 = 25².
2. m = 9, so m² = 81 and n = 41. The triple is (40, 9, 41). Check: 1600 + 81 = 1681 = 41².
3. m = 11, so m² = 121 and n = 61. The triple is (60, 11, 61). Check: 3600 + 121 = 3721 = 61².
4. m = 13, so m² = 169 and n = 85. The triple is (84, 13, 85). Check: 7056 + 169 = 7225 = 85².
5. m = 15, so m² = 225 and n = 113. The triple is (112, 15, 113). Check: 12544 + 225 = 12769 = 113².

2.5 Question 2: Are the generated triples non-primitive?

Statement. Does the odd-square method yield non-primitive Baudhayana triples? The hint says one of the smaller sides is one less than the hypotenuse.

Answer. No. Every triple produced directly by this method is primitive.

Steps.

1. The method produces (n - 1, m, n), and two of its sides, n - 1 and n, are consecutive integers.
2. Consecutive integers have no common factor greater than 1, so the greatest common factor of the whole triple is 1.
3. A triple is primitive exactly when its greatest common factor is 1, so the generated triple is primitive.
4. Non-primitive triples appear only after scaling, for example 2 × (24, 7, 25) = (48, 14, 50). Scaling is a separate step, not part of the odd-square construction.

2.5 Question 3: Primitive triples the method misses

Statement. Can this method miss some primitive triples? If so, name examples.

Answer. Yes. For example (8, 15, 17), (12, 35, 37) and (20, 21, 29) are primitive but are not produced by the odd-square method.

Steps.

1. The method always gives a triple in which the hypotenuse minus one leg equals 1, because the legs are n - 1 and m and the hypotenuse is n.
2. In (8, 15, 17) the differences are 17 - 15 = 2 and 17 - 8 = 9, so neither leg is one less than the hypotenuse.
3. In (12, 35, 37) the difference between the hypotenuse and the longer leg is 2, again not 1.
4. In (20, 21, 29) the differences are 8 and 9, so the pattern fails too.
5. So the odd-square construction is a useful generator, not a complete one. It gives the family where two sides are consecutive.

4.2 Figure it Out: further applications (pages 52 and 53)

These questions apply the theorem to squares, rhombi, rectangles, lattice grids and an equilateral triangle, and check the logical ideas behind triples.

2.7 Question 1: Diagonal of a square of side 5 cm

Statement. Find the diagonal of a square with side length 5 cm.

Answer. 5√2 cm, about 7.07 cm.

Steps.

1. The diagonal divides the square into two right triangles with legs 5 cm and 5 cm.
2. d² = 5² + 5² = 25 + 25 = 50.
3. d = √50 = 5√2. Since 7.07² = 49.9849, the value is about 7.07 cm.
4. As a formula, the diagonal of a square of side s is s√2.

2.7 Question 2: Missing sides in six right triangles

Statement. Find the missing side in each of the following right triangles, shown in the chapter's figure: (a) legs 7 and 9; (b) legs 4 and 10; (c) hypotenuse 41 and one leg 40; (d) legs 10 and √200; (e) legs 10 and √150; (f) hypotenuse 45 and one leg 27.

Answer. (a) √130, about 11.40. (b) √116 = 2√29, about 10.77. (c) 9. (d) √300 = 10√3, about 17.32. (e) √250 = 5√10, about 15.81. (f) 36.

Steps.

1. (a) c² = 49 + 81 = 130, so c = √130. Since 11.4² = 129.96, the value is about 11.40.
2. (b) c² = 16 + 100 = 116, so c = √116 = 2√29. Since 10.77² = 115.99, the value is about 10.77.
3. (c) The missing length is a leg: b² = 41² - 40² = 1681 - 1600 = 81, so b = 9.
4. (d) c² = 100 + 200 = 300, so c = √300 = 10√3. Since 17.32² = 299.98, the value is about 17.32.
5. (e) c² = 100 + 150 = 250, so c = √250 = 5√10. Since 15.81² = 249.96, the value is about 15.81.
6. (f) The missing length is a leg: b² = 45² - 27² = 2025 - 729 = 1296, so b = 36. This is the scaled triple 9 × (3, 4, 5) = (27, 36, 45).

2.7 Question 3: Side of a rhombus from its diagonals

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

Answer. 37 units.

Steps.

1. The diagonals of a rhombus bisect each other at right angles, so each diagonal is halved: 12 and 35.
2. These half diagonals are the legs of a right triangle whose hypotenuse is a side of the rhombus.
3. side² = 12² + 35² = 144 + 1225 = 1369.
4. side = √1369 = 37 units. This is the triple (12, 35, 37) scaled by the factor 2 giving diagonals 24 and 70.

2.7 Question 4: Is the hypotenuse the longest side?

Statement. Show that in a right triangle the hypotenuse is the longest side.

Answer. Yes, always.

Steps.

1. In a right triangle c² = a² + b² with both legs of positive length, so c² > a² and c² > b².
2. Taking positive square roots gives c > a and c > b.
3. A triangle cannot have zero or negative sides, so the strict inequalities always hold. The hypotenuse is the longest side, and the right angle is opposite it.

2.7 Question 5: Primitive or scaled?

Statement. True or false: every Baudhayana triple is primitive or a whole-number multiple of a primitive triple.

Answer. True.

Steps.

1. Take any triple (a, b, c) and let g be the greatest common factor of a, b and c.
2. If g = 1 the triple is primitive.
3. If g > 1, divide all three entries by g. The new numbers are integers, and (a/g)² + (b/g)² = (c/g)² still holds because dividing both sides of a² + b² = c² by g² keeps the equality.
4. The divided triple has greatest common factor 1, so it is primitive, and the original is g times that primitive triple.
5. Hence every triple belongs to one of the two classes, and the statement is true.

2.7 Question 6: Five rectangles with integer sides and diagonal

Statement. Give five examples of rectangles whose side lengths and diagonal are all integers.

Answer. Five examples are 3 by 4 (diagonal 5), 5 by 12 (diagonal 13), 8 by 15 (diagonal 17), 7 by 24 (diagonal 25), and 20 by 21 (diagonal 29).

Steps. By the theorem, the sides and diagonal of a rectangle form a right triangle with the diagonal as hypotenuse, so every Baudhayana triple gives a rectangle.

1. 3² + 4² = 9 + 16 = 25 = 5².
2. 5² + 12² = 25 + 144 = 169 = 13².
3. 8² + 15² = 64 + 225 = 289 = 17².
4. 7² + 24² = 49 + 576 = 625 = 25².
5. 20² + 21² = 400 + 441 = 841 = 29².

2.7 Question 7: A square with area 24

Statement. Construct a square whose area equals the difference of the squares with side lengths 5 units and 7 units.

Answer. The required area is 49 - 25 = 24, so the side is √24 = 2√6 units.

Steps.

1. Make a right triangle with hypotenuse 7 and one leg 5.
2. The other leg squared is 7² - 5² = 49 - 25 = 24, so the other leg is √24.
3. Build a square on that leg. Its area is (√24)² = 24 square units, exactly the difference required.
4. The side is √24 = 2√6, about 4.90 units, since 4.9² = 24.01.

2.7 Question 8: Squares on a dot grid

Statement. (a) Using grid dots as vertices, can you make a square of area 2, 3, 4 and 5 square units? (b) If the grid extends forever, what integer areas are possible?

Answer. (a) Areas 2, 4 and 5 are possible; area 3 is not possible on a square grid. (b) The possible integer areas are the numbers that can be written as the sum of two square numbers: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, and so on.

Steps.

1. A square whose side joins two grid points can be described by the horizontal and vertical steps p and q along its side. By the theorem the side squared is p² + q², so the area is p² + q².
2. For area 2, take p = 1 and q = 1. The square is tilted at 45 degrees.
3. For area 3, we would need p² + q² = 3 with whole numbers p and q. The squares 0, 1, 4, 9, ... cannot add to 3, so no such square exists.
4. For area 4, take p = 2 and q = 0, which gives an ordinary 2 by 2 square.
5. For area 5, take p = 2 and q = 1.
6. In general, an integer area is possible exactly when it is the sum of two perfect squares. For example 8 = 4 + 4, 10 = 9 + 1, 13 = 9 + 4, while 3, 6, 7, 11, 12 and 15 cannot be written this way.

2.7 Question 9: Area of an equilateral triangle of side 6

Statement. Find the area of an equilateral triangle whose side is 6 units. Use the hint that an altitude bisects the opposite side.

Answer. 9√3 square units, about 15.59 square units.

Steps.

1. Draw an altitude from one vertex to the opposite side. It bisects the side, so the base is split into two segments of 3 units.
2. The altitude h, the half-side 3 and the side 6 form a right triangle with hypotenuse 6, so h² + 3² = 6².
3. h² = 36 - 9 = 27, so h = √27 = 3√3.
4. Area = (1/2) × base × height = (1/2) × 6 × 3√3 = 9√3.
5. Since √3 is about 1.732, the area is about 15.59 square units.