Exercise practice 1: The Baudhayana-Pythagoras Theorem

3.1 Figure it Out: the hypotenuse of an isosceles right triangle (pages 39 and 40)

These questions use one relation: in an isosceles right triangle with equal sides a and hypotenuse c, the square on c has double the area of the square on a, so c² = 2a². Write the exact value first, then place it between two decimals.

2.3 Question 1: Arranging four pieces into a double-area square

Statement. Two identical square papers are cut as shown in the chapter: pieces 1 and 2 come from the first square, pieces 3 and 4 from the second. Arrange the four pieces to form one square whose area is twice that of either original square.

Answer. Each original square is cut along its diagonal into two congruent isosceles right triangles. The four triangles can be arranged with their right angles meeting at the centre, so that the four hypotenuses form the sides of a new square. That new square has side equal to the diagonal of the original square, so its area is double.

Steps. Let the original square have side a and area a².

1. Cutting each square along its diagonal gives four congruent isosceles right triangles, each with legs a and hypotenuse a√2, and each of area a²/2.
2. Place the four triangles with their right angles at one point and their legs along the two diagonals of the new square. The outer boundary is formed by the four hypotenuses, each of length a√2.
3. The new square therefore has side a√2 and area (a√2)² = 2a², which is double the original area.
4. Check by area: 4 triangles of area a²/2 make 2a², exactly the area of the new square, so the pieces cover it with no gaps.

2.3 Question 2: Hypotenuse and decimal bounds

Statement. An isosceles right triangle has equal sides of length 3, 4, 6, 8 and 9 in turn. Find each hypotenuse and give decimal bounds for it with one or more digits after the point.

Answer. The exact hypotenuses are 3√2, 4√2, 6√2, 8√2 and 9√2, approximately 4.24, 5.66, 8.49, 11.31 and 12.73. The required bounds are:

• 4.2 < 3√2 < 4.3
• 5.6 < 4√2 < 5.7
• 8.4 < 6√2 < 8.5
• 11.3 < 8√2 < 11.4
• 12.7 < 9√2 < 12.8

Steps. Use c² = 2a², so c = a√2.

1. For a = 3, c² = 18. Now 4.2² = 17.64 and 4.3² = 18.49, and 17.64 < 18 < 18.49, so 4.2 < √18 < 4.3.
2. For a = 4, c² = 32. 5.6² = 31.36 and 5.7² = 32.49, so 5.6 < √32 < 5.7.
3. For a = 6, c² = 72. 8.4² = 70.56 and 8.5² = 72.25, so 8.4 < √72 < 8.5.
4. For a = 8, c² = 128. 11.3² = 127.69 and 11.4² = 129.96, so 11.3 < √128 < 11.4.
5. For a = 9, c² = 162. 12.7² = 161.29 and 12.8² = 163.84, so 12.7 < √162 < 12.8. The bounds come from squaring the two decimals just below and just above the exact value.

2.3 Question 3: Equal sides from a hypotenuse of 10

Statement. An isosceles right triangle has hypotenuse 10 units. Find its two equal sides.

Answer. Each equal side is √50 = 5√2, about 7.07.

Steps.

1. Use c² = 2a² with c = 10: 100 = 2a².
2. So a² = 50 and a = √50 = 5√2.
3. Check with the chapter's hint: the square built on the hypotenuse has area 100, and it is made of two squares of equal area a², so each is 50.
4. Since 7.07² = 49.9849 and 7.08² = 50.1264, the side lies between 7.07 and 7.08.

3.2 Figure it Out: using the theorem on right triangles (page 47)

Here the theorem is used directly. If the two legs are known, add their squares and take the square root. If the hypotenuse and one leg are known, subtract the squares before taking the root.

2.4 Question 1: Hypotenuse from legs 5 cm and 12 cm

Statement. A right-angled triangle has shorter sides 5 cm and 12 cm. Draw and measure its hypotenuse, then check with the theorem.

Answer. 13 cm.

Steps.

1. The legs are a = 5 and b = 12.
2. c² = a² + b² = 25 + 144 = 169.
3. c = √169 = 13 cm.
4. A measured drawing gives about 13 cm, and the theorem gives the exact value.

2.4 Question 2: Leg from hypotenuse 17 cm and one leg 8 cm

Statement. A right-angled triangle has one short side 8 cm and hypotenuse 17 cm. Find the third side, then check by drawing and measuring.

Answer. 15 cm.

Steps.

1. Here the unknown is a leg, so rearrange the theorem: b² = c² - a².
2. b² = 17² - 8² = 289 - 64 = 225.
3. b = √225 = 15 cm.
4. This is the (8, 15, 17) triple, so the answer is exact.

2.4 Question 3: Constructing squares of triple and five times the area

Statement. Using Baudhayana's constructions, how would you build a square whose area is three times that of a given square? How about five times?

Answer. For three times, make a right triangle whose legs are the side a of the square and the diagonal a√2; the square on its hypotenuse has area 3a². For five times, use legs a and 2a; the square on its hypotenuse has area 5a².

Steps.

1. The construction on the diagonal already gives a square of area 2a² from the original square of area a².
2. To triple the area, combine the original square and the doubled square by Baudhayana's method: their side lengths become the two legs of a right triangle. The legs are a and a√2, so the hypotenuse is √(a² + 2a²) = a√3.
3. The square on this hypotenuse has area 3a², which is three times the original.
4. For five times, use legs a and 2a, because a² + (2a)² = 5a². The square on the hypotenuse a√5 has area 5a².
5. A second route to five times is to combine the original square with a square of area 4a², whose side is 2a; this gives the same right triangle.

2.4 Question 4: Missing side in five right triangles

Statement. In each right triangle, c is the hypotenuse. Find the missing side: (a) a = 5, b = 7; (b) a = 8, b = 12; (c) a = 9, c = 15; (d) a = 7, b = 12; (e) a = 1.5, b = 3.5.

Answer. (a) c = √74, about 8.60. (b) c = √208 = 4√13, about 14.42. (c) b = 12. (d) c = √193, about 13.89. (e) c = √14.5, about 3.81.

Steps.

1. (a) c² = 25 + 49 = 74, so c = √74. Since 8.6² = 73.96 and 8.61² = 74.1321, the value is about 8.60.
2. (b) c² = 64 + 144 = 208 = 16 × 13, so c = 4√13. Since 14.4² = 207.36 and 14.5² = 210.25, the value is about 14.42.
3. (c) The missing length is a leg: b² = c² - a² = 225 - 81 = 144, so b = 12.
4. (d) c² = 49 + 144 = 193, so c = √193. Since 13.9² = 193.21 and 13.89² = 192.9321, the value is about 13.89.
5. (e) c² = 2.25 + 12.25 = 14.5, so c = √14.5. Since 3.8² = 14.44 and 3.9² = 15.21, the value is about 3.81. For (a), (b), (d) and (e) the answers are irrational, so a decimal approximation is the only exact-looking option; for (c) the answer is a whole number.