Chapter 2 exercise answers: The Baudhāyana-Pythagoras Theorem

Open NCERT p. 39One way to think about it

1. Cut each square along one diagonal, as in the figure. This gives four pieces, 1, 2, 3 and 4, and all four are the same right triangle: its two equal sides are the side of the square, and its long side (the hypotenuse) is the diagonal of the square.
2. Place the four triangles so that their right-angle corners all meet at one point, each triangle turned a quarter turn from the one before. The four right angles make 4 × 90° = 360°, so they fit exactly round that point, and neighbouring triangles meet along equal sides with no gap and no overlap.
3. The four long sides now form the outline, so all four sides of the outline are equal (each is a diagonal of the original square). Each outer corner is made of two 45° angles of the triangles, so it is 45° + 45° = 90°. The outline is therefore a square whose side is the diagonal of the original square.
4. This new square is made of all four triangles, that is, of both original squares. So its area = area of one square + area of the other = double the area of either square.

In shortYes. Put the four triangles together with their right-angle corners meeting at the centre and their long sides (the diagonals) on the outside. They form a square whose side is the diagonal of the original square, and its area is double the area of either square.

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(i) 3

1. hypotenuse² = 2 × 3² = 18.
2. 4.2² = 17.64 and 4.3² = 18.49, and 17.64 < 18 < 18.49.

AnswerHypotenuse = 3√2 units, between 4.2 and 4.3 units.

(ii) 4

1. hypotenuse² = 2 × 4² = 32.
2. 5.6² = 31.36 and 5.7² = 32.49, and 31.36 < 32 < 32.49.

AnswerHypotenuse = 4√2 units, between 5.6 and 5.7 units.

(iii) 6

1. hypotenuse² = 2 × 6² = 72.
2. 8.4² = 70.56 and 8.5² = 72.25, and 70.56 < 72 < 72.25.

AnswerHypotenuse = 6√2 units, between 8.4 and 8.5 units.

(iv) 8

1. hypotenuse² = 2 × 8² = 128.
2. 11.3² = 127.69 and 11.4² = 129.96, and 127.69 < 128 < 129.96.

AnswerHypotenuse = 8√2 units, between 11.3 and 11.4 units.

(v) 9

1. hypotenuse² = 2 × 9² = 162.
2. 12.7² = 161.29 and 12.8² = 163.84, and 161.29 < 162 < 163.84.

AnswerHypotenuse = 9√2 units, between 12.7 and 12.8 units.

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1. For an isosceles right triangle, hypotenuse² = 2 × (equal side)², so (equal side)² = 10² ÷ 2 = 50.
2. 50 = 25 × 2, so the equal side = √50 = 5√2 units.

AnswerEach of the other two sides is 5√2 units (≈7.07 units).

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1. By Baudhāyana's theorem, hypotenuse² = 5² + 12² = 25 + 144 = 169.
2. √169 = 13.

AnswerHypotenuse = 13 cm.

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1. By Baudhāyana's theorem, (third side)² = 17² − 8² = 289 − 64 = 225.
2. √225 = 15.

AnswerThird side = 15 cm.

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Open NCERT p. 47One way to think about it

1. Let the given square have side a, so its area is a².
2. Triple the area: draw the diagonal of the given square. By the doubling construction, the square on this diagonal has area 2a², double the given square.
3. Draw a right angle. Along one arm mark off the side of the given square (length a); along the other arm mark off the diagonal. Join the two marked points.
4. By Baudhāyana's theorem, the square on this joining line (the hypotenuse) has area a² + 2a² = 3a². That square is three times the given square.
5. Five times the area: draw a right angle and mark off the side of the given square (a) on one arm and twice that side (2a) on the other. Join the two marked points.
6. By Baudhāyana's theorem, the square on this hypotenuse has area a² + (2a)² = a² + 4a² = 5a², five times the given square.
7. Other combinations work too — for example, arms equal to the sides of the triple square (area 3a²) and the double square (area 2a²) also give 3a² + 2a² = 5a². Any right triangle whose two arm-squares add up to the required area gives a correct construction.

In shortTriple: make a right triangle whose two arms are the side of the given square and its diagonal; the square on the hypotenuse has area a² + 2a² = 3a². Five times: make a right triangle with arms a and 2a; the square on the hypotenuse has area a² + 4a² = 5a².

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(i) a = 5, b = 7

1. Given: a = 5, b = 7. Find: c
2. Using a² + b² = c²:
3. 5² + 7² = c²
4. 25 + 49 = c²
5. 74 = c²
6. c = √74

Answerc = √74

(ii) a = 8, b = 12

1. Given: a = 8, b = 12. Find: c
2. Using a² + b² = c²:
3. 8² + 12² = c²
4. 64 + 144 = c²
5. 208 = c²
6. c = √208 = √(16 × 13) = 4√13

Answerc = 4√13

(iii) a = 9, c = 15

1. Given: a = 9, c = 15. Find: b
2. Using a² + b² = c²:
3. 9² + b² = 15²
4. 81 + b² = 225
5. b² = 144
6. b = 12

Answerb = 12

(iv) a = 7, b = 12

1. Given: a = 7, b = 12. Find: c
2. Using a² + b² = c²:
3. 7² + 12² = c²
4. 49 + 144 = c²
5. 193 = c²
6. c = √193

Answerc = √193

(v) a = 1.5, b = 3.5

1. Given: a = 1.5 = 3/2, b = 3.5 = 7/2. Find: c
2. Using a² + b² = c²:
3. (3/2)² + (7/2)² = c²
4. 9/4 + 49/4 = c²
5. 58/4 = c²
6. c = √(58/4) = √58/2

Answerc = √58/2

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Open NCERT p. 50Checked by computerAnswers can differ: one example

1. 49 is the 25th odd number and 49 = 7², so 24² + 7² = 25² → triple (7, 24, 25).
2. 81 is the 41st odd number and 81 = 9², so 40² + 9² = 41² → triple (9, 40, 41).
3. 121 is the 61st odd number and 121 = 11², so 60² + 11² = 61² → triple (11, 60, 61).
4. 169 is the 85th odd number and 169 = 13², so 84² + 13² = 85² → triple (13, 84, 85).
5. 225 is the 113th odd number and 225 = 15², so 112² + 15² = 113² → triple (15, 112, 113).

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

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1. This method always makes a triple of the form (n−1, m, n): two of its three numbers, n−1 and n, are consecutive integers.
2. Any common factor of all three numbers would have to divide n − (n−1) = 1, so the only possible common factor is 1.
3. So every triple this method produces is already primitive.

AnswerNo — this method only ever produces primitive Baudhāyana triples.

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Open NCERT p. 50Checked by computerAnswers can differ: one example

1. This method always produces triples of the form (n−1, m, n), where the hypotenuse n is exactly 1 more than the leg n−1.
2. (8, 15, 17) is primitive (no common factor above 1), but 17 − 15 = 2 and 17 − 8 = 9 — its hypotenuse is not 1 more than either leg.
3. So (8, 15, 17) cannot be produced by this method.

AnswerYes — for example, (8, 15, 17) is a primitive triple this method cannot produce.

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Open NCERT p. 52Checked by computer

1. The diagonal is the hypotenuse of an isosceles right triangle with both equal sides 5 cm: diagonal² = 5² + 5² = 50.
2. √50 = 5√2.

AnswerDiagonal = 5√2 cm (≈7.07 cm).

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Triangle with sides 7 and 9

1. The right-angle mark is between the sides 7 and 9, so both are shorter sides and the missing side is the hypotenuse c.
2. c² = 7² + 9² = 49 + 81 = 130, so c = √130.

Answer√130 units

Triangle with sides 4 and 10

1. The right-angle mark is between the sides 4 and 10, so both are shorter sides and the missing side is the hypotenuse c.
2. c² = 4² + 10² = 16 + 100 = 116, so c = √116 = √(4 × 29) = 2√29.

Answer2√29 units

Triangle with sides 40 and 41

1. The side 41 faces the right angle, so it is the hypotenuse; 40 is a shorter side.
2. (missing side)² = 41² − 40² = 1681 − 1600 = 81, so the missing side is 9.

Answer9 units

Triangle with sides 10 and √200

1. The side √200 faces the right angle, so it is the hypotenuse; 10 is a shorter side.
2. (missing side)² = (√200)² − 10² = 200 − 100 = 100, so the missing side is 10.

Answer10 units

Triangle with sides 10 and √150

1. The right-angle mark is between the sides 10 and √150, so both are shorter sides and the missing side is the hypotenuse c.
2. c² = 10² + (√150)² = 100 + 150 = 250, so c = √250 = √(25 × 10) = 5√10.

Answer5√10 units

Triangle with sides 27 and 45

1. The side 45 faces the right angle, so it is the hypotenuse; 27 is a shorter side.
2. (missing side)² = 45² − 27² = 2025 − 729 = 1296, so the missing side is 36.

Answer36 units

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Open NCERT p. 53Checked by computer

1. The diagonals of a rhombus bisect each other at right angles.
2. This creates four congruent right triangles, each with legs equal to half of each diagonal.
3. Half-diagonal 1: 24/2 = 12 units
4. Half-diagonal 2: 70/2 = 35 units
5. Each side of the rhombus is the hypotenuse of a right triangle with legs 12 and 35.
6. Using Baudhāyana's Theorem: s² = 12² + 35² = 144 + 1225 = 1369
7. s = √1369 = 37 units

AnswerThe sidelength of the rhombus is 37 units.

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Open NCERT p. 53One way to think about it

1. In a right triangle, let the two legs be a and b, and the hypotenuse be c.
2. By Baudhāyana's Theorem: a² + b² = c²
3. Since a and b are positive, we have a² + b² > a² and a² + b² > b²
4. Therefore c² > a² and c² > b²
5. Taking positive square roots: c > a and c > b
6. Thus, the hypotenuse is always the longest side of a right triangle.

In shortYes, the hypotenuse is always the longest side of a right triangle. This follows directly from the Baudhāyana-Pythagoras theorem: since c² = a² + b² where a and b are positive, we have c² > a² and c² > b², which means c > a and c > b.

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1. Take any Baudhāyana triple (a, b, c), so a² + b² = c². Let f be the highest common factor of a, b and c.
2. If f = 1, the three numbers have no common factor greater than 1, so the triple is primitive.
3. If f > 1, divide each number by f. Dividing a² + b² = c² by f² gives (a/f)² + (b/f)² = (c/f)², so (a/f, b/f, c/f) is again a Baudhāyana triple of whole numbers.
4. This new triple has no common factor greater than 1: if its three numbers shared a factor g > 1, then a, b and c would share the factor f × g, which is bigger than their highest common factor f.
5. So (a/f, b/f, c/f) is primitive, and (a, b, c) is this primitive triple scaled by f. For example, (9, 12, 15) = 3 × (3, 4, 5).

AnswerTrue. Every Baudhāyana triple is either primitive or a scaled version of a primitive triple: divide it by the highest common factor of its three numbers.

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Open NCERT p. 53Checked by computerAnswers can differ: one example

1. (3, 4, 5): sides 3 and 4, diagonal 5, since 3² + 4² = 5².
2. (6, 8, 10): sides 6 and 8, diagonal 10, since 6² + 8² = 10².
3. (5, 12, 13): sides 5 and 12, diagonal 13, since 5² + 12² = 13².
4. (8, 15, 17): sides 8 and 15, diagonal 17, since 8² + 15² = 17².
5. (7, 24, 25): sides 7 and 24, diagonal 25, since 7² + 24² = 25².

AnswerRectangles (3,4,5), (6,8,10), (5,12,13), (8,15,17) and (7,24,25) — sidelengths and diagonal, all integers.

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Open NCERT p. 53One way to think about it

1. The required area is 7² − 5² = 49 − 25 = 24 sq. units, so we need a square whose side, squared, is 24.
2. Draw a segment AB of length 5 units. At B, draw a line perpendicular to AB.
3. With centre A and radius 7 units, draw an arc cutting that perpendicular line at C. Triangle ABC has its right angle at B, so AC = 7 units is its hypotenuse and AB = 5 units is a shorter side.
4. By Baudhāyana's theorem, AB² + BC² = AC², so BC² = 7² − 5² = 49 − 25 = 24.
5. Construct a square on BC. Its area is BC² = 24 sq. units, the difference of the areas of the two squares. (Its side is √24 = 2√6 units, between 4.8 and 4.9 units, since 4.8² = 23.04 and 4.9² = 24.01.)

In shortMake a right triangle with a shorter side of 5 units and hypotenuse 7 units (right angle at one end of the 5-unit side, then an arc of radius 7 from its other end). The third side is √24 units, and the square on it has area 7² − 5² = 24 sq. units.

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(i) can you create a square that has an area of

1. Go from one corner of the square to the next by a steps across and b steps up (whole numbers). That side is the hypotenuse of a right triangle with shorter sides a and b, so by Baudhāyana's theorem the square's area is (side)² = a² + b². The next side goes b steps back and a steps up, so all four corners land on dots.
2. (a) 1 across and 1 up: area = 1² + 1² = 2. A tilted square of area 2 can be drawn.
3. (b) Area 3 would need a² + b² = 3. Using only 0 and 1 gives 0, 1 or 2, and any number 2 or more has a square of at least 4. So no square of area 3 can be drawn.
4. (c) 2 across and 0 up: an upright square of side 2, area 2² + 0² = 4.
5. (d) 2 across and 1 up: area = 2² + 1² = 5. A tilted square of area 5 can be drawn.

Answer(a) Yes (b) No (c) Yes (d) Yes.

(ii) Suppose the grid extends indefinitely.

1. Every square with its corners on dots has area a² + b² for some whole numbers a and b, and every such number can be drawn by going a across and b up.
2. So the possible areas are exactly the numbers that are a sum of two squares (one of them may be 0): 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, …
3. Numbers such as 3, 6, 7, 11, 12, 14 and 15 are not a sum of two squares, so they cannot be areas. Not every whole number is possible.

AnswerNo, not every whole number. The possible areas are the numbers a² + b² with whole numbers a and b: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, …; areas such as 3, 6, 7 and 11 are impossible.

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1. Let the triangle be ABC with every side 6 units. Draw the altitude AD from A, perpendicular to BC.
2. Triangles ABD and ACD both have a right angle at D, their hypotenuses AB and AC are both 6, and they share the side AD. By Baudhāyana's theorem, BD² = 6² − AD² and CD² = 6² − AD², so BD = CD. The altitude bisects BC, and BD = CD = 3 units.
3. In triangle ABD: AD² = AB² − BD² = 6² − 3² = 36 − 9 = 27, so the height AD = √27 = 3√3 units.
4. Area = ½ × base × height = ½ × 6 × 3√3 = 9√3 sq. units.

AnswerArea = 9√3 sq. units (about 15.6 sq. units).

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