CBSE Class 8 Mathematics The Baudhayana Pythagoras Theorem Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. The figure shows a square of sidelength 4 units together with the square drawn on one of its diagonals, exactly as Baudhāyana describes in Śulba-Sūtra Verse 1.9.
(i) A student says, “To double the area, double the sidelength.” Show with the given square why this is wrong.
(ii) Explain, by drawing east–west and north–south lines through the figure, why the square on the diagonal has double the area of the original square.
(iii) Find the area of the square on the diagonal and the exact length of the diagonal. Give bounds for this length with one digit after the decimal point.
Answer:
(i) The given square has side 4 units, so its area is 4 × 4 = 16 sq. units. Doubling the side gives side 8 units and area 8 × 8 = 64 sq. units.
64 = 4 × 16, so the area becomes four times, not two times. Area is side × side, so multiplying the side by 2 multiplies the area by 2 × 2 = 4.
(ii) Draw the east–west (horizontal) line along the top side and the north–south (vertical) line along the left side of the square, extended through the dashed square.

The original square is cut by its diagonal into two small triangles (triangle 1 and the lower-right half). The dashed square on the diagonal is cut by these lines into four such triangles (1, 2, 3, 4).
All these triangles are congruent, because each has two perpendicular sides of 4 units with the right angle between them. Since 4 triangles is twice 2 triangles, the new square has exactly double the area.
(iii) Area of the square on the diagonal = 2 × 16 = 32 sq. units.
If d is the diagonal, d² = 32, so d = √32 = 4√2 units.
5.6² = 31.36 and 5.7² = 32.49, and 31.36 < 32 < 32.49. Hence 5.6 < d < 5.7.
In simple words: Doubling the side gives 4 times the area, but the square on the diagonal is made of 4 triangles instead of 2, so its area is 32, double 16.

Teacher's Note:
1. Counting congruent triangles is an easy way to compare areas.
2. The diagonal √32 is not a whole number, but its square is.

Question 2. A square paper of sidelength 10 cm is folded inwards so that each crease passes through the midpoints of the sides, leaving an inner figure PQRS.
(i) Explain why PQRS is a square.
(ii) Explain why its area is half the area of the original paper.
(iii) Find the area and the exact sidelength of PQRS.
(iv) Will a square of half the sidelength have half the area? How many such squares are needed to fill the original paper?
Answer:
(i) The four creases join the midpoints of the sides and cut off four corner triangles. Each is a right triangle with both shorter sides 5 cm, so the four triangles are congruent and the four sides of PQRS are equal.
In each corner triangle the two acute angles are 45° each, so at every vertex of PQRS the angle left is 180° − 45° − 45° = 90°. Equal sides and right angles make PQRS a square.
(ii) Each corner triangle has area \( \frac{1}{2} \) × 5 × 5 = 12.5 cm², and four of them give 50 cm². The paper is 100 cm², so PQRS takes the other 50 cm², which is exactly half.
(iii) Area of PQRS = 50 cm². If s is its side, s² = 50, so s = √50 = 5√2 cm, which lies between 7.0 cm and 7.1 cm (7.0² = 49, 7.1² = 50.41).
(iv) No. A square of side 5 cm has area 25 cm², which is one-quarter of 100 cm². Four such squares are needed to fill the original paper.
In simple words: The midpoint square PQRS has area 50 cm² and side 5√2 cm, while a square of half the side has only a quarter of the area.

Teacher's Note:
1. Equal sides alone are not enough; the right angles must be shown too.
2. Half the area and half the side are different things.

Question 3. Let an isosceles right triangle have equal sides of length a and hypotenuse of length c.
(i) Using the square built on the hypotenuse and the square built on a side, derive the relation c² = 2a².
(ii) Use it to find the hypotenuse when a = 10 units, giving the exact value and bounds with one digit after the decimal point.
(iii) Use it to find the equal sides when c = √162 units.
Answer:
(i) Two copies of the triangle, placed together along the hypotenuse, form a square of side a and area a². The hypotenuse is the diagonal of this square.
By Baudhāyana’s rule, the square on the diagonal has double the area of the square. The square on the hypotenuse has area c × c = c². Therefore c² = 2 × a² = 2a².
(ii) c² = 2 × 10² = 200, so c = √200 = 10√2 units.
14.1² = 198.81 and 14.2² = 201.64. Since 198.81 < 200 < 201.64, 14.1 < c < 14.2.
(iii) c² = 162. Using c² = 2a²: 162 = 2a², so a² = \( \frac{162}{2} \) = 81 and a = √81 = 9 units.
In simple words: The square on the hypotenuse is double the square on a side, so c² = 2a²; this gives c = 10√2 for a = 10 and a = 9 for c = √162.

Teacher's Note:
1. The isosceles right triangle is half of a square.
2. Use the relation forwards and backwards.

Question 4. Baudhāyana’s Śulba-Sūtra (Verse 1.13) lists several triples of whole numbers that give the sides of a right triangle.
(i) Define a Baudhāyana triple and write any four triples listed in the chapter.
(ii) Verify by calculation that (12, 35, 37) is a Baudhāyana triple.
(iii) Define a primitive triple. Show that (9, 12, 15) is not primitive and name the primitive triple it comes from.
Answer:
(i) A Baudhāyana triple is a triple (a, b, c) of positive integers that can be the sidelengths of a right triangle, that is, a² + b² = c² with c the hypotenuse.
Four triples from the Śulba-Sūtra: (3, 4, 5), (5, 12, 13), (8, 15, 17) and (7, 24, 25). They are also called Baudhāyana–Pythagoras triples or Pythagorean triples.
(ii) 12² = 144 and 35² = 1225, so 12² + 35² = 1369. Also 37² = 1369.
The two sides are equal, so (12, 35, 37) is a Baudhāyana triple.
(iii) A triple is primitive if its three numbers have no common factor greater than 1.
In (9, 12, 15) all three numbers are divisible by 3, so it is not primitive. Dividing by 3 gives (3, 4, 5), and 3² + 4² = 25 = 5².
So (9, 12, 15) is the scaled version of the primitive triple (3, 4, 5), with k = 3.
In simple words: A Baudhāyana triple satisfies a² + b² = c²; (12, 35, 37) is one, and (9, 12, 15) = 3 × (3, 4, 5) is not primitive.

Teacher's Note:
1. Verify a triple by squaring and adding.
2. Divide by the common factor to find the primitive triple.

Question 5. The figure shows a right-angled triangle with squares drawn on all three of its sides.
(i) State Baudhāyana’s rule (Verse 1.12) for combining two different squares into one, in your own words.
(ii) Describe the construction: given two squares of sides a and b, how do you obtain the square whose area is the sum of their areas?
(iii) Apply the construction to squares of sides 3 cm and 4 cm: find the area and the sidelength of the combined square.
(iv) Explain how this agrees with the earlier result for two squares of the same size.
Answer:
(i) The square built on the diagonal (hypotenuse) of a right-angled figure has area equal to the sum of the areas of the squares built on its two perpendicular sides. So two different squares can always be replaced by one square of the same total area.
(ii) Draw a right-angled triangle whose two perpendicular sides have lengths a and b. Build a square on its hypotenuse. That square has area equal to a² + b².
(Verse 2.1 gives a practical way: join the two squares, mark off a rectangular part of the larger square using a side of the smaller one, and take the diagonal of that rectangle as the required side.)
(iii) Areas are 9 cm² and 16 cm². The combined square has area 9 + 16 = 25 cm², so its side is √25 = 5 cm, the hypotenuse of the 3–4–5 right triangle.
(iv) If a = b, the rule gives c² = a² + a² = 2a², exactly the earlier result for doubling a square, where the new side is the diagonal of the given square. So the general rule contains the special case.
In simple words: The square on the hypotenuse equals the two smaller squares together: 9 + 16 = 25, so the combined square has side 5 cm.

Teacher's Note:
1. In the figure, the big square on c has the same area as the two blue squares on a and b together.
2. Doubling a square is the case a = b.

Question 6. Use the Baudhāyana–Pythagoras theorem to solve the following.
(i) Find the exact length of the diagonal of a square of sidelength 15 units, and give bounds for it with one digit after the decimal point.
(ii) Find the diagonal of a rectangular field measuring 36 m by 77 m.
(iii) Find the sidelength of a rhombus whose diagonals are 14 units and 48 units.
(iv) In each part, state clearly which lengths act as the two shorter sides and which as the hypotenuse.
Answer:
(i) The two sides meet at a right angle, so the diagonal is the hypotenuse: d² = 15² + 15² = 450, so d = √450 = 15√2 units.
21.2² = 449.44 and 21.3² = 453.69, so 21.2 < d < 21.3.
(ii) d² = 36² + 77² = 1296 + 5929 = 7225, so d = √7225 = 85 m. The triple used is (36, 77, 85).
(iii) The diagonals bisect each other at right angles, so the half-diagonals are \( \frac{14}{2} \) = 7 and \( \frac{48}{2} \) = 24 units. Side = √(7² + 24²) = √(49 + 576) = √625 = 25 units.
(iv) In (i) the two sides of the square, each 15, are the shorter sides and the diagonal is the hypotenuse.
In (ii) the length 77 and the breadth 36 are the shorter sides and the diagonal 85 is the hypotenuse.
In (iii) the half-diagonals 7 and 24 are the shorter sides and the side of the rhombus, 25, is the hypotenuse.
In simple words: The square’s diagonal is 15√2 (between 21.2 and 21.3), the field’s diagonal is 85 m, and the rhombus side is 25 units.

Teacher's Note:
1. Find the right angle first, then the hypotenuse is opposite it.
2. 85² = 7225 can be checked as 85 × 85.

Question 7. The chapter generates Baudhāyana triples from odd square numbers.
(i) Write the rule connecting the sum of the first n odd numbers with n², and hence obtain (n − 1)² + (2n − 1) = n².
(ii) Taking the odd square 121, find n and write down the triple produced. Verify it.
(iii) Repeat with the odd square 81.
(iv) In every triple produced this way, how are the larger shorter side and the hypotenuse related? What does this tell you about whether such triples are primitive?
Answer:
(i) The sum of the first n odd numbers is n², and the nth odd number is 2n − 1: 1 + 3 + 5 + … + (2n − 3) + (2n − 1) = n².
The terms up to (2n − 3) are the first n − 1 odd numbers, whose sum is (n − 1)². Hence (n − 1)² + (2n − 1) = n².
If 2n − 1 is itself a square, we get a square plus a square equal to a square: a Baudhāyana triple.
(ii) 2n − 1 = 121 gives n = 61. Then 60² + 11² = 61². Check: 3600 + 121 = 3721 = 61². The triple is (11, 60, 61).
(iii) 2n − 1 = 81 gives n = 41. Then 40² + 9² = 41². Check: 1600 + 81 = 1681 = 41². The triple is (9, 40, 41).
(iv) The larger shorter side is n − 1 and the hypotenuse is n, so they are consecutive integers and differ by 1.
Any common factor of the three numbers must divide n − (n − 1) = 1, so it can only be 1. Hence this method always gives primitive triples.
In simple words: Each odd square 2n − 1 gives the triple with sides √(2n − 1), n − 1 and n, such as (11, 60, 61) and (9, 40, 41), and these are always primitive.

Teacher's Note:
1. Two consecutive numbers never share a factor bigger than 1.
2. This method misses triples like (8, 15, 17).

Question 8. Read the following and answer.
(i) Which property of square numbers made Fermat ask his famous question?
(ii) Write the equation Fermat considered and state exactly what he claimed about it.
(iii) What did Fermat write in the margin of his book, and why is the statement called Fermat’s Last Theorem?
(iv) Describe briefly how Andrew Wiles came to prove it.
Answer:
(i) Baudhāyana triples show that there are infinitely many squares that are the sum of two squares, for example 3² + 4² = 5² and 5² + 12² = 13². Fermat, a 17th-century French mathematician, wondered whether the same could happen with cubes, fourth powers and so on.
(ii) He considered xⁿ + yⁿ = zⁿ with x, y, z natural numbers, and claimed that it has no solution at all when n > 2.
(iii) In the margin of a book on properties of positive integers he wrote that he had found “a truly marvellous proof” but the margin was too small to contain it. No proof by him was ever found, and as the last of his claims left unsettled, it came to be called Fermat’s Last Theorem.
(iv) In 1963, ten-year-old Andrew Wiles read Eric Bell’s book The Last Problem about this puzzle and its 300 years of failed attempts. He decided to prove it, and finally succeeded in 1994.
In simple words: Squares can add up to a square, so Fermat claimed higher powers never can; he left no proof, and Andrew Wiles finally proved it in 1994.

Teacher's Note:
1. n = 2 has many solutions; n > 2 has none.
2. Wiles’s story shows the power of a childhood question.

Question 9. The figure shows a square of side b and a smaller square of side a joined along a line, with two cuts drawn. The two cuts separate the figure into three pieces, and these three pieces can be rearranged to form a single square.
(i) Show that each of the two cut-off triangles has perpendicular sides a and b, and that both cuts have the same length c.
(ii) Explain why, when the three pieces are rearranged, the four-sided figure obtained has all four sides equal to c and all four angles equal to 90°. (Use the fact that the two acute angles of a right triangle add up to 90°.)
(iii) By comparing the total area before and after the rearrangement, prove that a² + b² = c².
(iv) Why is this theorem called the Baudhāyana–Pythagoras theorem?
Answer:
(i) The marked point is at distance b from the left end of the base. Cut 1 runs from the top-left corner of the small square to the marked point: its horizontal part is b and its vertical part is a, so it cuts off a right triangle with perpendicular sides a and b.
Cut 2 runs from the marked point to the top-right corner of the big square: its horizontal part is (a + b) − b = a and its vertical part is b, giving a second right triangle with perpendicular sides a and b.
The two triangles are congruent, so both cuts have the same length; call it c.
(ii) The pieces are placed so that every side of the new figure is the hypotenuse of a copy of the same right triangle, so all four sides equal c.
At each corner, one acute angle x of a triangle meets the other acute angle 90° − x of a congruent triangle along a straight line. The corner angle is 180° − (x + 90° − x) = 180° − 90° = 90°. So the figure is a square of side c and area c².
(iii) Before cutting, the figure is the two squares joined, with total area a² + b². Cutting and moving pieces does not change the total area, and after rearranging they exactly cover a square of area c².
Therefore a² + b² = c², where c is the hypotenuse of a right triangle with shorter sides a and b.
(iv) Baudhāyana stated this result in his Śulba-Sūtra around 800 BCE, the first time in history it appears in this general form. The Greek mathematician Pythagoras (c. 500 BCE), about two centuries later, also studied it. The joint name honours both.
In simple words: The two cuts are hypotenuses of right triangles with legs a and b; rearranging the three pieces gives a square of side c, so a² + b² = c².

Teacher's Note:
1. Cutting and rearranging keeps the area the same.
2. The angle fact x + (90° − x) = 90° gives the right angles.

Question 10. The dots of a square grid are one unit apart, and the grid extends as far as you like in every direction.
(i) Explain how a tilted square can be drawn with its vertices on grid dots, and show that a tilt of 1 right and 1 up gives a square of area 2 square units.
(ii) Find the area of the square whose side goes 1 right and 2 up. Hence write the general rule for the area of such a tilted square.
(iii) Using your rule, explain why a square of area 3 square units cannot be drawn on this grid, while squares of area 4 and 5 square units can.
(iv) Construct a square whose area equals the difference of the areas of squares of sides 7 units and 5 units. State its exact sidelength and give bounds with one digit after the decimal point.
Answer:
(i) Start at a dot and reach the next vertex by going 1 right and 1 up. Repeating the same move, turned a quarter turn each time, brings you back to the start, so all four vertices land on dots and the figure is a square.
Its side is the hypotenuse of a right triangle with shorter sides 1 and 1, so side² = 1² + 1² = 2. Its area is 2 square units. (Check: it fits in a 2 × 2 box of area 4, minus four corner triangles of area \( \frac{1}{2} \) each: 4 − 2 = 2.)
(ii) For a side going 1 right and 2 up, area = 1² + 2² = 5 square units.
In general, if the side goes a dots across and b dots up, the area of the square is a² + b².
(iii) Area 4: take a = 2, b = 0, an ordinary upright square of side 2. Area 5: take a = 1, b = 2.
Area 3 is not possible: the only squares not more than 3 are 0 and 1, and 0 + 0, 0 + 1, 1 + 1 give 0, 1, 2, never 3. So no whole numbers a, b give a² + b² = 3.
(iv) Required area = 7² − 5² = 49 − 25 = 24 sq. units. Draw a right triangle with hypotenuse 7 units and one shorter side 5 units; its third side t satisfies t² + 25 = 49, so t² = 24. The square built on this third side is the required one.
Its side is √24 = 2√6 units. Since 4.8² = 23.04 and 4.9² = 24.01, the side lies between 4.8 and 4.9 units.
In simple words: A tilted grid square with side going a across and b up has area a² + b², so areas 2, 4, 5 are possible but 3 is not; the difference square has area 24 and side √24.

Teacher's Note:
1. Every tilted grid square is a Baudhāyana picture.
2. For a difference of squares, use the hypotenuse as the bigger side.

Case Study Based Questions

Question 11. A ladder against a wall. A painter leans a straight ladder against the vertical wall of a house. The foot of the ladder rests on level ground and the wall meets the ground at a right angle. The ladder is 13 m long, and its foot is placed 5 m away from the wall. The painter wants to know exactly how high up the wall the top of the ladder reaches, because the window he must paint is at a height of 12 m. Later he shifts the foot of the same ladder further out, to a distance of 12 m from the wall, to reach a lower window.
(i) In the right-angled triangle formed by the wall, the ground and the ladder, which side is the hypotenuse?
(ii) Write the Baudhāyana–Pythagoras relation for this triangle.
(iii) How high up the wall does the top of the ladder reach in the first position?
(iv) How high does it reach after the foot is shifted to 12 m from the wall?
Answer:
(i) The ladder is the hypotenuse, because it lies opposite the right angle where the wall meets the ground.
(ii) If h is the height reached and d the distance of the foot from the wall, then d² + h² = (length of ladder)².
(iii) 5² + h² = 13², so 25 + h² = 169, h² = 144 and h = 12 m. The top reaches exactly the window at 12 m.
(iv) 12² + h² = 13², so 144 + h² = 169, h² = 25 and h = 5 m. The sides (5, 12, 13) form a Baudhāyana triple, which is why both answers are whole numbers.
In simple words: The 13 m ladder is the hypotenuse; with its foot 5 m out it reaches 12 m high, and with its foot 12 m out it reaches 5 m.

Teacher's Note:
1. The ladder length stays the same; only the two legs swap.
2. Moving the foot out lowers the top.

Question 12. The size of a television screen. Television screens are always described by the length of their diagonal, never by their length and breadth. A shopkeeper has two rectangular screens on display. The first screen measures 48 cm by 20 cm. The second screen measures 48 cm by 36 cm. A customer wants the screen with the longer diagonal, but the price labels show only the length and the breadth of each screen. The shopkeeper explains that the two sides of a rectangular screen meet at a right angle, so the diagonal can be worked out from the two sides.
(i) Why does the diagonal of a rectangular screen act as the hypotenuse of a right-angled triangle?
(ii) Find the diagonal of the first screen.
(iii) Find the diagonal of the second screen.
(iv) Which screen should the customer buy, and by how many centimetres is its diagonal longer?
Answer:
(i) The length and breadth of a rectangle meet at a right angle. The diagonal joins the opposite corners, so it is the side opposite that right angle, the hypotenuse of a right triangle whose shorter sides are the length and the breadth.
(ii) d² = 48² + 20² = 2304 + 400 = 2704, so d = √2704 = 52 cm.
(iii) d² = 48² + 36² = 2304 + 1296 = 3600, so d = √3600 = 60 cm.
(iv) The second screen, as 60 cm > 52 cm. Its diagonal is longer by 60 − 52 = 8 cm.
In simple words: The diagonals are 52 cm and 60 cm, so the second screen is bigger by 8 cm.

Teacher's Note:
1. (48, 20, 52) = 4 × (12, 5, 13) and (48, 36, 60) = 12 × (4, 3, 5).
2. Compare diagonals, not just one side.

Question 13. Running across a ground. A school ground is rectangular, measuring 80 m along one side and 150 m along the other. During a fitness test, Meera starts at one corner of the ground and runs along two adjacent sides to reach the opposite corner. Ravi starts at the same corner at the same moment but runs straight across the ground along the diagonal to the same opposite corner. Both run at the same steady speed, and the ground is flat, so the diagonal path is a straight line.
(i) What distance does Meera cover?
(ii) Find the length of the diagonal that Ravi runs.
(iii) How much shorter is Ravi’s path than Meera’s?
(iv) The three numbers 80, 150 and your answer to (ii) form a Baudhāyana triple. Which primitive triple is it a scaled version of?
Answer:
(i) Meera runs 80 + 150 = 230 m.
(ii) The diagonal is the hypotenuse: d² = 80² + 150² = 6400 + 22500 = 28900, so d = √28900 = 170 m.
(iii) Ravi’s path is shorter by 230 − 170 = 60 m.
(iv) Dividing 80, 150 and 170 by 10 gives (8, 15, 17), and 8² + 15² = 289 = 17². So (80, 150, 170) is the scaled version of (8, 15, 17) with k = 10.
In simple words: Meera runs 230 m, Ravi runs the 170 m diagonal, saving 60 m; the triple is 10 × (8, 15, 17).

Teacher's Note:
1. The straight diagonal is always shorter than two sides.
2. Remove the common factor 10 to spot the triple.

Question 14. Flying a kite. On a windy afternoon, Arun flies a kite from an open field. He lets out the full 100 m of his string, and the string is pulled perfectly straight and tight by the wind. His friend Sana stands directly below the kite and measures the distance from Arun’s hand to her feet as 60 m along the level ground. Arun holds the string at ground level. The kite, Arun’s hand and Sana’s feet are the three corners of a right-angled triangle, with the right angle at Sana’s feet.
(i) Which length in this situation is the hypotenuse?
(ii) At what height above the ground is the kite flying?
(iii) If Arun pulls in the string until only 80 m is out, and the horizontal distance becomes 48 m, find the new height of the kite.
(iv) Name the primitive Baudhāyana triple behind both answers.
Answer:
(i) The string is the hypotenuse, since it lies opposite the right angle at Sana’s feet.
(ii) 60² + h² = 100², so 3600 + h² = 10000, h² = 6400 and h = 80 m.
(iii) 48² + h² = 80², so 2304 + h² = 6400, h² = 4096 and h = 64 m.
(iv) (60, 80, 100) = 20 × (3, 4, 5) and (48, 64, 80) = 16 × (3, 4, 5). Both come from (3, 4, 5).
In simple words: The string is the hypotenuse; the kite is 80 m high at first and 64 m later, both from the (3, 4, 5) triple.

Teacher's Note:
1. Subtract squares to find a leg.
2. 64 × 64 = 4096.

Question 15. A ramp at the library. A library has its entrance 0.7 m above the level of the footpath. To make the entrance accessible, a straight ramp is built from a point on the footpath to the top of the entrance step. The horizontal distance from the foot of the ramp to the base of the step is 2.4 m. The ramp, the vertical rise and the horizontal run form a right-angled triangle. The carpenter must cut a wooden plank for the sloping surface and also fix a steel rod along the horizontal run, so he needs both lengths exactly. A second, gentler ramp is later planned for the same rise of 0.7 m with a horizontal run of 3.5 m.
(i) Sketch the right-angled triangle and mark the rise, the run and the ramp.
(ii) Find the length of the first ramp.
(iii) The plank costs ₹300 per metre. Find the cost of the plank for the first ramp.
(iv) For the second ramp, is the slanting length more or less than 3.6 m? Justify by squaring.
Answer:
(i) The sketch is a right triangle with a vertical side of 0.7 m (the rise), a horizontal side of 2.4 m (the run) with the right angle between them, and the sloping ramp joining their ends.

(ii) L² = 0.7² + 2.4² = 0.49 + 5.76 = 6.25, so L = √6.25 = 2.5 m. (The sides are (7, 24, 25) divided by 10.)
(iii) Cost = 2.5 × 300 = ₹750.
(iv) L² = 0.7² + 3.5² = 0.49 + 12.25 = 12.74. Now 3.6² = 12.96, and 12.74 < 12.96, so the slanting length is less than 3.6 m.
In simple words: The first ramp is 2.5 m long and costs ₹750; the second ramp’s length squared is 12.74, less than 3.6² = 12.96.

Teacher's Note:
1. Decimals work in the theorem just like whole numbers.
2. Comparing squares avoids finding an exact square root.

Question 16. Śulba-Sūtra ropes on a field. Builders of Vedic altars, and surveyors even today, make an exact right angle on the ground without any instrument. They take a long rope and tie knots so that it is divided into three parts of lengths 3, 4 and 5 units. Three people hold the rope at the two ends and at the two knots and pull it into a triangle. The corner between the parts of lengths 3 and 4 then turns out to be exactly a right angle. A surveyor marking out a rectangular plot tries other sets of knot lengths to see which ones also work.
(i) Show by calculation why the 3–4–5 rope gives a right angle.
(ii) Will a rope knotted into parts of 6, 8 and 10 units work? Justify.
(iii) Will a rope knotted into parts of 5, 6 and 7 units work? Justify.
(iv) Write two more knot sets from Baudhāyana’s own list that the surveyor could use.
Answer:
(i) 3² + 4² = 9 + 16 = 25 = 5². So (3, 4, 5) is a Baudhāyana triple and the rope triangle is right-angled, with the right angle between the parts 3 and 4.
(ii) Yes. 6² + 8² = 36 + 64 = 100 = 10². (6, 8, 10) is (3, 4, 5) scaled by 2, so it also gives a right angle.
(iii) No. 5² + 6² = 25 + 36 = 61, but 7² = 49. Since 61 ≠ 49, the lengths do not form a Baudhāyana triple and the corner is not a right angle.
(iv) Any two from Baudhāyana’s list, for example (5, 12, 13) and (8, 15, 17). (He also lists (7, 24, 25), (12, 35, 37) and (15, 36, 39).)
In simple words: A rope works only when its parts form a Baudhāyana triple: 3–4–5 and 6–8–10 work, 5–6–7 does not.

Teacher's Note:
1. The right angle is always opposite the longest part.
2. Any scaled triple also works.

Question 17. A tree in the storm. During a storm, a tall straight tree standing on level ground snaps at a point above the ground. The upper part does not break off completely; it bends over and its top touches the ground at a distance of 15 m from the foot of the tree. The broken upper part, now acting as a slanting piece, measures 17 m from the point of the break to the top. The lower part of the trunk is still standing upright, making a right angle with the ground. A forest officer must record the full height the tree had before the storm.
(i) Which length in this picture is the hypotenuse of the right-angled triangle?
(ii) Find the height of the part of the trunk still standing.
(iii) What was the full height of the tree before it snapped?
(iv) If the top had touched the ground 6 m from the foot instead, with the same standing part, how long would the broken piece be?
Answer:
(i) The broken slanting piece of 17 m is the hypotenuse, because the standing trunk makes a right angle with the ground.
(ii) If h is the standing part, h² + 15² = 17², so h² = 289 − 225 = 64 and h = 8 m.
(iii) Full height = standing part + broken part = 8 + 17 = 25 m.
(iv) With standing part 8 m and ground distance 6 m, the broken piece p satisfies p² = 8² + 6² = 64 + 36 = 100, so p = 10 m.
In simple words: The 17 m broken piece is the hypotenuse, so the standing part is 8 m and the tree was 25 m tall.

Teacher's Note:
1. Add both parts for the original height.
2. (8, 15, 17) and (6, 8, 10) are both triples.

Question 18. Doubling a square altar. An old text describes a square fire-altar of area 9 square metres. A priest is asked to build a new square altar of exactly double the area, using only a rope and pegs. His first idea is to double each side of the old altar. His teacher stops him and quotes Baudhāyana: the diagonal of a square produces a square of double the area of the original square. So the priest stretches a rope along the diagonal of the old altar and uses that length as the side of the new altar.
(i) What is the sidelength of the old altar?
(ii) What area would the priest’s first idea have produced? Why was it wrong?
(iii) What is the area of the correct new altar, and what is the exact length of its side?
(iv) Give bounds for this sidelength with one digit after the decimal point.
Answer:
(i) Area = 9 m², so the sidelength is √9 = 3 m.
(ii) Doubling each side gives side 6 m and area 36 m². That is 4 times the old area, not 2 times, because area is side × side, so doubling the side multiplies the area by 2 × 2 = 4.
(iii) The correct altar has area 2 × 9 = 18 m². Its side is the diagonal of the old altar: (diagonal)² = 3² + 3² = 18, so the side is √18 = 3√2 m.
(iv) 4.2² = 17.64 and 4.3² = 18.49. Since 17.64 < 18 < 18.49, the side lies between 4.2 m and 4.3 m.
In simple words: The old altar has side 3 m; doubling the side would give 36 m², but the diagonal gives the correct 18 m² altar with side 3√2 m.

Teacher's Note:
1. This is exactly the problem that Verse 1.9 solves.
2. √18 is about 4.24.

Question 19. Wires for a flagpole. A vertical flagpole of height 24 m stands on flat ground. To keep it steady, a first steel wire is stretched tight from the top of the pole to a peg driven into the ground 7 m from the base of the pole. A second wire is stretched from a hook fixed 15 m up the pole to another peg 20 m from the base, on the opposite side. Wire is sold only in whole metres and costs ₹90 per metre. The engineer also has a spare wire of length exactly 26 m and wants to run it from the very top of the pole to a third peg on the ground.
(i) Find the length of the first wire and the length of the second wire.
(ii) Find the total cost of these two wires.
(iii) How far from the base of the pole must the third peg be driven so that the 26 m wire is exactly tight from the top?
(iv) Could a wire of length 20 m ever reach the top of the pole from a peg on the ground? Give a reason.
Answer:
(i) First wire: L₁² = 24² + 7² = 576 + 49 = 625, so L₁ = 25 m.
Second wire: L₂² = 15² + 20² = 225 + 400 = 625, so L₂ = 25 m. (The triples are (7, 24, 25) and (15, 20, 25) = 5 × (3, 4, 5).)
(ii) Total wire = 25 + 25 = 50 m. Cost = 50 × 90 = ₹4500.
(iii) Let the peg be d m from the base. 24² + d² = 26², so 576 + d² = 676, d² = 100 and d = 10 m.
(iv) No. The wire would be the hypotenuse, which is always the longest side of a right triangle (c² = a² + b² is greater than a², so c > a). A wire from the ground to the top must be longer than 24 m, and 20 m is shorter.
In simple words: Both wires are 25 m (₹4500 in all), the 26 m wire needs a peg 10 m away, and a 20 m wire is too short for a 24 m pole.

Teacher's Note:
1. Different right triangles can share the same hypotenuse.
2. The hypotenuse must be longer than the pole.

Question 20. A four-sided plot. A plot of land ABCD is to be fenced and levelled. The surveyor measures AB = 15 m, BC = 20 m, CD = 24 m and DA = 7 m, and finds that the corner at B is a right angle. He then stretches a rope from A to C across the plot and measures that length too. Looking at his four measurements and the length AC, he realises that the corner at D must also be a right angle, without measuring any angle at all. Fencing costs ₹250 per metre and levelling costs ₹40 per square metre.
(i) Find the length of the diagonal AC.
(ii) Show that the corner at D is also a right angle.
(iii) Find the total area of the plot.
(iv) Find the total cost of fencing the boundary and levelling the plot.
Answer:
(i) In △ABC the corner at B is a right angle, so AC is the hypotenuse. AC² = 15² + 20² = 225 + 400 = 625, so AC = 25 m.
(ii) In △ACD, DA = 7 m, CD = 24 m and AC = 25 m. Now 7² + 24² = 49 + 576 = 625 = 25².
So (7, 24, 25) is a Baudhāyana triple, and these lengths form a right triangle with AC as hypotenuse. The right angle is opposite AC, that is, at D.
(iii) Area of △ABC = \( \frac{1}{2} \) × 15 × 20 = 150 m². Area of △ACD = \( \frac{1}{2} \) × 7 × 24 = 84 m². Total area = 150 + 84 = 234 m².
(iv) Perimeter = 15 + 20 + 24 + 7 = 66 m. Fencing cost = 66 × 250 = ₹16,500. Levelling cost = 234 × 40 = ₹9,360.
Total cost = 16500 + 9360 = ₹25,860.
In simple words: AC = 25 m, and 7² + 24² = 25² makes the angle at D a right angle; the plot has area 234 m² and costs ₹25,860 in all.

Teacher's Note:
1. The converse: if the sides fit a² + b² = c², the triangle is right-angled.
2. Split the plot along AC into two right triangles.