Questions: The Baudhayana-Pythagoras Theorem

Q1. What question did Baudhayana ask, and what was his answer?

Answer (English): Baudhayana asked how to build a square holding twice the area of a given one. His verse 1.9 says that building a square on the diagonal gives one with twice the area, so draw the square on the diagonal of the given square. Doubling the side does not work, because that multiplies the area by four.

Hinglish: Baudhayana ne poochha ki diye gaye square ka double area wala square kaise banaye. Unka jawab tha ki square ki diagonal par naya square banao, kyunki diagonal wala square original ka do guna area rakhta hai. Side ko double karna kaam nahi karta, kyunki usse area chaar guna ho jata hai.

Q2. Why is the area of the square on the diagonal double the original area?

Answer (English): The diagonal cuts the original square into two congruent isosceles right triangles. The square drawn on the diagonal contains four triangles congruent to those two. Two small triangles make the original square and four make the new square, so the new area is double.

Hinglish: Diagonal original square ko do barabar isosceles right triangles mein kaat deta hai. Diagonal par bane square mein aise hi char triangles aate hain. Original square do triangles se banta hai aur naya square char se, isliye naya area double hota hai. Extra horizontal aur vertical lines se char triangles ke barabar hone ki pushti ho jati hai.

Q3. Why is √2 not a fraction?

Answer (English): Suppose √2 = m/n with whole numbers m and n. Squaring gives 2n² = m². In a perfect square every prime appears an even number of times. The left side has the prime 2 an odd number of times, one from the factor 2 plus an even count from n², while the right side has it an even number of times. That is impossible, so no such fraction exists. Euclid gave this proof in the Elements.

Hinglish: Maan lo √2 = m/n, jahan m aur n poore numbers hain. Squaring se 2n² = m² milta hai. Perfect square mein har prime even number of times aata hai. Left side mein 2 odd times hai, right side mein even times, jo possible nahi. Isliye aisa fraction nahi ho sakta. Yeh proof Euclid ne apni book Elements mein diya tha.

Q4. State the Baudhayana-Pythagoras theorem and give one use.

Answer (English): In a right-angled triangle with legs a and b and hypotenuse c, a² + b² = c². One use is to find a distance that cannot be measured directly. For a 3 cm by 4 cm right triangle, 3² + 4² = 25, so the hypotenuse is 5 cm. It turns a geometry question into a calculation.

Hinglish: Right triangle mein agar perpendicular sides a aur b hain aur hypotenuse c hai, to a² + b² = c². Iska ek use hai aisi doori nikaalna jo seedha naapni mushkil ho. 3 cm aur 4 cm wale right triangle mein 9 + 16 = 25, isliye hypotenuse 5 cm. Theorem geometry ke sawaal ko ek chhote calculation mein badal deta hai.

Q5. How do you find a leg when the hypotenuse and the other leg are known?

Answer (English): Rearrange the theorem as a² = c² - b² and then take the square root. If the hypotenuse is 13 and one leg is 5, the other leg is √(169 - 25) = √144 = 12. The hypotenuse is the longest side, so it takes the place of c.

Hinglish: Theorem ko a² = c² - b² likh kar square root le lo. Agar hypotenuse 13 aur ek leg 5 hai to doosri leg √(169 - 25) = √144 = 12 hoti hai. Hypotenuse hamesha sabse badi side hoti hai, isliye use c ki jagah rakhna hota hai, jodne wali jagah par nahi.

Q6. What is a Baudhayana triple, and what does primitive mean?

Answer (English): A Baudhayana triple is a triple of positive integers (a, b, c) with a² + b² = c². Examples are (3, 4, 5), (8, 15, 17) and (5, 12, 13). A triple is primitive when its greatest common factor is 1. So (3, 4, 5) is primitive, while (6, 8, 10) is not, because it is 2 × (3, 4, 5).

Hinglish: Baudhayana triple positive integers ka aisa group (a, b, c) hai jisme a² + b² = c² ho, jaise (3,4,5) aur (5,12,13). Agar teeno numbers ka common factor 1 ho to triple primitive kehlata hai. Isliye (3,4,5) primitive hai, lekin (6,8,10) nahi, kyunki wo (3,4,5) ka double hai.

Q7. If (a, b, c) is a triple, why is (ka, kb, kc) also a triple?

Answer (English): Multiply the equation a² + b² = c² by k². The left side becomes k²a² + k²b² = (ka)² + (kb)², and the right side becomes k²c² = (kc)². The equation still holds, so (ka, kb, kc) satisfies the theorem and is a triple. This is why the (3, 4, 5) triple gives the infinite family (3k, 4k, 5k).

Hinglish: a² + b² = c² ko k² se multiply karo. Left side (ka)² + (kb)² ban jata hai aur right side (kc)². Equation phir bhi sach hai, isliye (ka, kb, kc) bhi triple hai. Isi wajah se (3,4,5) se (3k, 4k, 5k) wali infinite family milti hai.

Q8. How are new triples generated from odd squares?

Answer (English): Use the identity (n - 1)² + (2n - 1) = n². When the nth odd number 2n - 1 is itself a perfect square m², the identity becomes (n - 1)² + m² = n², which is a triple (n - 1, m, n). For example 9 = 2 × 5 - 1 is an odd square, so n = 5 and the triple is (4, 3, 5). The next odd square 25 gives n = 13 and the triple (12, 5, 13).

Hinglish: Identity (n - 1)² + (2n - 1) = n² use karo. Jab nth odd number 2n - 1 khud perfect square m² ho, to (n - 1)² + m² = n² ban jata hai, yaani triple (n - 1, m, n). Jaise 9 = 2 × 5 - 1 odd square hai, isliye n = 5 aur triple (4,3,5). Agla odd square 25 se n = 13 aur triple (12,5,13) milta hai.

Q9. What is Fermat's Last Theorem, and who proved it?

Answer (English): Fermat's Last Theorem says that for n greater than 2, the equation xⁿ + yⁿ = zⁿ has no positive integer solutions. Fermat wrote in a book margin that he had a proof, but none was found for over three centuries. Andrew Wiles, who had read about the problem at age ten, proved it in 1994.

Hinglish: Fermat's Last Theorem kehti hai ki n > 2 ke liye xⁿ + yⁿ = zⁿ ka koi positive integer solution nahi hota. Fermat ne kitaab ke margin mein likha tha ki unke paas proof hai, lekin teen sau saal tak koi proof nahi mila. Andrew Wiles ne, jo das saal ki umar mein is problem ke baare mein padh chuke the, 1994 mein ise prove kiya.

Q10. Use the theorem to find the depth of water in the lotus problem.

Answer (English): The stem stands 1 unit above the water and its tip touches the water 3 units from the base. If the depth is x, the stem is x + 1 units long, and the base, the tip and the new touch point form a right triangle with legs 3 and x and hypotenuse x + 1. Then 9 + x² = x² + 2x + 1, so 8 = 2x and x = 4. The water is 4 units deep.

Hinglish: Lotus ka stem paani se 1 unit upar hai aur jhukne par tip base se 3 units door paani ko chhooti hai. Agar depth x hai to stem x + 1 lambi hai, aur base, tip aur touch point se right triangle banta hai. Isliye 9 + x² = (x + 1)², jisse 9 = 2x + 1 aur x = 4. Paani ki depth 4 units hai.

Quick check

You have revised the theorem and its applications. The quiz below has five multiple choice questions on the theorem, triples and the constructions from the chapter. Answer without checking the notes, then read each explanation, especially for any question you missed. Aim for at least three correct answers.