CG Board Class 9 Mathematics Chapter 1 Solutions: History of Mathematics (गणित का इतिहास) – 2026-27

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Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 1 Solutions: History of Mathematics (गणित का इतिहास) in Bilingual Format (English & Hindi Medium).

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SECTION / EXERCISE

Exercise 1.1 / 1.1

Official Textbook Exercises

End-of-Chapter Exercises (Solutions)

Question 1

Bharti Krishnateertha: Vedic Mathematics Varahamihira: Panch…


📝 Question:

  1. Bharti Krishnateertha: Vedic Mathematics
  2. Varahamihira: Panch Siddhanta
  3. Brahmagupta: Brahmasphoot Siddhanta
  4. Bhaskaracharya: Siddhanta Shiromani
  5. Aaryabhatt: Aaryabhatiya
  6. The word used for the sky and zero was *Kham.
  7. Tallakshana was used to denote the number 1053.
  8. The alphanumeric system was used by the Mathematician Aaryabhatt in his book Aaryabhatiya.
  9. *Beej Ganit / Algebra was also known as Implicit Mathematics.
  10. Vedic Mathematics have the 16 Sutras.
  • English: The modern decimal number system uses 10 digits (0-9) where every digit has both a face value and a place value. This place value system makes computations simple, scientific, and globally applicable.
  • English:
  1. Concept of zero is found in ancient Vedas using the word ‘Kham’ (Yajurveda).
  2. Mentioned by Sanskrit grammarians Panini (500 B.C.), Pingal (200 B.C.), and Rishi Grutsamad.
  3. First physical symbol representing zero was found in the Bakhshali Manuscript (300-400 A.D.).
  • English: When letters or alphabets are used to represent numbers, it is called a coded/alphanumeric system. Aryabhata used this system extensively in Aaryabhatiya to express large numerical values concisely.
  • English: In a rectangle, the area of the square drawn on the diagonal is equal to the sum of the areas of the squares drawn on its two sides. In a right angled triangle, Hypotenuse2 = Base2 + Perpendicular2. Baudhayan described this theorem centuries before Pythagoras.
  • English: Aryabhata calculated that a circle of diameter 20000 has a circumference of approximately 62832. Thus, π = (62832)/(20000) = 3.1416, which is accurate to 4 decimal places.
  • English: Created by Swami Bharti Krishna Teertha (1884-1960 A.D.). The book contains 16 main Sutras and 13 Upsutras spread across 40 chapters for rapid mental calculations.
  • English: Obtaining a single digit number by repeatedly adding all digits of a number is called its Digit sum (Bijank).
  • Answer: 6, 2, 7, 9, 7, 3, 1, 2, 4, 4
  1. 15: 1 + 5 = 6
  2. 38: 3 + 8 = 11 → 1 + 1 = 2
  3. 88: 8 + 8 = 16 → 1 + 6 = 7
  4. 99: 9 + 9 = 18 → 1 + 8 = 9
  5. 412: 4 + 1 + 2 = 7
  6. 867: 8 + 6 + 7 = 21 → 2 + 1 = 3
  7. 4852: 4 + 8 + 5 + 2 = 19 → 1 + 9 = 10 → 1 + 0 = 1
  8. 9875: 9 + 8 + 7 + 5 = 29 → 2 + 9 = 11 → 1 + 1 = 2
  9. 24601: 2 + 4 + 6 + 0 + 1 = 13 → 1 + 3 = 4
  10. 48956701: 4 + 8 + 9 + 5 + 6 + 7 + 0 + 1 = 40 → 4 + 0 = 4
  1. 23 × 32
  • Step 1: 3 × 2 = 6
  • Step 2: (2 × 2) + (3 × 3) = 4 + 9 = 13 (write 3, carry 1)
  • Step 3: 2 × 3 = 6 + 1 = 7
    1. 57 × 99
    • LHS = 57 – 1 = 56; RHS = 99 – 56 = 43
      1. 13 × 17: 1 × 2 | 3 × 7 = 221
      2. 22 × 28: 2 × 3 | 2 × 8 = 616
      3. 34 × 36: 3 × 4 | 4 × 6 = 1224
      4. 91 × 99: 9 × 10 | 1 × 9 = 9009
      5. 35 × 35: 3 × 4 | 5 × 5 = 1225
      6. 42 × 48: 4 × 5 | 2 × 8 = 2016
      7. 72 × 78: 7 × 8 | 2 × 8 = 5616
      8. 93 × 97: 9 × 10 | 3 × 7 = 9021
      9. 104 × 106: 10 × 11 | 4 × 6 = 11024
      10. 105 × 105: 10 × 11 | 5 × 5 = 11025
      11. 203 × 207: 20 × 21 | 3 × 7 = 42021
      12. 405 × 405: 40 × 41 | 5 × 5 = 164025
      13. 502 × 508: 50 × 51 | 2 × 8 = 255016
      14. 603 × 607: 60 × 61 | 3 × 7 = 366021
      15. 704 × 706: 70 × 71 | 4 × 6 = 497024
      16. 905 × 905: 90 × 91 | 5 × 5 = 819025
      17. 193 × 197: 19 × 20 | 3 × 7 = 38021
      18. 292 × 298: 29 × 30 | 2 × 8 = 87016
      19. 392 × 398: 39 × 40 | 2 × 8 = 156016
      20. 495 × 495: 49 × 50 | 5 × 5 = 245025
      1. 13 × 13 (Base 10): Dev +3, +3 → 13+3 | 3 × 3 = 169
      2. 104 × 102 (Base 100): Dev +04, +02 → 104+2 | 08 = 10608
      3. 105 × 106 (Base 100): Dev +05, +06 → 105+6 | 30 = 11130
      4. 98 × 94 (Base 100): Dev -02, -06 → 98-6 | 12 = 9212
      5. 122 × 102 (Base 100): Dev +22, +02 → 122+2 | 44 = 12444
      6. 96 × 107 (Base 100): Dev -04, +07 → 96+7 | -28 = 10300 – 28 = 10272
      7. 1012 × 1004 (Base 1000): Dev +012, +004 → 1012+4 | 048 = 1016048
      8. 998 × 974 (Base 1000): Dev -002, -026 → 998-26 | 052 = 972052
      9. 1016 × 998 (Base 1000): Dev +016, -002 → 1016-2 | -032 = 1014000 – 32 = 1013968
      • 152 = 1 × 2 | 25 = 225
      • 252 = 2 × 3 | 25 = 625
      • 352 = 3 × 4 | 25 = 1225
      • 452 = 4 × 5 | 25 = 2025
      • 552 = 5 × 6 | 25 = 3025
      • 752 = 7 × 8 | 25 = 5625
      • 852 = 8 × 9 | 25 = 7225
      • 952 = 9 × 10 | 25 = 9025
      • 1052 = 10 × 11 | 25 = 11025
      • 1152 = 11 × 12 | 25 = 13225

      Find squares using (a | b)2 = a2 | 2ab | b2 (Anurupyena method) /

      Solve using Yavadhunam Tavadhunikritya Vargam Cha Yojayet /

      1. √(9409): Right pair 09 → 3 or 7; Left pair 94 → 9; 952 = 9025 < 9409 → 97
      2. √(7569): Right pair 69 → 3 or 7; Left pair 75 → 8; 852 = 7225 < 7569 → 87
      3. √(8281): Right pair 81 → 1 or 9; Left pair 82 → 9; 952 = 9025 > 8281 → 91
      4. √(3249): Right pair 49 → 3 or 7; Left pair 32 → 5; 552 = 3025 < 3249 → 57

      Multiply algebraic expressions using Urdhwatiryak Vidhi /

      1. (4x + 1)(x + 5):
      • Step 1: 1 × 5 = 5
      • Step 2: (4x × 5) + (1 × x) = 20x + x = 21x
      • Step 3: 4x × x = 4x2
💡 Solution:

  • Answer / उत्तर:
  • हिंदी: आधुनिक दशमलव संख्या प्रणाली में दस अंकों (0-9) का प्रयोग होता है, जहाँ प्रत्येक अंक का वास्तविक मान और स्थानीय मान होता है। यह स्थानीय मान पद्धति गणनाओं को अत्यंत सरल, वैज्ञानिक और सर्वमान्य बनाती है।
  • Answer / उत्तर:
  • हिंदी:
  1. शून्य की अवधारणा यजुर्वेद में ‘खम्’ शब्द के रूप में पाई जाती है।
  2. इसका श्रेय महर्षि गृत्समद, पाणिनि (500 ई.पू.) और पिंगल (200 ई.पू.) को दिया जाता है।
  3. शून्य के प्रतीक का प्रथम लिखित प्रमाण बख्शाली पाण्डुलिपि (300-400 ई.) में मिलता है।
  • Answer / उत्तर:
  • हिंदी: जब अंकों को व्यक्त करने के लिए अक्षरों या वर्णमाला का उपयोग किया जाता है, तो इसे वर्णांक (Alphanumeric) प्रणाली कहते हैं। आर्यभट्ट ने अपनी पुस्तक आर्यभटीय में बड़ी संख्याओं को संक्षिप्त रूप में दर्शाने के लिए इस पद्धति का प्रयोग किया था।
  • Answer / उत्तर:
  • हिंदी: किसी दीर्घचतुर्भुज (आयत) के विकर्ण पर बना वर्ग दोनों भुजाओं पर बने वर्गों के योगफल के बराबर होता है। समकोण त्रिभुज में कर्ण^2 = आधार^2 + लम्ब^2। बौधायन ने पाइथागोरस से सदियों पूर्व इस प्रमेय का विस्तृत वर्णन दिया था।
  • Answer / उत्तर:
  • हिंदी: आर्यभट्ट ने बताया कि 20000 व्यास वाले वृत्त की परिधि लगभग 62832 होती है। अतः π = (62832)/(20000) = 3.1416, जो दशमलव के 4 स्थानों तक शुद्ध है।
  • Answer / उत्तर:
  • हिंदी: इसके रचयिता स्वामी भारती कृष्ण तीर्थ (1884-1960 ई.) हैं। इसमें मानसिक गणित और तीव्र गणनाओं के लिए 16 मुख्य सूत्र और 13 उपसूत्र वर्णित हैं, जो 40 अध्यायों में विभाजित हैं।
  • Answer: 2, 3, 5, 7
  1. 241: Digit sum = 7. Required addition = 9 – 7 = 2
  2. 861: Digit sum = 15 → 6. Required addition = 9 – 6 = 3
  3. 4441: Digit sum = 13 → 4. Required addition = 9 – 4 = 5
  4. 83504: Digit sum = 20 → 2. Required addition = 9 – 2 = 7
  • English: Digit sums are used to quickly verify the correctness of addition, subtraction, multiplication, and division operations without full recalculation.
  • Answer: 736
    1. 44 × 52
    • Step 1: 4 × 2 = 8
    • Step 2: (4 × 2) + (4 × 5) = 8 + 20 = 28 (write 8, carry 2)
    • Step 3: 4 × 5 = 20 + 2 = 22
    • Answer: 2288
    1. 92 × 37
    • Step 1: 2 × 7 = 14 (write 4, carry 1)
    • Step 2: (9 × 7) + (2 × 3) = 63 + 6 = 69 + 1 = 70 (write 0, carry 7)
    • Step 3: 9 × 3 = 27 + 7 = 34
    • Answer: 3404
    1. 55 × 55
    • Step 1: 5 × 5 = 25 (write 5, carry 2)
    • Step 2: (5 × 5) + (5 × 5) = 25 + 25 = 50 + 2 = 52 (write 2, carry 5)
    • Step 3: 5 × 5 = 25 + 5 = 30
    • Answer: 3025
    1. 123 × 321
    • Step 1: 3 × 1 = 3
    • Step 2: (2 × 1) + (3 × 2) = 8
    • Step 3: (1 × 1) + (3 × 3) + (2 × 2) = 14 (write 4, carry 1)
    • Step 4: (1 × 2) + (2 × 3) = 8 + 1 = 9
    • Step 5: 1 × 3 = 3
    • Answer: 39483
    1. 414 × 232
    • Step 1: 4 × 2 = 8
    • Step 2: (1 × 2) + (4 × 3) = 14 (write 4, carry 1)
    • Step 3: (4 × 2) + (4 × 2) + (1 × 3) = 19 + 1 = 20 (write 0, carry 2)
    • Step 4: (4 × 3) + (1 × 2) = 14 + 2 = 16 (write 6, carry 1)
    • Step 5: 4 × 2 = 8 + 1 = 9
    • Answer: 96048
    1. 504 × 618
    • Step 1: 4 × 8 = 32 (write 2, carry 3)
    • Step 2: (0 × 8) + (4 × 1) = 4 + 3 = 7
    • Step 3: (5 × 8) + (4 × 6) + (0 × 1) = 64 + 0 = 64 (write 4, carry 6)
    • Step 4: (5 × 1) + (0 × 6) = 5 + 6 = 11 (write 1, carry 1)
    • Step 5: 5 × 6 = 30 + 1 = 31
    • Answer: 311472
    1. 812 × 453
    • Step 1: 2 × 3 = 6
    • Step 2: (1 × 3) + (2 × 5) = 13 (write 3, carry 1)
    • Step 3: (8 × 3) + (2 × 4) + (1 × 5) = 37 + 1 = 38 (write 8, carry 3)
    • Step 4: (8 × 5) + (1 × 4) = 44 + 3 = 47 (write 7, carry 4)
    • Step 5: 8 × 4 = 32 + 4 = 36
    • Answer: 367836
  • Answer: 5643
    1. 4378 × 9999
    • LHS = 4378 – 1 = 4377; RHS = 9999 – 4377 = 5622
    • Answer: 43775622
    1. 87 × 999
    • Multiplicand = 087; LHS = 086; RHS = 999 – 086 = 913
    • Answer: 86913
    1. 345 × 99999
    • Multiplicand = 00345; LHS = 00344; RHS = 99999 – 00344 = 99655
    • Answer: 34499655
    1. 48 × 9
    • Condition 3: LHS = 48 – 1 = 47; Combine = 479
    • Subtract LHS: 479 – 47 = 432
    1. 9457 × 999
    • Condition 3: LHS = 9457 – 1 = 9456; Combine = 9456999
    • Subtract LHS: 9456999 – 9456 = 9447543
    1. 342 (a=3, b=4):
    • 32 | 2(3)(4) | 42 = 9 | 24 | 16 = 1156
    1. 192 (a=1, b=9):
    • 12 | 2(1)(9) | 92 = 1 | 18 | 81 = 361
    1. 542 (a=5, b=4):
    • 52 | 2(5)(4) | 42 = 25 | 40 | 16 = 2916
    1. 642 (a=6, b=4):
    • 62 | 2(6)(4) | 42 = 36 | 48 | 16 = 4096
    1. 922 (a=9, b=2):
    • 92 | 2(9)(2) | 22 = 81 | 36 | 4 = 8464
    • 122 (Base 10, Dev +2): (12+2) | 22 = 144 (Note: Official back-of-book key prints 114 due to a typographical transposition, but mathematically 122 = 144).
    • 142 (Base 10, Dev +4): (14+4) | 42 = 18 | 16 = 196
    • 1022 (Base 100, Dev +02): (102+2) | 022 = 10404
    • 1052 (Base 100, Dev +05): (105+5) | 052 = 11025
    • 1082 (Base 100, Dev +08): (108+8) | 082 = 11664
    • 942 (Base 100, Dev -06): (94-6) | (-06)2 = 8836
    • 9962 (Base 1000, Dev -004): (996-4) | (-004)2 = 992016
  • Answer: 4x2 + 21x + 5
    1. (4x + 2y)(3x + 3y):
    • Step 1: 2y × 3y = 6y2
    • Step 2: (4x × 3y) + (2y × 3x) = 12xy + 6xy = 18xy
    • Step 3: 4x × 3x = 12x2
    • Answer: 12x2 + 18xy + 6y2
    1. (x – 3y)(x + 3y):
    • Step 1: (-3y)(3y) = -9y2
    • Step 2: (x)(3y) + (-3y)(x) = 0
    • Step 3: x × x = x2
    • Answer: x2 – 9y2
    1. (x + 4)(x + 4):
    • Step 1: 4 × 4 = 16
    • Step 2: 4x + 4x = 8x
    • Step 3: x × x = x2
    • Answer: x2 + 8x + 16 (Note: Official back-of-book key misprints the middle coefficient as 4x, giving x2 + 4x + 16, but the complete algebraic derivation gives x2 + 8x + 16).
    1. (x2 + 2x + 1)(x2 + 3x + 4):
    • Vertical right: 1 × 4 = 4
    • Cross 2 terms: (2x × 4) + (1 × 3x) = 11x
    • Cross 3 terms: (x2 × 4) + (1 × x2) + (2x × 3x) = 11x2
    • Cross 2 left terms: (x2 × 3x) + (2x × x2) = 5x3
    • Vertical left: x2 × x2 = x4
    • Answer: x4 + 5x3 + 11x2 + 11x + 4
    1. (2x2 + 3y – 4)(3x2 + 4y + 5):
    • Term 1: 2x2 × 3x2 = 6x4
    • Term 2: (2x2 × 4y) + (3y × 3x2) = 17x2y
    • Term 3: (2x2 × 5) + (-4 × 3x2) + (3y × 4y) = 10x2 – 12x2 + 12y2 = -2x2 + 12y2
    • Term 4: (3y × 5) + (-4 × 4y) = 15y – 16y = -y
    • Term 5: -4 × 5 = -20
    • Answer: 6x4 + 17x2y – 2x2 + 12y2 – y – 20

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