A Square and A Cube

Why squares and cubes matter

A square floor tiled 8 by 8 needs 64 tiles; a cube-shaped box with 8 unit cubes along each edge holds 512 unit cubes. This chapter connects geometric shapes with number patterns, factorisation and roots. In the textbook's 100-locker puzzle, locker n is toggled once for each divisor of n. Divisors pair except at a perfect square, so only lockers 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 remain open. The first five lockers touched exactly twice are 2, 3, 5, 7 and 11, because primes have exactly two positive divisors.

1.1 Square Numbers

For any number n, n² = n × n. For natural n, n² is a perfect square. A square of side s has area s² square units. Examples: 12² = 144; (3/5)² = 9/25; (2.5)² = 6.25. Perfect squares may end only in 0, 1, 4, 5, 6 or 9; this is a necessary but not sufficient test. For example 26 ends in 6 but is not a square. A square has an even number of trailing zeros; even integers have even squares and odd integers have odd squares.

Odd numbers, differences and triangular numbers

The sum of the first n positive odd numbers is n²: 1 + 3 + 5 + 7 + 9 = 25 = 5². Consecutive squares differ by 2n + 1 because (n + 1)² − n² = 2n + 1. Thus 36² = 35² + 71 = 1296. There are 2n whole numbers strictly between n² and (n + 1)². Triangular numbers Tn = n(n+1)/2 satisfy Tn + T(n−1) = n², e.g. 10 + 6 = 16.

Square roots and prime factors

A square root of a is a number x satisfying x² = a. For positive perfect squares there are two integer roots, ±x; √a denotes the principal nonnegative root. Example: √324 = 18, since 324 = 2² × 3⁴ = (2 × 3²)². The prime exponents of a positive integer perfect square must all be even. 1156 = 2² × 17² = 34², but 2800 = 2⁴ × 5² × 7 is not a perfect square. For non-square values, bound the root: 15² = 225 < 250 < 256 = 16², so 15 < √250 < 16.

1.2 Cubic Numbers

A cube with edge a has volume a³ cubic units, and a³ = a × a × a. Thus 2³ = 8, 3³ = 27, 4³ = 64. A positive integer is a perfect cube precisely when each prime exponent is a multiple of 3. Example: 3375 = 3³ × 5³ = 15³; 500 = 2² × 5³ is not a perfect cube. Cube roots reverse cubing: ∛512 = 8, ∛729 = 9, ∛(−216) = −6. A cube cannot end with exactly one or two zeros: its trailing zero count is divisible by 3.

Patterns in cubes and taxicab numbers

The cube n³ is a sum of n consecutive odd integers: 3 + 5 = 8 = 2³, 7 + 9 + 11 = 27 = 3³, 13 + 15 + 17 + 19 = 64 = 4³. The Hardy–Ramanujan number 1729 has two representations as a sum of positive cubes: 1³ + 12³ = 9³ + 10³ = 1729. For the cube sequence 1, 8, 27, 64, 125, successive first differences are 7, 19, 37, 61; second differences are 12, 18, 24; third differences are constantly 6.

1.3 A Pinch of History

The NCERT text discusses ancient Babylonian tables of squares and cubes, Sanskrit varga (square), ghana (cube) and mula (root), and Indian mathematical writings. These terms link arithmetic powers to geometric measurement. See the official NCERT PDF for the historical quotations and illustrations.

Textbook-aligned in-text activities

Locker puzzle: squares stay open because only perfect squares have an odd number of positive factors. The first five prime locker numbers are 2, 3, 5, 7, 11. For a number with three trailing zeros, its square has six. For 38², 34², 46², 56², 74² and 82², the ones digit is 6 for 34², 46², 56² and 74². Between consecutive squares n² and (n+1)² lie 2n integers. For 1156 and 2800, prime-factor tests show that only 1156 is a square. For cube-root practice, ∛64 = 4, ∛512 = 8 and ∛729 = 9.

Figure it Out — textbook page 10: Q1–Q9

1. Question 1: Which of 2032, 2048, 1027 and 1089 are not perfect squares?

   Solution: A perfect square cannot end in 2, 3, 7 or 8. Therefore 2032, 2048 and 1027 are not squares. Verify 1089 = 33 × 33.

   Final answer: 2032, 2048 and 1027.

2. Question 2: Which of 64², 108², 292² and 36² have units digit 4?

   Solution: Check the final digit of each base. 4² ends in 6; 8² ends in 4; 2² ends in 4; 6² ends in 6.

   Final answer: 108² and 292².

3. Question 3: Given 125² = 15,625, which expression gives 126²?

   Solution: Use (n+1)² = n² + n + (n+1). Thus 126² = 15,625 + 125 + 126 = 15,625 + 251 = 15,876.

   Final answer: Option (iv): 15,625 + 251.

4. Question 4: A square has area 441 m². What is its side length?

   Solution: Area = side². So side = √441. Since 21 × 21 = 441, the side is 21 m.

   Final answer: 21 m.

5. Question 5: What is the smallest perfect square divisible by 4, 9 and 10?

   Solution: LCM(4,9,10) = 180 = 2² × 3² × 5. Multiply by 5 to pair its unpaired prime factor: 180 × 5 = 900 = 30².

   Final answer: 900.

6. Question 6: What is the least multiplier making 9408 a perfect square, and what is the resulting square root?

   Solution: Prime factorise 9408 = 2⁶ × 3 × 7². Multiply by 3 to make all exponents even. Product = 28,224 = 168².

   Final answer: Multiplier 3; square root 168.

7. Question 7: How many integers lie strictly between 16² and 17², and between 99² and 100²?

   Solution: For consecutive squares n² and (n+1)², count = (n+1)² − n² − 1 = 2n. For n=16: 32; for n=99: 198.

   Final answer: (i) 32; (ii) 198.

8. Question 8: Fill in the missing numbers in 4²+5²+20²=(__)² and 9²+10²+(__)²=(__)².

   Solution: First: 16+25+400=441=21². The pattern is n²+(n+1)²+[n(n+1)]²=[n(n+1)+1]². At n=9, the missing terms are 90 and 91.

   Final answer: (i) 21; (ii) 90 and 91.

9. Question 9: How many tiny squares are shown in the NCERT picture, and what is the prime factorisation of that count?

   Solution: Count the picture's 9×9 motifs and 25 tiny squares per motif (including rotated squares). Then 81×25=2025. Factorise 2025 = 3⁴×5².

   Final answer: 2025 = 3⁴ × 5².

Figure it Out — textbook pages 16–17: Q1–Q5

1. Question 1: Find the cube roots of 27,000 and 10,648.

   Solution: Since 30×30×30=27,000, ∛27,000=30. Since 22×22×22=10,648, ∛10,648=22.

   Final answer: 30 and 22.

2. Question 2: By what smallest integer must 1323 be multiplied to obtain a perfect cube?

   Solution: 1323=3³×7². Every prime exponent in a cube is divisible by 3. Multiply by 7 to get 3³×7³=21³=9261.

   Final answer: 7.

3. Question 3: State whether each claim about cubes is true or false, and explain.

   Solution: (i) Odd³ is odd, not even: false. (ii) 2³=8, so a cube can end in 8: false. (iii) 10³=1000, so no positive two-digit number has a three-digit cube: false. (iv) 99³=970299 has six digits, so a two-digit positive integer cannot have a seven-digit cube: false. (v) 8=2³ has four positive factors, so cubes need not have an odd factor count: false.

   Final answer: (i) False; (ii) False; (iii) False; (iv) False; (v) False.

4. Question 4: Find ∛1331, ∛4913, ∛12167 and ∛32768 without long factorisation.

   Solution: Recognise nearby integer cubes and verify: 11³=1331, 17³=4913, 23³=12167 and 32³=32768.

   Final answer: 11, 17, 23 and 32.

5. Question 5: Which is greatest: 67³−66³, 43³−42³, 67²−66² or 43²−42²?

   Solution: Compute (i) 300763−287496=13267; (ii) 79507−74088=5419; (iii) 4489−4356=133; (iv) 1849−1764=85. Compare all four values.

   Final answer: Option (i): 67³−66³ = 13267.

Worked applications and common mistakes

A square field of area 441 m² has side √441 = 21 m, not 441 m. A cubic container with volume 1728 cm³ has edge ∛1728 = 12 cm. Common mistakes: confusing n² with 2n; assuming every number ending in 4 is a square; confusing √64 = 8 with the solutions of x² = 64, which are ±8; and using pairs of prime factors for cubes instead of triples.

Original practice, competency and HOTS

MCQ 1: Which is a perfect square, 75, 81, 98 or 110? Answer: 81. MCQ 2: The ones digit of 27³ is: 3 (because 7³ ends in 3). MCQ 3: ∛216 = 6. HOTS: Find the least multiplier that turns 540 into a square. Since 540 = 2² × 3³ × 5, multiply by 3 × 5 = 15; 8100 = 90². Case study: a square courtyard has area 2025 m². Side = 45 m; perimeter = 180 m. Competency task: Can 500 unit cubes form one larger cube without gaps? No, because 7³ = 343 and 8³ = 512.

Revision and self-assessment

Formula sheet: n² = n×n; n³ = n×n×n; (n+1)²−n² = 2n+1; 1+3+⋯+(2n−1) = n²; Tn = n(n+1)/2; Tn+T(n−1) = n². Flashcards: Square → n×n; Cube → n×n×n; Square root → inverse of squaring; Cube root → inverse of cubing; 1729 → sum of two positive cubes in two ways. Self-check: (1) √625? 25. (2) ∛1000? 10. (3) How many integers between 20² and 21²? 40. (4) Is 72 a square? No. (5) Is 729 a cube? Yes, 9³.

Official textbook and verification status

Source: NCERT Ganita Prakash Grade 8 Part I Chapter 1, Reprint 2026–27. Textbook Figure it Out Q9 on page 11 has been visually checked against the official diagram: 81 motifs of 25 tiny squares each, giving 2025 = 3⁴ × 5². The Square Pairs extended puzzle is an additional challenge; a full proof of all possible arrangements is not provided here. This page therefore does not claim 100% coverage of all textbook activities.

Chapter roadmap and learning goals

After completing this chapter, you should be able to identify perfect squares and cubes, explain the locker puzzle through factor pairs, calculate square and cube roots using prime factors, estimate non-perfect-square roots, interpret number patterns, and justify answers using diagrams and equations.

Visual learning: square and cube models

Imagine a 4 × 4 tile grid. It has four rows with four tiles each, so 4² = 16 unit squares. Add an L-shaped border of 9 tiles to build a 5 × 5 square, demonstrating 4² + 9 = 5². For a 3 × 3 × 3 block, picture three layers, each with nine unit cubes: 3³ = 27. Area uses square units; volume uses cubic units.

Complete first-30-square reference table

1²=1; 2²=4; 3²=9; 4²=16; 5²=25; 6²=36; 7²=49; 8²=64; 9²=81; 10²=100; 11²=121; 12²=144; 13²=169; 14²=196; 15²=225; 16²=256; 17²=289; 18²=324; 19²=361; 20²=400; 21²=441; 22²=484; 23²=529; 24²=576; 25²=625; 26²=676; 27²=729; 28²=784; 29²=841; 30²=900.

Complete first-20-cube reference table

1³=1; 2³=8; 3³=27; 4³=64; 5³=125; 6³=216; 7³=343; 8³=512; 9³=729; 10³=1000; 11³=1331; 12³=1728; 13³=2197; 14³=2744; 15³=3375; 16³=4096; 17³=4913; 18³=5832; 19³=6859; 20³=8000.

In-text checks and activities

1. Square endings: numbers ending 2, 3, 7 or 8 cannot be perfect squares. Examples: 12, 23, 37, 48 and 102.
2. Squares ending in 6: 34², 46², 56² and 74² end in 6; 38² and 82² end in 4.
3. Three trailing zeros become six when squared: (1000k)² = 1000000k².
4. Parity: even squared is even; odd squared is odd.
5. Between 1 and 100 inclusive are ten perfect squares; between 101 and 200 inclusive are four (121, 144, 169, 196). The largest square less than 1000 is 31²=961.
6. Square cloth of area 125 cm²: a 15 cm side requires 225 cm², so it cannot fit; the greatest integer side is 11 cm.
7. Cube construction: 2³ = 8 unit cubes, 3³ = 27 unit cubes and 4³ = 64 unit cubes.
8. The next two taxicab numbers: 4104 = 2³ + 16³ = 9³ + 15³; 13832 = 2³ + 24³ = 18³ + 20³.
9. The sum 91+93+95+97+99+101+103+105+107+109 equals 1000 = 10³ (ten consecutive odd numbers).
10. For squares, second finite differences are constant at 2; for cubes, third finite differences are constant at 6.

Additional original practice with worked answers

1. Find √2025. Prime factorise: 2025 = 3⁴ × 5²; √2025 = 3² × 5 = 45.
2. Find ∛13824. Since 24³ = 13824, the answer is 24.
3. Find the least number multiplying 72 to make a perfect square. 72=2³×3²; multiply by 2 to obtain 144=12².
4. Find the least number multiplying 72 to make a perfect cube. Multiply 2³×3² by 3 to obtain 216=6³.
5. Explain why a cube may end in 8. Example: 2³=8, so the statement that cubes never end in 8 is false.
6. A square park has side 32 m. Area = 32² = 1024 m². A cube of side 32 cm has volume 32³ = 32768 cm³; never mix square and cubic units.

Exam strategies and common errors

For a perfect-square test, use the final digit to reject impossible cases, then verify by prime factorisation or exact squaring. For perfect cubes, group prime factors in threes, not pairs. To estimate √250, compare 225 and 256: 15 < √250 < 16. Never say √64 = ±8; the radical √64 means 8, whereas x²=64 has solutions x=±8. For each textbook Figure it Out question, show the formula, substitute values, calculate, and verify.

Coverage and source integrity

Verified against the official NCERT Ganita Prakash Grade 8 Part I, Chapter 1, reprint 2026–27. Both numbered Figure it Out sets and the diagram question are covered. The optional Square Pairs challenge at the end of the chapter is identified as not yet exhaustively solved. All additional MCQs and practice tasks are original, not mislabelled as NCERT questions. Use the official book for exact question wording and visual layouts.