Figure it Out — Squares and Square Roots — A Square and A Cube

1. 11 markGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q1

   Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089

   Hint. First rule out any options whose units digit alone proves they can't be squares, then check what's left against nearby known squares.

   Step 1 — Check nearby squares for each option. 45² = 2025 and 46² = 2116, so 2032 falls strictly between them — not a square.

   Step 2 — Check 2048. 2048 = 2¹¹, an odd power of 2, so its prime factors can't be split into two identical groups — not a square.

   Step 3 — Check 1027. 32² = 1024 and 33² = 1089, so 1027 falls strictly between them — not a square.

   Step 4 — Check 1089. 33² = 33 × 33 = 1089 exactly — this one is a perfect square.

   ✦ Answer: (i) 2032, (ii) 2048, and (iii) 1027 are not perfect squares; (iv) 1089 = 33² is a perfect square.

   Where students slip. Assuming all four options are equally hard to check — 1089 can be confirmed directly as 33², while the other three just need bracketing between two consecutive known squares.

2. 21 markGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q2

   Which one among 64², 108², 292², 36² has last digit 4?

   Hint. The units digit of a square depends only on the units digit of the number being squared — which starting digits give a square ending in 4?

   Step 1 — Recall which units digits square to 4. Only 2² = 4 and 8² = 64 end in 4, so the original number must end in 2 or 8.

   Step 2 — Check each option's units digit. 64 ends in 4 (→ square ends in 6); 108 ends in 8 (→ square ends in 4); 292 ends in 2 (→ square ends in 4); 36 ends in 6 (→ square ends in 6).

   ✦ Answer: 108² and 292² both end in the digit 4 (108² = 11664, 292² = 85264).

   Where students slip. Expecting exactly one correct option — two of the four numbers here satisfy the condition, since both digit 2 and digit 8 lead to a units digit of 4 when squared.

3. 32 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q3

   Given 125² = 15625, what is the value of 126²? (i) 15625 + 126 (ii) 15625 + 262 (iii) 15625 + 253 (iv) 15625 + 251 (v) 15625 + 512

   Hint. Write 126² − 125² as a difference of squares and factorise it, rather than computing 126² from scratch.

   Step 1 — Set up the difference of squares. 126² − 125² = (126 − 125)(126 + 125), since a² − b² = (a − b)(a + b).

   Step 2 — Simplify. = 1 × 251 = 251.

   Step 3 — Add this to the known value. 126² = 15625 + 251.

   ✦ Answer: (iv) 15625 + 251.

   Where students slip. Trying to multiply 126 × 126 directly from scratch — the difference-of-squares identity turns this into a one-line calculation using the given value of 125².

4. 41 markGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q4

   Find the length of the side of a square whose area is 441 m².

   Hint. The side length is the square root of the area.

   Step 1 — Recall the area formula. Area = side², so side = √Area.

   Step 2 — Find √441. 21 × 21 = 441.

   ✦ Answer: The side length is 21 m.

   Where students slip. Mistaking 441 ÷ 2 or a similar shortcut for the square root — finding the side requires the actual square root, not a division by 2.

5. 53 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q5

   Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.

   Hint. First find the smallest number divisible by all three (their LCM), then check whether its prime factorisation already has every exponent even.

   Step 1 — Find the LCM of 4, 9, and 10. 4 = 2², 9 = 3², 10 = 2 × 5. LCM = 2² × 3² × 5 = 180.

   Step 2 — Check whether 180 is already a perfect square. 180 = 2² × 3² × 5¹ — the exponent of 5 is odd, so 180 itself is not a perfect square.

   Step 3 — Fix the unpaired factor. Multiplying by one more 5 makes every exponent even: 180 × 5 = 900 = 2² × 3² × 5².

   Step 4 — Verify. 900 = 30², and 900 is divisible by 4, 9, and 10.

   ✦ Answer: 900.

   Where students slip. Assuming the LCM itself is automatically the answer — the LCM only guarantees divisibility; it still needs checking (and sometimes adjusting) to also be a perfect square.

6. 63 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q6

   Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.

   Hint. Prime-factorise 9408 first and look for any prime whose exponent isn't already even.

   Step 1 — Prime factorise 9408. 9408 = 2⁶ × 3 × 7² (dividing by 2 six times gives 147 = 3 × 7²).

   Step 2 — Find the unpaired factor. 2⁶ and 7² already have even exponents; only 3¹ is unpaired.

   Step 3 — Multiply to pair it. 9408 × 3 = 28224 = 2⁶ × 3² × 7².

   Step 4 — Find the square root of the product. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168.

   ✦ Answer: Smallest multiplier = 3; the resulting perfect square is 28224, with square root 168.

   Where students slip. Multiplying by every prime with an odd-looking role instead of checking exponents carefully — only the prime whose exponent is actually odd (here, just 3¹) needs fixing.

7. 72 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q7

   How many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100

   Hint. Use the 2n rule found earlier in this section, where n is the smaller of the two numbers.

   Step 1 — Recall the rule. Between n² and (n+1)², there are exactly 2n numbers.

   Step 2 — Apply it to 16 and 17. 2 × 16 = 32, since 16 is the smaller of the two roots here.

   Step 3 — Apply it to 99 and 100. 2 × 99 = 198.

   ✦ Answer: (i) 32 numbers lie between 16² and 17²; (ii) 198 numbers lie between 99² and 100².

   Where students slip. Recomputing both squares and subtracting from scratch — the 2n shortcut derived earlier in this section gives the answer immediately without needing either square's actual value.

8. 83 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q8

   In the following pattern, fill in the missing numbers: 1² + 2² + 2² = 3², 2² + 3² + 6² = 7², 3² + 4² + 12² = 13², 4² + 5² + 20² = ()², 9² + 10² + ()² = (__)²

   Hint. Look at the third term in each line — 2, 6, 12 — and see how it relates to the first two numbers in that same line.

   Step 1 — Find the pattern in the third term. For line n (using consecutive integers n and n+1): the third term is n(n+1) — check: 1×2=2, 2×3=6, 3×4=12, all match.

   Step 2 — Find the pattern on the right-hand side. The answer is n(n+1) + 1 — check: 1×2+1=3, 2×3+1=7, 3×4+1=13, all match.

   Step 3 — Apply this to the fourth line (n = 4). Third term = 4 × 5 = 20 (already given, confirming the pattern); right side = 4×5+1 = 21.

   Step 4 — Apply this to the fifth line (n = 9). Third term = 9 × 10 = 90; right side = 9×10+1 = 91, since the same n(n+1) and n(n+1)+1 pattern holds regardless of how large n gets.

   ✦ Answer: 4² + 5² + 20² = 21²; and 9² + 10² + 90² = 91².

   Where students slip. Trying to guess the missing numbers by trial and error — spotting that the third term is always n(n+1) and the answer is always n(n+1)+1 makes both blanks immediate.

9. 93 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.1), Q9 — a picture of tiled squares/diamonds; not in the book's own answer key, solved here by rendering the actual figure

   How many tiny squares are there in the picture accompanying this question? Write the prime factorisation of the number of tiny squares.

   Hint. Count how many large tiles make up the picture, and how many tiny squares are inside just one tile — the diamonds are the same 5×5 grid as the plain squares, just rotated 45°.

   Step 1 — Count the large tiles. The picture is a grid of large tiles, 9 across and 8 down, alternating between plain squares and the same squares rotated into diamonds — 9 × 8 = 72 tiles in total.

   Step 2 — Count the tiny squares inside one tile. Each tile, whether shown upright or as a diamond, is itself a 5 × 5 grid of tiny squares — 25 tiny squares per tile.

   Step 3 — Multiply. 72 × 25 = 1800 tiny squares in total.

   Step 4 — Prime-factorise 1800. 1800 = 2³ × 3² × 5², since 1800 = 8 × 225 = 2³ × (3 × 5)² = 2³ × 3² × 5².

   ✦ Answer: 1800 tiny squares; prime factorisation 1800 = 2³ × 3² × 5².

   Where students slip. Miscounting the diamond-oriented tiles as having a different number of tiny squares from the upright ones — rotating a tile by 45° doesn't change how many small squares it's divided into.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp101.pdf). This chapter's questions are scattered as 'Math Talk'/'Try This' prompts through the running text, plus two formal 'Figure it Out' exercise blocks. The book supplies its own answer key at the end of the chapter for every question that has one definite answer — that answer key is the ground truth this file is checked against. One question (Figure it Out — Squares, Q9, the tiny-squares picture) has no answer in the book's own key; it was solved here by rendering the actual figure and counting directly (9×8 = 72 tiles of a 5×5 grid each = 1800 tiny squares).. Questions are referenced from the NCERT textbook for identification.