In-text — Patterns in Tiles and Areas — We Distribute Yet Things Multiply

1. 13 marksGanita Prakash Cl-8 Part 1, in-text, page 153

   How many square tiles are there in each figure of the sequence? How many are there in Step 4? What about Step 10? Write an algebraic expression for the number of tiles in Step n.

   Hint. Each figure is a large square with a smaller square removed from the middle — count both.

   Step 1 — See the structure. Each figure is a large square with a smaller square missing from the centre, so the tile count is a difference of two squares — which is exactly why this activity sits in this chapter.

   Step 2 — Read off the first three steps.

   Step	Outer square	Inner square	Tiles
   1	3² = 9	1² = 1	3² − 1² = 8
   2	4² = 16	2² = 4	4² − 2² = 12
   3	5² = 25	3² = 9	5² − 3² = 16

   The outer side runs 3, 4, 5 and the inner side runs 1, 2, 3 — always two apart.

   Step 3 — Extend the pattern. Step 4: 6² − 4² = 36 − 16 = 20 Step 10: 12² − 10² = 144 − 100 = 44

   Step 4 — Write the general expression. At Step n the inner side is n and the outer side is n + 2, so the number of tiles is (n + 2)² − n²

   Step 5 — A second method that simplifies it. Expanding with Identity 1C, a² − b² = (a + b)(a − b), with a = n + 2 and b = n: (n + 2)² − n² = [(n + 2) + n][(n + 2) − n] = (2n + 2)(2) = 4n + 4

   This explains the counts jumping by exactly 4 each time: 8, 12, 16, 20, … Check Step 10: 4(10) + 4 = 44 ✓

   A third way to see it, without algebra: the border is 4 strips of n tiles plus 4 corner tiles, giving 4n + 4 directly.

   ✦ Step 4 has 6² − 4² = 20 tiles and Step 10 has 12² − 10² = 44. In general Step n has (n + 2)² − n², which simplifies by Identity 1C to 4n + 4.

2. 23 marksGanita Prakash Cl-8 Part 1, in-text, page 154

   Write an expression for the area of the dashed region in the figure (a p by s rectangle with a strip of width r removed along two sides). Use more than one method. Then substitute p = 6, r = 3.5, s = 9 and calculate the area.

   Hint. Either subtract the removed strips from the whole rectangle, or find the dimensions of the dashed rectangle directly.

   Method 1 — subtract the strips from the whole. Whole rectangle = ps. Remove a strip of width r along one side (area pr) and another along the adjacent side (area sr). But the corner square of side r has been taken away twice, so add it back once: Area = ps − pr − sr + r²

   Method 2 — find the dashed rectangle's own dimensions. Removing a strip of width r from each of two adjacent sides leaves a rectangle measuring (p − r) by (s − r): Area = (p − r)(s − r)

   The two agree, since expanding the second gives (p − r)(s − r) = ps − pr − sr + r² ✓ This is exactly Identity 1 at work — and it is why the r² term appears in Method 1, which is the step most often forgotten.

   Substituting p = 6, r = 3.5, s = 9.

   Using Method 2 (fewer operations): (p − r)(s − r) = (6 − 3.5)(9 − 3.5) = 2.5 × 5.5 = 13.75 square units

   Checking with Method 1: ps − pr − sr + r² = 54 − 21 − 31.5 + 12.25 = 54 − 52.5 + 12.25 = 13.75 ✓

   Why the book asks for two methods. They cross-check each other, and the algebraic agreement between them is the distributive property made visible as areas — the theme of the whole chapter.

   ✦ Area = ps − pr − sr + r², equivalently (p − r)(s − r). With p = 6, r = 3.5, s = 9 this gives 13.75 square units.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp106.pdf). The chapter develops the distributive property into the three standard identities — 1A (a+b)², 1B (a−b)², 1C (a+b)(a−b) — via a multiplication-grid model, then applies them to fast mental multiplication and to area/tile patterns. Every expansion here was independently re-expanded term by term and every numeric answer recomputed before comparison with the book's printed answer key. TWO NOTES: (1) the twelve 'Mind the Mistake, Mend the Mistake' items on page 150 have NO answers in the printed key — each has been worked out from first principles here, including identifying which four of the twelve are in fact already correct, and this is stated openly in the solution; (2) the circle-pattern activity in §6.4 ('This Way or That Way') is omitted because the circle counts cannot be recovered from the text without the printed figure.. Questions are referenced from the NCERT textbook for identification.