NCERT Solutions Ganita Prakash (Part 1) Chapter 6 .4 This Way or That Way, All Ways Lead to the Bay — In-text Questions

Book page 1536 Updated on2026-09-05

Q1.

How many square tiles are there in each figure?

Answer

Figure	Number of tiles	Seen as a difference of squares
Step 1	8	3² – 1²
Step 2	12	4² – 2²
Step 3	16	5² – 3²

The counts go up by 4 each time: 8, 12, 16, …

Q2.

How many are there in Step 4 of the sequence? What about Step 10?

Answer

Step 4: 6² – 4² = 36 – 16 = 20 tiles
Step 10: 12² – 10² = 144 – 100 = 44 tiles

Identity 1C makes both instant:

(n + 2)² – n² = ((n + 2) + n)((n + 2) – n)
= (2n + 2) × 2 = 4n + 4
Step 4: 4 × 4 + 4 = 20 ✓ Step 10: 4 × 10 + 4 = 44 ✓

Q3.

Write an algebraic expression for the number of tiles in Step n. Share your methods with the class. Can you find more than one method to arrive at the answer?

Answer

Several ways of seeing the frame, all landing on the same expression.

Way of seeing it	Expression	Simplified
Big square minus hole	(n + 2)² – n²	4n + 4
Four sides of length n + 1, no overlap	4(n + 1)	4n + 4
Two full rows + two short columns	2(n + 2) + 2n	4n + 4
Four sides of n tiles + four corners	4n + 4	4n + 4

Number of tiles at Step n = (n + 2)² – n² = 4n + 4 = 4(n + 1)

Check: n = 1 → 8 ✓, n = 2 → 12 ✓, n = 3 → 16 ✓, n = 10 → 44 ✓

Why it happens: the “four sides of length n + 1” view explains the neat answer. Walk round the frame and give each corner tile to the side on its left; every side then owns n + 1 tiles and nothing is double-counted.

Q4.

Find the area of the (interior) shaded region in the figure below. All four rectangles have the same dimensions.

Answer

Two students see it two ways, and both are right.

Tadang's method. The whole figure is a square of side (m + n), and the four rectangles each measure m by n.

Shaded area = (m + n)² – 4mn

Yusuf's method. The shaded part is itself a square. Along the top edge the big side m + n is covered by one rectangle's short side m followed by its long side n; the shaded square starts where that m ends and stops where the next rectangle's m begins, so its side is (m + n) – m – m = n – m.

Shaded area = (n – m)²

Four equal m × n rectangles set in a pinwheel round a square hole of side n – m.

Area of the shaded region = (m + n)² – 4mn = (n – m)²

Q5.

By expanding both expressions, check that (m + n)² – 4mn = (n – m)².

Answer

(m + n)² – 4mn = (m² + 2mn + n²) – 4mn
= m² – 2mn + n²

(n – m)² = n² – 2nm + m²
= m² – 2mn + n²

The two expansions are the same expression, so the two methods agree — as they must, since they measure one region.

Test at m = 3, n = 7: (10)² – 4 × 21 = 100 – 84 = 16, and (7 – 3)² = 16 ✓

Did you know? Rearranged, this says (m + n)² – (n – m)² = 4mn, another special case of Identity 1C: the difference of those two squares is (2n)(2m).

Q6.

Find out the area of the region with slanting lines in the figure. All three rectangles have the same dimensions (Fig. 1).

Answer

Each of the three rectangles measures x by y. The slanting region lies between the top and bottom bars, on either side of the middle bar.

Anusha's method. ABCD is the square of side x lying between the two bars; EFGH is the middle bar, x tall and y wide.

Required area = Area(ABCD) – Area(EFGH)
= x² – xy

Vaishnavi's method. The whole outline PQSR has width x and height QS = y + x + y = x + 2y.

Area of PQSR = x(x + 2y)
Required area = x(x + 2y) – 3xy
= x² + 2xy – 3xy = x² – xy

Aditya's method. The slanting region is two equal rectangles, one on each side of the middle bar, each x tall and (x – y)⁄2 wide.

Required area = 2 × x × (x – y)⁄2 = x(x – y) = x² – xy

Area of the slanting region = x² – xy = x(x – y)

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