NCERT Solutions Ganita Prakash Chapter 1 Figure it Out — Relation to Number Sequences

Book page 11 & 12 Updated on2026-09-05

Q1.
Count the number of sides in each shape in the sequence of Regular Polygons. Which number sequence do you get? What about the number of corners in each shape in the sequence of Regular Polygons? Do you get the same number sequence? Can you explain why this happens?
Answer
ShapeTriangleQuadri­lateralPentagonHexagonHeptagonOctagonNonagonDecagon
Sides345678910
Corners345678910

Both rows give the same sequence — the counting numbers starting from 3: 3, 4, 5, 6, 7, 8, 9, 10, …

Why the two counts are always equal: in a closed figure made of straight lines, walk once around the boundary. Every side ends at a corner, and at every corner exactly two sides meet. So sides and corners come in a perfect one-to-one pairing — each side can be matched with the corner at its end. Therefore number of sides = number of corners (vertices) in any closed figure, not just regular ones.
Q2.
Count the number of lines in each shape in the sequence of Complete Graphs. Which number sequence do you get? Can you explain why?
Answer
Complete graphK2K3K4K5K6K7
Dots234567
Lines136101521

We get 1, 3, 6, 10, 15, 21, … — the triangular numbers.

Why it happens: go from one graph to the next. When you add a new dot to K2 (which has 2 dots), the new dot must be joined to those 2 dots, so 2 new lines appear. Adding the next dot brings 3 new lines, then 4, and so on. So the totals are
1,   1 + 2 = 3,   1 + 2 + 3 = 6,   1 + 2 + 3 + 4 = 10 …
which are exactly the running sums of the counting numbers — the triangular numbers.
Another way to see it: in Kn each of the n dots is joined to the other (n − 1) dots. That counts every line twice (once from each end), so the number of lines is n × (n − 1) ÷ 2. For K6: 6 × 5 ÷ 2 = 15. ✔
Q3.
How many little squares are there in each shape of the sequence of Stacked Squares? Which number sequence does this give? Can you explain why?
Answer
1491625
Stacked Squares — 1, 4, 9, 16 and 25 little squares.
1,   4,   9,   16,   25,   36 …

We get the square numbers.

Why it happens: the nth shape is a big square made of n rows, and each row holds n little squares. So the total is n × n = n². For example the 4th shape has 4 rows of 4 little squares = 16.
Q4.
How many little triangles are there in each shape of the sequence of Stacked Triangles? Which number sequence does this give? Can you explain why? (Hint: In each shape in the sequence, how many triangles are there in each row?)
Answer
1491625
Stacked Triangles — count the upward-pointing and downward-pointing little triangles together: 1, 4, 9, 16, 25.
1,   4,   9,   16,   25 …

Again we get the square numbers.

Why it happens (follow the hint): count the little triangles row by row.
Row 1 → 1 triangle
Row 2 → 3 triangles (2 pointing up, 1 pointing down)
Row 3 → 5 triangles (3 up, 2 down)
Row 4 → 7 triangles (4 up, 3 down)
The rows contain 1, 3, 5, 7, 9 … — the odd numbers. And we already know from page 7 that adding the odd numbers gives the squares:
1 + 3 + 5 + 7 = 16 = 4 × 4
That is why a stacked triangle of n rows contains n² little triangles.
Q5.
To get from one shape to the next shape in the Koch Snowflake sequence, one replaces each line segment ‘—’ by a ‘speed bump’. As one does this more and more times, the changes become tinier and tinier with very very small line segments. How many total line segments are there in each shape of the Koch Snowflake? What is the corresponding number sequence?
Answer

The starting shape is an equilateral triangle with 3 line segments. Each step replaces every single segment by a speed bump made of 4 smaller segments, so the count is multiplied by 4 every time.

Stage1st2nd3rd4th5th
Line segments31248192768
As a product33 × 43 × 4 × 43 × 4 × 4 × 43 × 4 × 4 × 4 × 4

The number sequence is 3, 12, 48, 192, 768, …, that is 3 × the powers of 4. (This sequence is not in Table 1.)

Why: one segment → 4 segments. So if a stage has k segments, the next stage has 4k. Starting from 3, the counts are 3, 3×4, 3×4², 3×4³, …
Was this helpful? Report an error