NCERT Solutions for Class 6th Maths Chapter 6 Regular polygons · Split and rejoin — In-text Questions

Book page 136 Updated on2026-09-19

Q1.

Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalise your understanding for the perimeter of other regular polygons.

Answer

Regular shapes are all around us. Measure one side with a scale, then multiply.

Object	Regular shape	One side (typical)	Perimeter
Carrom board	Square	74 cm	4 × 74 = 296 cm
₹1 coin face	Circle-like (not a polygon)	—	measure with a thread
Bathroom floor tile	Square	30 cm	4 × 30 = 120 cm
Stop-sign board	Regular octagon	25 cm	8 × 25 = 200 cm
Honeycomb cell	Regular hexagon	6 mm	6 × 6 = 36 mm
Set-square (60°)	—	—	add the three sides

The generalisation:

Perimeter of a regular polygon = n × s
where n = number of sides and s = length of one side

Check it against the formulas you already know: n = 3 gives 3 × s (equilateral triangle), n = 4 gives 4 × s (square), n = 6 gives 6 × s (regular hexagon). ✔

Q2.

Split and rejoin: A rectangular paper chit of dimension 6 cm × 4 cm is cut as shown into two equal pieces. These two pieces are joined in different ways. For example, the arrangement a. has a perimeter of 28 cm. Find out the length of the boundary (i.e., the perimeter) of each of the other arrangements below.

The 6 cm × 4 cm chit with the cut shown, arrangement a., and arrangements b., c. and d., page 136.

Answer

The 6 cm × 4 cm chit is cut into two equal pieces, each 6 cm × 2 cm. Every arrangement uses both pieces, so only the length of the joined edge changes.

Perimeter of one piece = 2 × (6 + 2) = 16 cm
Two separate pieces = 2 × 16 = 32 cm
When they are joined along a common edge of length L, that edge is hidden twice:
Perimeter = 32 − 2 × L

Arrangements b, c and d of the two 6 cm × 2 cm pieces, and (last) an arrangement whose perimeter is 22 cm.

Arrangement	Shape formed	Length of the joined edge	Perimeter
a. (given)	12 cm × 2 cm rectangle	2 cm	32 − 4 = 28 cm
b.	L-shape	2 cm	32 − 4 = 28 cm
c.	T-shape	2 cm	32 − 4 = 28 cm
d.	Z / step shape	3 cm	32 − 6 = 26 cm

Check b by walking round it: 8 + 6 + 2 + 4 + 6 + 2 = 28 cm ✔ Check d: 2 + 3 + 2 + 6 + 2 + 3 + 2 + 6 = 26 cm ✔

Why the area never changes: in every arrangement the same two pieces are used, so the area stays 6 × 4 = 24 sq cm. Only the perimeter changes — a beautiful example of “same area, different perimeter”.

Q3.

Arrange the two pieces to form a figure with a perimeter of 22 cm.

The 6 cm × 4 cm chit with the cut shown, arrangement a., and arrangements b., c. and d., page 136.

Answer

Use the rule found above and work backwards.

32 − 2 × L = 22
2 × L = 32 − 22 = 10
L = 5 cm

So the two pieces must touch along 5 cm. Place the two 6 cm × 2 cm pieces side by side with their long edges together, but slide one of them 1 cm up. Then 5 cm of the 6 cm edges touch (the last figure in the picture above).

Walking round: 2 + 1 + 2 + 6 + 2 + 1 + 2 + 6 = 22 cm ✔

Tip: if you join the long edges completely (6 cm), you get a 6 cm × 4 cm rectangle again, with perimeter 32 − 12 = 20 cm — the smallest possible. Sliding the pieces apart increases the perimeter step by step.

Was this helpful?