NCERT Solutions for Class 6th Maths Chapter 6 Making it ‘More’ or ‘Less’ — In-text Questions
Book page 145 & 146 Updated on2026-09-19
Q1.
Observe these two figures. Is there any similarity or difference between the two?
The two figures printed on page 145.
Answer
Similarity: both figures are made of exactly 9 unit squares, so both have an area of 9 square units.
Difference: their boundaries are very different lengths.
Nine unit squares packed into a 3 × 3 square (perimeter 12 units) and spread into a “C” shape (perimeter 20 units).
First figure (3 × 3 square) → perimeter = 4 × 3 = 12 units
Second figure (the “C” shape) → perimeter = 20 units
What this teaches:equal areas do not mean equal perimeters. The more squares hide their sides against each other, the shorter the boundary becomes.
Q2.
Using 9 unit squares, solve the following. 1. What is the smallest perimeter possible?
Answer
The smallest possible perimeter is 12 units, and it comes from the 3 × 3 square.
Perimeter of the 3 × 3 square = 4 × 3 = 12 units
Why nothing can beat it: nine separate squares have 9 × 4 = 36 sides. Every time two squares are joined, two of those sides disappear from the boundary. The 3 × 3 square has 12 joins — the most possible for 9 squares — so 36 − 2 × 12 = 12 units of boundary are left. No other arrangement can make more joins.
Q3.
2. What is the largest perimeter possible?
Answer
The largest possible perimeter is 20 units. One such figure is a straight strip of 9 squares (a 1 × 9 rectangle).
Perimeter of the 1 × 9 rectangle = 2 × (9 + 1) = 20 units
Why 20 is the maximum: the squares must form one connected figure, so at least 8 joins are needed to link 9 squares together. Each join removes 2 units of boundary, so the perimeter is at most
36 − 2 × 8 = 20 units
Any figure with exactly 8 joins — the straight strip, an L, a T, a plus sign — has this maximum perimeter of 20 units.
Q4.
3. Make a figure with a perimeter of 18 units.
Answer
Work out how many joins are needed:
36 − 2 × (joins) = 18 → joins = 9
So we need a figure with 9 joins. Put 5 squares in a bottom row and 4 squares in a top row, sliding the top row so that only 2 squares sit directly above the bottom row.
A step-shaped figure of 9 unit squares: 4 + 3 joins inside the rows and 2 joins between the rows — 9 joins in all, perimeter 18 units.
Walking round the boundary: 5 + 1 + 2 + 1 + 4 + 1 + 3 + 1 = 18 units ✔
Area is still 9 square units.
Q5.
4. Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning?
Answer
Perimeter
Joins needed
How many different shapes?
12 units
12
Only one — the 3 × 3 square
18 units
9
Many different shapes
20 units
8
Many different shapes
Reasoning:
Perimeter 12 is unique. It needs 12 joins, the greatest number 9 squares can possibly make, and only the tightly packed 3 × 3 square achieves that. Move even one square and a join is lost, so the perimeter grows.
Perimeter 20 needs only 8 joins — the least a connected figure can have. The 1 × 9 strip, an L-shape (6 + 4), a T-shape and a plus sign all have exactly 8 joins, so all of them have perimeter 20.
Perimeter 18 needs 9 joins, and there are several ways to arrange that — for example 5 + 4 in two rows, or a 2 × 4 block with one square hanging off a corner.
Rule to remember: perimeter = 4 × (number of squares) − 2 × (number of joins). It answers every question of this kind in one line.
Q6.
Let’s do something tricky now! We have a figure below having perimeter 24 units. Without calculating all over again, observe, think and find out what will be the change in the perimeter if a new square is attached as shown on the right.
The figure of unit squares on the left, and on the right the same figure with the new square (dashed) put in the place shown in the book.
Answer
There will be no change at all — the perimeter stays 24 units.
The figure of 11 unit squares (perimeter 24 units). The new square slots into the corner, touching the figure along two of its sides.
Look at where the new square goes. It fits into a corner, so two of its sides get hidden against the figure.
The new square brings 4 new sides.
2 of them are hidden, and they also hide 2 sides of the old figure.
Change = 4 − 2 × 2 = 0
New perimeter = 24 + 0 = 24 units
The quick rule: if a new square touches the figure along k sides, the perimeter changes by 4 − 2k. So k = 1 → +2, k = 2 → no change, k = 3 → −2. The area, of course, always increases by 1 square unit.
Q7.
Experiment placing this new square at different places and think what the change in perimeter will be. Can you place the square so that the perimeter: a) increases; b) decreases; c) stays the same?
Answer
Yes — all three are possible. Use the rule change = 4 − 2k, where k is the number of sides along which the new square touches the figure.
Where you place it
Sides touching (k)
Change
New perimeter
a. Sticking out on a flat edge
1
+2
26 units
c. In a corner (an “inner” corner)
2
0
24 units
b. In a slot with squares on three sides
3
−2
22 units
a) To increase the perimeter — attach it to a flat outer edge, where it touches on only one side. The perimeter becomes 26 units.
b) To decrease the perimeter — drop it into a notch that already has squares on three sides. The perimeter becomes 22 units.
c) To keep the perimeter the same — fit it into an inner corner, touching on exactly two sides, as shown in the book's figure.
Try This: in the given figure, find the notch surrounded on three sides. Placing the new square there makes the figure bigger in area but smaller in perimeter — a surprising but perfectly logical result.