Using five squares of side 1 cm, joined edge to edge, build two shapes whose perimeters are different. Give each shape's perimeter.

Step-by-step solution

1. 1

   See why the area cannot decide the perimeter. Every arrangement uses the same five squares, so every shape has area $5\ \text{cm}^2$. Perimeter, however, depends on how the squares touch — this is the point of the exercise: equal area does not force equal perimeter.

2. 2

   Derive the counting rule. Five separate squares would expose $5\times 4=20$ unit edges. Every time two squares are joined, one edge of each is hidden, removing $2$ from the total:

   $\text{perimeter}=20-2\times(\text{number of shared edges})$

   So the shape with fewer internal contacts has the larger perimeter.

3. 3

   Shape 1 — a straight row of five squares. Laid in a line, each square touches only its immediate neighbours, giving $4$ shared edges:

   $P=20-2(4)=12\ \text{cm}$

   Checking directly: the row is $5$ cm by $1$ cm, so $P=2(5+1)=12$ cm.

4. 4

   Shape 2 — three squares in a row with two stacked on top. Take cells at $(0,0),(1,0),(2,0)$ on the bottom and $(0,1),(1,1)$ on top. The contacts are $(0,0)\!-\!(1,0)$, $(1,0)\!-\!(2,0)$, $(0,1)\!-\!(1,1)$, $(0,0)\!-\!(0,1)$ and $(1,0)\!-\!(1,1)$ — five shared edges:

   $P=20-2(5)=10\ \text{cm}$

5. 5

   Confirm shape 2 by walking its boundary. Starting at the bottom-left corner and going clockwise: up $2$, right $2$, down $1$, right $1$, down $1$, left $3$. That is $2+2+1+1+1+3=10$ cm, matching the formula.

6. 6

   State the answer and the underlying principle.

   $\boxed{\text{straight row: }12\ \text{cm};\quad 3\text{-plus-}2\text{ block: }10\ \text{cm}}$

   An L-shape of five squares (three across, two up one side) also has only $4$ shared edges, so it gives $12$ cm — the same as the row. Compactness, not visual "bendiness", is what lowers the perimeter.