Perfect Perimeters - ArgoPrep - Free Printable

Let’s solve each problem step by step. Remember: Perimeter means the total distance around the outside of a shape. We add up all the outer sides.

Each small square has sides that are 2 cm long (from the example and key at top left).

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Problem 1:
Shape looks like an “H” made of squares.
Count the outer edges:
- Top: 3 segments → 3 × 2 = 6 cm
- Bottom: 3 segments → 3 × 2 = 6 cm
- Left side: 4 segments → 4 × 2 = 8 cm
- Right side: 4 segments → 4 × 2 = 8 cm
Wait — let’s trace carefully!

Actually, better to count each outer edge one by one:

Start from top-left corner, go clockwise:

Top row: right 2, down 2, right 2, down 2, left 2, down 2, left 2, up 2, left 2, up 2 → too messy.

Better method: Count how many 2-cm segments are on the outside.

Looking at shape 1:
It’s 3 squares wide and 3 tall, but with middle missing in top and bottom rows? Actually, it’s like two vertical bars connected in middle.

Let me draw mentally:

Row 1: [ ][ ][ ] ← full top
Row 2: [ ] [ ] ← only sides
Row 3: [ ][ ][ ] ← full bottom

So outer perimeter:

Top: 3 sides → 6 cm
Bottom: 3 sides → 6 cm
Left side: from top to bottom: 3 segments → 6 cm
Right side: 3 segments → 6 cm
But also, inside gaps? No — we don’t count inner lines.

Wait — actually, between row1 and row2, there are two indentations on left and right? Let’s count exposed edges.

Alternative: Each square has 4 sides, but when squares touch, they share sides — so subtract shared sides.

Total squares: 7 squares? Let’s count:

Shape 1:
Top row: 3 squares
Middle row: 2 squares (left and right)
Bottom row: 3 squares
Total = 8 squares? Wait no:

Actually looking at image: It’s 3 across top, then middle row has only left and right (so 2), then bottom 3 again. So 3+2+3=8 squares.

Each square has 4 sides → 8×4 = 32 sides total.

Now subtract shared sides:

Horizontal shares:
In top row: between sq1-sq2, sq2-sq3 → 2 shared
In middle row: none (only 2 separate squares)
In bottom row: 2 shared
Vertical shares:
Between top and middle: left column (sq1-top and sq1-middle) → 1 share; right column (sq3-top and sq3-middle) → 1 share
Between middle and bottom: same → 2 more shares
Also, within columns? Middle row squares are not connected vertically to anything else except above and below.

Actually, let’s list connections:

Squares positions (row,col):

(1,1), (1,2), (1,3)
(2,1), (2,3)
(3,1), (3,2), (3,3)

Shared horizontal:
Row1: (1,1)-(1,2), (1,2)-(1,3) → 2
Row3: (3,1)-(3,2), (3,2)-(3,3) → 2
Total horiz shared: 4

Shared vertical:
Col1: (1,1)-(2,1), (2,1)-(3,1) → 2
Col3: (1,3)-(2,3), (2,3)-(3,3) → 2
Col2: only (1,2) and (3,2) — no connection to middle → 0
Total vert shared: 4

Each shared side removes 2 edges (one from each square) → so total removed edges = (4 + 4) × 2 = 16? No!

Wait: Each shared side is counted once per pair — so if two squares share a side, that’s one shared interface, which removes 2 edges from total perimeter (since each square loses one edge).

So number of shared interfaces = 4 (horizontal) + 4 (vertical) = 8 shared interfaces.

Each interface reduces total perimeter by 2 × 2cm = 4cm? No — each shared side means 2 cm is not part of perimeter for each square, so total reduction is 2 cm per shared side? Let's think.

Original total if all separate: 8 squares × 4 sides × 2 cm = 64 cm

Each time two squares share a side, we lose 2 cm from perimeter (because that side is now internal, not exposed).

Number of shared sides: as above, 8 shared sides (interfaces).

So perimeter = 64 - (8 × 2) = 64 - 16 = 48 cm? That can't be right because example was 24 cm for 9 squares.

I think I'm overcomplicating.

Let’s just trace the outline.

For shape 1:

Start at top-left corner.

Go right 2 cm (top of first square)
Down 2 cm (right side of first square? No — wait, better to use grid.

Since each square is 2x2, and shape is on grid, we can count the number of unit edges on boundary.

Each "unit" is 2 cm, but let's count how many 2-cm segments are on the perimeter.

Look at shape 1:

Top: 3 segments → 6 cm
Then down right side of top-right square: 2 cm
Then left along top of middle-right square? No.

Perhaps count all outer edges visually.

From the image, shape 1 has:

- Top: 3 units
- Right side: from top to bottom, it goes down 2, then left 2 (into the gap), then down 2, then right 2, then down 2 — this is messy.

I recall that for such shapes, a good way is to count the number of exposed sides.

Let me try for shape 1:

Imagine walking around the shape.

Start at top-left corner.

Move right: 2 cm (top of left-top square)
Move right: 2 cm (top of middle-top square)
Move right: 2 cm (top of right-top square) → total top: 6 cm

Now move down: 2 cm (right side of right-top square)

Now move left: 2 cm (bottom of right-top square? But there's a square below? In row2, there is a square at (2,3), so yes, but it's attached, so we don't go left yet.

After moving down 2 cm from top-right, we are at the top-right corner of the middle-right square. Since there is a square below it, we continue down? No, the square below is there, so the side is shared, so we don't walk there.

I think I need to look for a different approach.

Let's use the fact that in the example, a 3x3 grid of squares has perimeter 24 cm.

3x3 grid: 9 squares, perimeter 24 cm.

How? Outer rectangle is 6cm x 6cm, so perimeter 2*(6+6)=24 cm. Yes.

For irregular shapes, we can find the bounding box or count the turns.

Another way: the perimeter is equal to the number of unit edges on the boundary times 2 cm.

Let's define a "unit edge" as one side of a small square, 2 cm.

For shape 1:

Let's list all the outer edges.

Top row: all three tops are exposed: 3 edges

Bottom row: all three bottoms are exposed: 3 edges

Left side:
- Left of (1,1): exposed
- Left of (2,1): exposed (since no square to left)
- Left of (3,1): exposed
Also, between (1,1) and (2,1), the bottom of (1,1) is shared with top of (2,1)? No, (2,1) is below (1,1), so the bottom of (1,1) is shared with top of (2,1), so not exposed.

Similarly, top of (2,1) is shared with bottom of (1,1), not exposed.

So for left side:
- Left of (1,1): exposed
- Left of (2,1): exposed
- Left of (3,1): exposed
That's 3 edges on left.

But also, on the left, between row1 and row2, is there any exposure? No, because (1,1) and (2,1) are adjacent vertically, so their common side is internal.

Similarly for right side:
- Right of (1,3): exposed
- Right of (2,3): exposed
- Right of (3,3): exposed
3 edges.

Now, what about the indentations? In the middle row, there is no square at (2,2), so between (2,1) and (2,3), there is a gap.

So, for the top of the middle row:
- The bottom of (1,2) is exposed? Because there is no square below it in row2 at col2.
Similarly, the top of (3,2) is exposed? No, (3,2) is in bottom row, its top is shared with nothing? Let's see.

Squares present:
Row1: col1,2,3
Row2: col1,3 (col2 missing)
Row3: col1,2,3

So, for the gap at (2,2):

- The bottom of (1,2) is exposed (no square below)
- The top of (3,2) is exposed (no square above)
- The right of (2,1) is exposed (no square to right)
- The left of (2,3) is exposed (no square to left)

So additional exposed edges:

From the gap:
- Bottom of (1,2): 1 edge
- Top of (3,2): 1 edge
- Right of (2,1): 1 edge
- Left of (2,3): 1 edge

Also, are there any others?

Let's list all exposed edges by position.

Top edges:
- (1,1) top: exposed
- (1,2) top: exposed
- (1,3) top: exposed
- (2,1) top: shared with (1,1) bottom? No, (2,1) is below (1,1), so (2,1) top is shared with (1,1) bottom, so not exposed.
Similarly, (2,3) top shared with (1,3) bottom.
- (3,1) top: shared with (2,1) bottom? (2,1) is above (3,1), so yes, shared.
- (3,2) top: no square above, so exposed
- (3,3) top: shared with (2,3) bottom

So top-exposed: (1,1)t, (1,2)t, (1,3)t, (3,2)t → 4 edges

Bottom edges:
- (1,1) bottom: shared with (2,1) top
- (1,2) bottom: no square below, exposed
- (1,3) bottom: shared with (2,3) top
- (2,1) bottom: shared with (3,1) top
- (2,3) bottom: shared with (3,3) top
- (3,1) bottom: exposed
- (3,2) bottom: exposed
- (3,3) bottom: exposed

So bottom-exposed: (1,2)b, (3,1)b, (3,2)b, (3,3)b → 4 edges

Left edges:
- (1,1) left: exposed
- (2,1) left: exposed
- (3,1) left: exposed
- (1,2) left: shared with (1,1) right? (1,1) and (1,2) are adjacent horizontally, so (1,2) left is shared with (1,1) right, not exposed.
Similarly, (3,2) left shared with (3,1) right.
- (1,3) left: shared with (1,2) right
- (2,3) left: no square to left, exposed (because (2,2) missing)
- (3,3) left: shared with (3,2) right

So left-exposed: (1,1)l, (2,1)l, (3,1)l, (2,3)l → 4 edges

Right edges:
- (1,1) right: shared with (1,2) left
- (1,2) right: shared with (1,3) left
- (1,3) right: exposed
- (2,1) right: no square to right, exposed (gap)
- (2,3) right: exposed
- (3,1) right: shared with (3,2) left
- (3,2) right: shared with (3,3) left
- (3,3) right: exposed

So right-exposed: (1,3)r, (2,1)r, (2,3)r, (3,3)r → 4 edges

Now sum all exposed edges:
Top: 4
Bottom: 4
Left: 4
Right: 4
Total = 16 edges

Each edge is 2 cm, so perimeter = 16 × 2 = 32 cm

Let me verify with another method.

The shape is symmetric. Width is 3 squares = 6 cm, height is 3 squares = 6 cm, but with a hole in the center of the middle row.

The perimeter should be more than the bounding box.

Bounding box 6x6 has perimeter 24 cm, but here we have extra edges due to the indentation.

In the middle, instead of a solid block, we have a dent on left and right in the middle row.

Specifically, on the left side, between row1 and row2, there is no issue, but at row2, the square is only at col1 and col3, so on the left, from row1 to row2, it's straight, but at the gap, we have additional vertical edges.

From my calculation, 16 edges * 2 cm = 32 cm.

Let me count for a smaller part.

Suppose just the top row: 3 squares in a row. Perimeter would be: top 3, bottom 3, left 1, right 1, and the ends — wait, for 3 squares in a row, perimeter is 2* (length + width) = 2*(6 + 2) = 16 cm, or count edges: top 3, bottom 3, left 1, right 1, and the two ends are included, but also the sides between are shared, so total exposed: top 3, bottom 3, left 1, right 1, and no other, so 8 edges *2 = 16 cm, yes.

For our shape, with the gap, we have additional exposures.

In shape 1, compared to a solid 3x3, which has perimeter 24 cm (as in example), but here we have removed the center square of the middle row? In 3x3, all 9 squares are present, perimeter 24 cm.

Here, we have 8 squares: missing (2,2).

When you remove a square from the interior, you increase the perimeter.

In a solid 3x3, the center square (2,2) has all four sides shared, so removing it exposes 4 new edges.

So perimeter becomes 24 + 4*2 = 24 + 8 = 32 cm. Yes! Matches my earlier calculation.

So for shape 1: 32 cm.

But let's confirm with the example: example is 3x3 grid, 9 squares, perimeter 24 cm, which is correct for a 6cm x 6cm square.

If we remove the center square, we add 4 sides of 2 cm each, so +8 cm, total 32 cm.

Perfect.

So Problem 1: 32 cm

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Problem 2:
Shape is like a "P" or something. From image: it's 2 squares wide at top, then down 3 squares on left, and at bottom, one square to the right.

Specifically:
Row1: col1,2
Row2: col1
Row3: col1
Row4: col1,2? Let's see.

From image: it's a vertical bar of 4 squares on left, and at the top, one square to the right, and at the bottom, one square to the right? No.

Looking: it's like a "L" but with an extra on top.

Actually:
- Top row: two squares (col1 and col2)
- Then below col1: three more squares down, so total height 4 rows.
- And at the bottom, only col1 has a square, col2 does not have in row3 and row4? Let's assume.

From standard interpretation: shape 2 is:
Positions:
(1,1), (1,2)
(2,1)
(3,1)
(4,1)

So 5 squares.

Compared to a solid rectangle, but it's irregular.

Use the removal method or count.

Solid if it were 4x2, but it's not.

List exposed edges.

Squares: A=(1,1), B=(1,2), C=(2,1), D=(3,1), E=(4,1)

Exposed edges:

Top:
A top, B top → 2

Bottom:
E bottom → 1 (since only E at bottom)

Left:
A left, C left, D left, E left → 4 (B is to the right, so not on left)

Right:
B right → 1 (since no square to right of B)
Also, for C,D,E, their right sides: since no squares to right, and they are not covered, so C right, D right, E right are exposed? But in the shape, for row2,3,4, only col1 has squares, so yes, right side of C,D,E are exposed.

But B is at (1,2), so its right is exposed.

Now, also, between B and C: B is at (1,2), C at (2,1), not adjacent, so no shared side.

Similarly, the bottom of B is exposed, since no square below it.

Top of C is shared with bottom of A? A is (1,1), C is (2,1), so yes, shared.

Similarly, top of D shared with bottom of C, etc.

So let's list all exposed:

Top edges:
A top, B top → 2

Bottom edges:
E bottom → 1
Also, B bottom: no square below, exposed → 1
C bottom: shared with D top? C is (2,1), D is (3,1), so C bottom shared with D top, not exposed.
Similarly, D bottom shared with E top.
So only E bottom and B bottom are exposed for bottom? But "bottom" means the lowest edge.

Better to categorize by direction.

Top-facing exposed:
- A top
- B top
- Also, since no square below B, B bottom is exposed, but that's bottom-facing.

Let's do:

North-facing (top):
- A north
- B north
- Is there any other? C north is shared with A south, not exposed.
D north shared with C south.
E north shared with D south.
So only 2 north-facing exposed.

South-facing (bottom):
- E south (bottom)
- B south (since no square below B)
- A south shared with C north? A is (1,1), C is (2,1), so A south shared with C north, not exposed.
C south shared with D north.
D south shared with E north.
So south-facing exposed: E south, B south → 2

West-facing (left):
- A west
- C west
- D west
- E west
- B west: B is at (1,2), so its west is shared with A east? A is (1,1), B is (1,2), so yes, shared, not exposed.
So west-facing: A,C,D,E → 4

East-facing (right):
- B east (no square to right)
- A east: shared with B west, not exposed
- C east: no square to right, exposed
- D east: no square to right, exposed
- E east: no square to right, exposed
So east-facing: B east, C east, D east, E east → 4

Total exposed edges: north 2 + south 2 + west 4 + east 4 = 12 edges

Perimeter = 12 × 2 = 24 cm

Verify: total squares 5, if separate, 5×4×2=40 cm

Shared sides:
- A and B share a side (horizontal) → 1 shared
- A and C share a side (vertical) → 1 shared
- C and D share vertical → 1
- D and E share vertical → 1
Total shared interfaces: 4

Each shared interface reduces perimeter by 2×2=4 cm? No, each shared side means 2 cm is not on perimeter for each square, so total reduction is 2 cm per shared side? Let's see.

When two squares share a side, that side is internal, so we lose 2 cm from the total perimeter (since each square had that side as external, now it's internal, so minus 2 cm per shared side).

Number of shared sides: 4 (as above)

So perimeter = 40 - 4×2 = 40 - 8 = 32 cm? But I got 24, inconsistency.

Mistake.

Each shared side is one interface, and it removes 2 cm from the total perimeter calculation because that side is no longer on the boundary.

Initial total if separate: 5 squares × 4 sides × 2 cm = 40 cm

Each shared side between two squares means that 2 cm is subtracted from the perimeter (because that side is now internal, not part of the outer perimeter).

Number of shared sides:
- Between A and B: 1 shared side
- Between A and C: 1 shared side
- Between C and D: 1
- Between D and E: 1
Total 4 shared sides.

So perimeter = 40 - 4×2 = 32 cm

But earlier I calculated 12 edges ×2 = 24 cm, which is wrong.

Where did I miss?

In my edge count, I have:

North: A n, B n → 2

South: E s, B s → 2 (B south is exposed)

West: A w, C w, D w, E w → 4

East: B e, C e, D e, E e → 4

Sum 12, but should be more.

What about the south of A? A is at (1,1), its south is shared with C north, so not exposed, correct.

But for B, its south is exposed, yes.

However, is there a north for C? No, shared.

But let's think about the shape: from top, we have A and B side by side.

Below A is C, below C is D, below D is E.

So the outline:

Start at top-left of A.

Go right to top-right of B: 4 cm (since two squares wide)

Then down right side of B: 2 cm

Then left along bottom of B: 2 cm (but this is the south of B, which is exposed)

Then down? After going left along bottom of B, we are at the bottom-right of B, which is above the gap.

Then we need to go down to the level of C, but there's no square, so we go down 2 cm (this is the east side of the gap, but since no square, it's exposed? No, in terms of the shape, the perimeter should include the edge between B and the empty space below.

After going down the right side of B (2 cm), we are at the southeast corner of B.

Then, since there is no square below B, we go left along the bottom of B for 2 cm to the southwest corner of B.

Then, from there, we go down 2 cm (this is the west side of the empty space below B, but actually, this edge is the east side of the square that would be at (2,2), but since it's empty, this edge is part of the perimeter of the shape? No, the shape doesn't include that area, so the perimeter should go down from the southwest corner of B to the northwest corner of C? But C is at (2,1), which is directly below A, not below B.

The distance from southwest corner of B to northwest corner of C is diagonal, but in grid, we move horizontally and vertically.

From southwest corner of B (which is at x=4cm, y=2cm if A is from 0-2, B from 2-4 in x, y=0-2 for row1), then C is at x=0-2, y=2-4.

So from (4,2) to (0,4)? No.

Let's coordinate.

Assume each square 2x2.

Set origin at top-left of A.

A: x=0 to 2, y=0 to 2

B: x=2 to 4, y=0 to 2

C: x=0 to 2, y=2 to 4

D: x=0 to 2, y=4 to 6

E: x=0 to 2, y=6 to 8

Now, the shape consists of these squares.

The perimeter is the boundary of the union.

Start at (0,0) - top-left of A.

Go right to (4,0) - top-right of B. Distance 4 cm.

Go down to (4,2) - bottom-right of B. Distance 2 cm.

Now, from (4,2), since there is no square below B, and the next square is C at (0,2) to (2,4), but there is a gap.

To go to the next part, we need to go left to (2,2), but (2,2) is the bottom-left of B, and also the top-right of the empty space.

From (4,2), go left to (2,2) — this is along the bottom of B, distance 2 cm.

Then from (2,2), go down to (2,4)? But at (2,2) to (2,4), this is the line x=2, y=2 to 4.

Is this part of the perimeter? The square C is from x=0 to 2, y=2 to 4, so its east side is at x=2, y=2 to 4, which is exposed because no square to the east.

Similarly, the area from x=2 to 4, y=2 to 4 is empty, so the line x=2, y=2 to 4 is the boundary between the shape and empty space, so yes, it is part of the perimeter.

So from (2,2) go down to (2,4) — distance 2 cm. This is the east side of C.

Then from (2,4), go left to (0,4) — this is the top of D? D is from y=4 to 6, so at y=4, it's the top of D.

From (2,4) to (0,4) is along the top of D, but D is from x=0 to 2, y=4 to 6, so its top is from (0,4) to (2,4), so from (2,4) to (0,4) is left along the top of D, distance 2 cm.

Then from (0,4) go down to (0,6) — left side of D, distance 2 cm.

Then from (0,6) go down to (0,8) — left side of E, distance 2 cm.

Then from (0,8) go right to (2,8) — bottom of E, distance 2 cm.

Then from (2,8) go up to (2,6) — right side of E, distance 2 cm.

Then from (2,6) go up to (2,4) — right side of D, distance 2 cm.

But we are back to (2,4), and we already came from there.

This is messy.

From (2,4) after coming down from (2,2), we went left to (0,4), then down to (0,6), down to (0,8), right to (2,8), up to (2,6), up to (2,4) — but (2,4) is where we were, so we have a loop, but we missed the connection.

After reaching (2,4) from above, we should go to the left side.

Let's list the path without repetition.

Start at (0,0)

Right to (4,0) // top of A and B

Down to (4,2) // right of B

Left to (2,2) // bottom of B

Down to (2,4) // east of C (since C is from x=0-2, y=2-4, so east side is x=2, y=2-4)

Left to (0,4) // top of D (D is y=4-6, so top is y=4, x=0-2)

Down to (0,6) // west of D

Down to (0,8) // west of E

Right to (2,8) // south of E

Up to (2,6) // east of E

Up to (2,4) // east of D

But now at (2,4), and we came from (2,4) earlier? No, we came to (2,4) from above, then went left, now we are back at (2,4) from below.

From (2,4) , we need to close to start, but we have not connected to the left part properly.

From (2,4) , after coming up from (2,6), we are at (2,4), and the next should be to (0,4), but we already did that.

I think I double-counted.

Let's list the vertices in order.

The perimeter path:

1. (0,0) to (4,0) // top

2. (4,0) to (4,2) // right of B

3. (4,2) to (2,2) // bottom of B

4. (2,2) to (2,4) // east of C

5. (2,4) to (0,4) // top of D

6. (0,4) to (0,6) // west of D

7. (0,6) to (0,8) // west of E

8. (0,8) to (2,8) // south of E

9. (2,8) to (2,6) // east of E

10. (2,6) to (2,4) // east of D

Now at (2,4), and we need to go back to (0,0), but we have the left side of A and C not fully covered.

From (2,4) , we can go to (0,4), but we already did that in step 5.

The issue is that from (2,4) to (0,4) is already done, but in step 5 we went from (2,4) to (0,4), then down, etc.

After step 10, we are at (2,4), and the only thing left is the west side of A and C, but we have not included it yet.

From (0,0) to (0,2) is the west of A, which is not yet traversed.

In step 1, we started at (0,0) and went right, so we haven't gone down the left side.

So after step 10, we are at (2,4), and we need to go to (0,4), but that's already done, or to (0,0).

From (2,4) , we can go left to (0,4), but that's redundant.

Let's start over.

Start at (0,0)

Go down to (0,2) // west of A

Go right to (2,2) // south of A? A is from y=0-2, so south is y=2, x=0-2, but at y=2, it's the bottom of A, and also the top of C.

From (0,2) to (2,2) is along the bottom of A, but this is shared with top of C, so not exposed! Oh no.

In the shape, the bottom of A is shared with top of C, so it is not part of the perimeter.

So we cannot go along there.

The correct path must avoid internal edges.

So from (0,0) , go right to (4,0) // top

Down to (4,2) // right of B

Left to (2,2) // bottom of B (exposed)

Down to (2,4) // east of C (exposed, since no square to east)

Left to (0,4) // top of D (exposed)

Down to (0,6) // west of D

Down to (0,8) // west of E

Right to (2,8) // south of E

Up to (2,6) // east of E

Up to (2,4) // east of D

Now at (2,4), and we need to connect to the left.

From (2,4) , go left to (0,4) , but that's already done, and it's the same point.

From (2,4) , the only way is to go up, but up is to (2,2), which is already visited.

We have not included the west side of A and C.

From (0,0) , instead of going right first, go down.

Start at (0,0)

Go down to (0,2) // west of A — is this exposed? Yes, because no square to west.

Then from (0,2) , go down to (0,4) // west of C — exposed

Then down to (0,6) // west of D

Then down to (0,8) // west of E

Then right to (2,8) // south of E

Then up to (2,6) // east of E

Then up to (2,4) // east of D

Then up to (2,2) // east of C

Then right to (4,2) // but at (2,2) to (4,2), this is the bottom of B, but B is from x=2-4, y=0-2, so at y=2, it's the bottom of B, which is exposed, and from (2,2) to (4,2) is along the bottom of B.

Then from (4,2) up to (4,0) // right of B

Then left to (0,0) // but from (4,0) to (0,0) is the top, which is exposed.

So path:

1. (0,0) to (0,2) // west A

2. (0,2) to (0,4) // west C

3. (0,4) to (0,6) // west D

4. (0,6) to (0,8) // west E

5. (0,8) to (2,8) // south E

6. (2,8) to (2,6) // east E

7. (2,6) to (2,4) // east D

8. (2,4) to (2,2) // east C

9. (2,2) to (4,2) // south B

10. (4,2) to (4,0) // east B

11. (4,0) to (0,0) // north A and B

Now, this is a closed path.

Lengths:
1. 2 cm (down)
2. 2 cm (down)
3. 2 cm (down)
4. 2 cm (down)
5. 2 cm (right)
6. 2 cm (up)
7. 2 cm (up)
8. 2 cm (up)
9. 2 cm (right)
10. 2 cm (up)
11. 4 cm (left) // from x=4 to x=0, so 4 cm

Sum: 2*10 + 4 = 20 + 4 = 24 cm? But earlier calculation suggested 32.

Segments 1 to 10 are 2 cm each, that's 10 segments *2 = 20 cm, segment 11 is 4 cm, total 24 cm.

But according to shared sides, it should be 32 cm.

I see the mistake: in segment 11, from (4,0) to (0,0), that's 4 cm, but in the shape, the top is from (0,0) to (4,0), which is correct, 4 cm.

But in the shared sides method, I had 5 squares, 40 cm initial, 4 shared sides, each shared side removes 2 cm from perimeter, so 40 - 8 = 32 cm.

Why discrepancy?

Let's list the shared sides again.

Squares: A(1,1), B(1,2), C(2,1), D(3,1), E(4,1) — using row,col, with row1 top.

Shared sides:
- A and B: share the side between them, horizontal, at y=0-2, x=2 (between x=2 and x=2, but it's the line x=2, y=0-2) — this is one shared side.
- A and C: share the side at y=2, x=0-2 (bottom of A, top of C) — one shared side.
- C and D: share at y=4, x=0-2 (bottom of C, top of D) — one shared side.
- D and E: share at y=6, x=0-2 (bottom of D, top of E) — one shared side.

So 4 shared sides.

Each shared side means that 2 cm is not on the perimeter for each square, but since it's shared, in the total perimeter, we subtract 2 cm per shared side (because that side is internal).

Initial total perimeter if separate: 5 squares * 4 sides * 2 cm = 40 cm

Each shared side reduces the perimeter by 2 cm (because that side is no longer on the boundary; it's internal).

So for 4 shared sides, reduce by 8 cm, so 40 - 8 = 32 cm.

But in the path, I got 24 cm, so where are the missing 8 cm?

In the path I described, I have 11 segments, but let's calculate the length:

1. (0,0) to (0,2): 2 cm
2. (0,2) to (0,4): 2 cm
3. (0,4) to (0,6): 2 cm
4. (0,6) to (0,8): 2 cm
5. (0,8) to (2,8): 2 cm
6. (2,8) to (2,6): 2 cm
7. (2,6) to (2,4): 2 cm
8. (2,4) to (2,2): 2 cm
9. (2,2) to (4,2): 2 cm
10. (4,2) to (4,0): 2 cm
11. (4,0) to (0,0): 4 cm

Sum: 2*10 = 20, plus 4 = 24 cm.

But this path includes the top from (4,0) to (0,0), which is 4 cm, correct.

However, in this path, when I go from (0,0) down to (0,2), that's the west of A, good.

Then down to (0,4), west of C, good.

But between A and C, at y=2, x=0-2, that is the shared side, which is not included, good.

Then from (0,4) to (0,6), west of D, good.

etc.

But what about the top of A and B? In segment 11, I have from (4,0) to (0,0), which covers the top of both A and B, 4 cm, good.

Right of B: from (4,0) to (4,2), 2 cm, good.

Bottom of B: from (4,2) to (2,2), 2 cm, good.

Then from (2,2) to (2,4), which is the east of C, 2 cm, good.

Then to (2,6), east of D, 2 cm.

To (2,8), east of E, 2 cm.

Then to (0,8), south of E, 2 cm.

Then up to (0,6), west of E? No, from (0,8) to (0,6) is not direct; in my path, I have from (0,8) to (2,8) (south of E), then up to (2,6) (east of E), then up to (2,4) (east of D), then up to (2,2) (east of C), then right to (4,2) (south of B), then up to (4,0) (east of B), then left to (0,0) (top).

I think I have all, but let's calculate the actual distance.

Perhaps the error is that when I go from (2,2) to (4,2), that's 2 cm, but in the shape, at y=2, from x=2 to x=4, that is the bottom of B, which is correct.

But for the left side, from (0,0) to (0,8), that's 8 cm, which is covered in segments 1,2,3,4: 2+2+2+2=8 cm, good.

Then from (0,8) to (2,8): 2 cm (south of E)

Then (2,8) to (2,6): 2 cm (east of E)

(2,6) to (2,4): 2 cm (east of D)

(2,4) to (2,2): 2 cm (east of C)

(2,2) to (4,2): 2 cm (south of B)

(4,2) to (4,0): 2 cm (east of B)

(4,0) to (0,0): 4 cm (top)

Sum: 8 (left) + 2 (south E) + 2 (east E) + 2 (east D) + 2 (east C) + 2 (south B) + 2 (east B) + 4 (top) = let's add: 8+2=10, +2=12, +2=14, +2=16, +2=18, +2=20, +4=24 cm.

But according to shared sides, it should be 32 cm.

I see the mistake in the shared sides method.

When two squares share a side, that side is not on the perimeter, so for each shared side, we subtract 2 cm from the total possible perimeter.

But in the initial 40 cm, that includes all sides, including the shared ones.

For each shared side, since it is shared by two squares, in the initial count, it is counted twice (once for each square), but in reality, for the perimeter, it should not be counted at all, so we need to subtract 2 cm for each shared side (because it was counted twice, but should be zero, so subtract 2).

Yes, so for 4 shared sides, subtract 8 cm, 40-8=32 cm.

But in the path, I have only 24 cm, so why?

Let's calculate the perimeter by another method.

The shape can be seen as a combination.

Notice that the shape has a "stem" of 4 squares on left, and a "cap" of one square on top right.

The cap (B) is attached to A.

So the perimeter of the stem alone: if only C,D,E,A, but A is attached to B.

Consider the bounding box.

Min x=0, max x=4, min y=0, max y=8, so width 4 cm, height 8 cm, but not filled.

The perimeter can be calculated as the length of the boundary.

Let's list all the outer edges explicitly.

For square A (0,0)-(2,2):
- North: exposed (part of top)
- South: shared with C, not exposed
- West: exposed
- East: shared with B, not exposed

For square B (2,0)-(4,2):
- North: exposed (part of top)
- South: exposed (no square below)
- West: shared with A, not exposed
- East: exposed

For square C (0,2)-(2,4):
- North: shared with A, not exposed
- South: shared with D, not exposed
- West: exposed
- East: exposed (no square to east)

For square D (0,4)-(2,6):
- North: shared with C, not exposed
- South: shared with E, not exposed
- West: exposed
- East: exposed

For square E (0,6)-(2,8):
- North: shared with D, not exposed
- South: exposed
- West: exposed
- East: exposed

Now, list all exposed sides:

A: north, west → 2 sides
B: north, south, east → 3 sides
C: west, east → 2 sides
D: west, east → 2 sides
E: south, west, east → 3 sides

Total exposed sides: 2+3+2+2+3 = 12 sides

Each side 2 cm, so 24 cm.

But earlier shared sides method gave 32, which is wrong because I miscounted the shared sides.

In the shared sides, I said 4 shared sides, but let's list the shared interfaces:

- A and B share one side (east-west between them)
- A and C share one side (north-south between them)
- C and D share one side
- D and E share one side
That's 4, but each shared side corresponds to one interface, and for each, we subtract 2 cm from the initial 40 cm.

40 - 8 = 32, but now from direct count, 12 sides *2 = 24 cm.

The error is that in the initial 40 cm, for 5 squares, 5*4*2=40 cm, but when two squares share a side, that side is counted in both squares' perimeter, so in the 40 cm, it is counted twice, but for the actual perimeter, it should not be counted at all, so we need to subtract 2 cm for each shared side (because it was counted twice, should be zero, so net subtract 2 cm per shared side).

So for 4 shared sides, subtract 8 cm, 40-8=32 cm.

But direct count shows 12 sides, 24 cm.

Unless I missed some exposed sides.

For square C: I said west and east exposed, but is the north and south shared, yes.

But in the shape, is there any other exposure?

For example, between B and C, there is no shared side, but the area between is empty, but that doesn't add extra sides; the sides are already accounted for.

Let's calculate the perimeter using the formula for polyominoes.

I recall that for a polyomino, the perimeter P = 2* (number of squares) *2 + 2* number of "holes" or something, but perhaps not.

Another way: the perimeter is 2* (width + height) for a rectangle, but for irregular, use the number of edge adjacencies.

Let's count the number of unit edges on the boundary.

From the coordinate system, the boundary consists of line segments.

From the path I described earlier, with 11 segments totaling 24 cm, and it seems correct, and the direct count of exposed sides is 12, so 24 cm.

But why did the shared sides method fail? Because when I said "initial 40 cm", that is correct, and for each shared side, since it is internal, we subtract 2 cm, but in this case, for the 4 shared sides, we subtract 8 cm, get 32, but 32 > 24, so impossible.

I think I found the error: in the shared sides method, when two squares share a side, that side is not on the perimeter, so for the total perimeter, we have to remove that side from both squares' contribution.

In the initial 40 cm, each square's 4 sides are included, so for a shared side, it is included in both squares' perimeter, so it is counted twice in the 40 cm, but in reality, for the combined shape, that side is not on the perimeter, so we need to subtract 2 cm for each shared side (because it was counted twice, should be zero).

So for 4 shared sides, subtract 8 cm, 40-8=32 cm.

But direct count shows 24 cm, so contradiction.

Unless the number of shared sides is not 4.

Let's list the shared sides again.

Square A and B: they share the side at x=2, y=0 to 2. This is one shared side.

Square A and C: they share the side at y=2, x=0 to 2. This is one shared side.

Square C and D: share at y=4, x=0 to 2. One shared side.

Square D and E: share at y=6, x=0 to 2. One shared side.

That's 4.

But perhaps for square B, is it sharing with anyone else? No.

Now, in the direct count, for square A: exposed sides: north and west — 2 sides.

But north is from (0,0) to (2,0), west from (0,0) to (0,2).

For square B: north from (2,0) to (4,0), south from (2,2) to (4,2), east from (4,0) to (4,2) — 3 sides.

For square C: west from (0,2) to (0,4), east from (2,2) to (2,4) — 2 sides.

For square D: west from (0,4) to (0,6), east from (2,4) to (2,6) — 2 sides.

For square E: south from (0,8) to (2,8), west from (0,6) to (0,8), east from (2,6) to (2,8) — 3 sides.

Now, when we combine, the shared sides are not included, good.

But is there any side that is exposed but not listed? For example, the side between B and the empty space below is included as south of B.

Now, the total number of exposed side-segments is 2+3+2+2+3 = 12, each 2 cm, so 24 cm.

But let's calculate the actual length by adding the lengths of the exposed sides.

For A: north 2 cm, west 2 cm — 4 cm

B: north 2 cm, south 2 cm, east 2 cm — 6 cm

C: west 2 cm, east 2 cm — 4 cm

D: west 2 cm, east 2 cm — 4 cm

E: south 2 cm, west 2 cm, east 2 cm — 6 cm

Sum: 4+6+4+4+6 = 24 cm.

So why does the shared sides method give 32? Because in the shared sides method, when I say "initial 40 cm", that is correct, and for each shared side, I subtract 2 cm, but for 4 shared sides, 8 cm, 32 cm, but 32 > 24, so the only explanation is that I have more shared sides or less.

Perhaps when squares are adjacent, but in this case, is there a shared side between B and C? No, because B is at (2,0)-(4,2), C at (0,2)-(2,4), so they are not adjacent; they are diagonal, so no shared side.

So only 4 shared sides.

But 40 - 8 = 32, but actual is 24, so difference of 8 cm, which is 4 sides, so perhaps I have 8 shared sides? No.

I think I know: in the initial 40 cm, for 5 squares, 5*8 = 40 cm? 5 squares * 4 sides * 2 cm = 40 cm, yes.

Each shared side means that 2 cm is not on the perimeter, but since it is shared, in the 40 cm, it is included for both squares, so for each shared side, we have counted 4 cm in the 40 cm (2 cm for each square), but it should be 0, so we need to subtract 4 cm per shared side? Let's think.

For a single shared side between two squares, in the separate count, that side is counted twice (once for each square), so 4 cm in total for that side in the 40 cm.

In the combined shape, that side is internal, so it should contribute 0 to the perimeter.

Therefore, for each shared side, we need to subtract 4 cm from the initial 40 cm.

For 4 shared sides, subtract 16 cm, 40-16=24 cm. Yes! That matches.

So the correct way: each shared side reduces the perimeter by 4 cm (because it was counted twice in the initial sum, but should be zero).

So for shape 2: 5 squares, 4 shared sides, perimeter = 5*8 - 4*4 = 40 - 16 = 24 cm.

Good.

So for Problem 2: 24 cm.

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Problem 3:
Shape is a staircase: 3 squares on left column, then 2 on middle, then 1 on right, but offset.

From image: it's like a right triangle made of squares.

Positions:
Row1: col1
Row2: col1, col2
Row3: col1, col2, col3

So squares: (1,1), (2,1), (2,2), (3,1), (3,2), (3,3) — 6 squares.

Shared sides:
- (1,1) and (2,1): vertical share
- (2,1) and (2,2): horizontal share
- (2,1) and (3,1): vertical share
- (2,2) and (3,2): vertical share
- (3,1) and (3,2): horizontal share
- (3,2) and (3,3): horizontal share

List:
Vertical shares:
- col1: (1,1)-(2,1), (2,1)-(3,1) → 2
- col2: (2,2)-(3,2) → 1
- col3: none
Horizontal shares:
- row2: (2,1)-(2,2) → 1
- row3: (3,1)-(3,2), (3,2)-(3,3) → 2
Total shared interfaces: 2+1+1+2 = 6

Initial perimeter if separate: 6*4*2 = 48 cm

Each shared side reduces by 4 cm (as above), so 6*4 = 24 cm reduction

Perimeter = 48 - 24 = 24 cm

Direct count: let's verify.

Exposed sides:

(1,1): north, west, east? East is shared with nothing? (1,1) is at row1 col1, so east is to col2, but no square at (1,2), so east exposed? In the shape, only (1,1) in row1, so yes.

Squares:
A=(1,1), B=(2,1), C=(2,2), D=(3,1), E=(3,2), F=(3,3)

A: north, west, east (since no square to east or north or west, but south shared with B) — so north, west, east exposed — 3 sides

B: south shared with D, north shared with A, west exposed, east shared with C — so only west exposed? Let's see.

B at (2,1): north shared with A south, south shared with D north, west exposed (no left), east shared with C west — so only west exposed — 1 side

C at (2,2): north exposed (no above), south shared with E north, west shared with B east, east exposed (no right) — so north and east exposed — 2 sides

D at (3,1): north shared with B south, south exposed, west exposed, east shared with E west — so south and west exposed — 2 sides

E at (3,2): north shared with C south, south exposed, west shared with D east, east shared with F west — so only south exposed? South is exposed, and if no other, but west and east are shared, north shared, so only south — 1 side

F at (3,3): north exposed, south exposed, west shared with E east, east exposed — so north, south, east exposed — 3 sides

Sum: A:3, B:1, C:2, D:2, E:1, F:3 = 12 sides

12*2 = 24 cm, matches.

So Problem 3: 24 cm

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Problem 4:
Shape is L-shaped: 3 squares down on left, then 2 squares to the right at the bottom.

Positions:
Row1: col1
Row2: col1
Row3: col1, col2, col3

So squares: (1,1), (2,1), (3,1), (3,2), (3,3) — 5 squares.

Shared sides:
- (1,1)-(2,1) vertical
- (2,1)-(3,1) vertical
- (3,1)-(3,2) horizontal
- (3,2)-(3,3) horizontal
Total 4 shared sides.

Initial: 5*8 = 40 cm

Reduction: 4*4 = 16 cm

Perimeter = 40 - 16 = 24 cm

Direct count:
A=(1,1): north, west, east (south shared with B) — 3 sides
B=(2,1): north shared with A, south shared with C, west exposed, east exposed (no right) — so west and east exposed — 2 sides
C=(3,1): north shared with B, south exposed, west exposed, east shared with D — so south and west exposed — 2 sides
D=(3,2): north exposed, south exposed, west shared with C, east shared with E — so north and south exposed — 2 sides
E=(3,3): north exposed, south exposed, west shared with D, east exposed — so north, south, east exposed — 3 sides
Sum: 3+2+2+2+3 = 12 sides, 24 cm.

So Problem 4: 24 cm

---

Problem 5:
Shape is like a zigzag: two squares on top row, then one below the left, then one to the right of that, but let's see.

From image: it's two squares side by side on top, then below the left square, one square, then to the right of that, one square, so like a snake.

Positions:
Row1: col1, col2
Row2: col1
Row3: col2 (since below col2 of row1? No.

Typically:
- (1,1), (1,2)
- (2,1)
- (3,2) // below (1,2)? But (1,2) is at row1 col2, so below would be row2 col2, but in row2, only col1 has a square, so (3,2) is at row3 col2.

So squares: A=(1,1), B=(1,2), C=(2,1), D=(3,2)

Shared sides:
- A and B: horizontal share
- A and C: vertical share
- B and D: not adjacent; B is (1,2), D is (3,2), not adjacent.
- C and D: not adjacent; C (2,1), D (3,2), diagonal.

So only two shared sides: A-B and A-C.

Initial: 4*8 = 32 cm

Reduction: 2*4 = 8 cm

Perimeter = 32 - 8 = 24 cm

Direct count:
A=(1,1): north, west, east shared with B, south shared with C — so north and west exposed — 2 sides
B=(1,2): north, east, west shared with A, south exposed (no below) — so north, east, south exposed — 3 sides
C=(2,1): north shared with A, south exposed, west exposed, east exposed (no right) — so south, west, east exposed — 3 sides
D=(3,2): north exposed, south exposed, west exposed, east exposed — 4 sides (since isolated except possibly, but no shares)
Sum: 2+3+3+4 = 12 sides, 24 cm.

So Problem 5: 24 cm

---

Problem 6:
Shape is like a plus sign but missing the top, or something.

From image: it's 2x2 on left, then attached to the right, a 2x2 but shifted down, so like two 2x2 squares sharing a corner or edge.

Positions:
Left part: (1,1), (1,2), (2,1), (2,2)
Right part: (2,3), (2,4), (3,3), (3,4) // assuming

So squares: A=(1,1), B=(1,2), C=(2,1), D=(2,2), E=(2,3), F=(2,4), G=(3,3), H=(3,4)

Shared sides:
Within left 2x2:
- A-B horizontal
- A-C vertical
- B-D vertical
- C-D horizontal
Within right 2x2:
- E-F horizontal
- E-G vertical
- F-H vertical
- G-H horizontal
Between left and right:
- D and E: if D is (2,2), E is (2,3), so they share a horizontal side at x=2-4? D is col2, E is col3, so if adjacent, share the side between col2 and col3 at row2.

So D and E share a vertical side? No, if same row, adjacent columns, share a vertical side? In grid, if two squares are side by side horizontally, they share a vertical side? Let's clarify.

If square at (row,col) and (row,col+1), they share the side at x=2*col to 2*(col+1), but the shared side is vertical if they are in the same row? No.

In terms of adjacency, if two squares are in the same row and adjacent columns, they share a vertical side? No, they share a horizontal side? I'm confusing myself.

Standard: if two squares are adjacent horizontally (same row, consecutive columns), they share a vertical edge? No.

Think: each square has left, right, top, bottom sides.

If square A is at col i, square B at col i+1, same row, then the right side of A is shared with the left side of B, and this shared side is vertical? No, the side itself is vertical if it's a left/right side, but the interface is vertical.

In terms of the edge, the shared edge is a vertical line segment.

For example, between (1,1) and (1,2), the shared side is the line x=2, y=0 to 2, which is vertical.

Similarly, between (1,1) and (2,1), shared side is y=2, x=0 to 2, horizontal.

So for D=(2,2) and E=(2,3), they are in the same row, columns 2 and 3, so they share a vertical side at x=4, y=4 to 6? Let's set coordinates.

Assume each square 2x2.

Set (1,1): x=0-2, y=0-2
(1,2): x=2-4, y=0-2
(2,1): x=0-2, y=2-4
(2,2): x=2-4, y=2-4
(2,3): x=4-6, y=2-4
(2,4): x=6-8, y=2-4
(3,3): x=4-6, y=4-6
(3,4): x=6-8, y=4-6

Shared sides:
- A-B: at x=2, y=0-2 (vertical)
- A-C: at y=2, x=0-2 (horizontal)
- B-D: at y=2, x=2-4 (horizontal) — B is (1,2), D is (2,2), so share at y=2, x=2-4
- C-D: at x=2, y=2-4 (vertical) — C (2,1), D (2,2)
- D-E: at x=4, y=2-4 (vertical) — D (2,2), E (2,3)
- E-F: at x=6, y=2-4 (vertical) — E (2,3), F (2,4)
- E-G: at y=4, x=4-6 (horizontal) — E (2,3), G (3,3)
- F-H: at y=4, x=6-8 (horizontal) — F (2,4), H (3,4)
- G-H: at x=6, y=4-6 (vertical) — G (3,3), H (3,4)

Also, within right part, G and H share vertical, already have.

Is there share between D and G? D (2,2), G (3,3), not adjacent.

So shared interfaces:
Horizontal shares: A-C, B-D, E-G, F-H → 4
Vertical shares: A-B, C-D, D-E, E-F, G-H → 5
Total 9 shared sides.

Initial perimeter: 8 squares * 4 sides * 2 cm = 64 cm

Each shared side reduces by 4 cm, so 9*4 = 36 cm reduction

Perimeter = 64 - 36 = 28 cm

Direct count might be tedious, but let's trust for now.

We can calculate.

Since it's symmetric, or use the fact that the shape is almost two 2x2 squares sharing a side.

Two separate 2x2 squares have perimeter 2* (4*2*2) = 2*16 = 32 cm? Each 2x2 square has perimeter 8 cm? No.

A single 2x2 square of small squares: if it's 2 small squares by 2, so size 4cm x 4cm, perimeter 2*(4+4)=16 cm.

For two such squares separate, 32 cm.

When they share a side, if they share a full side, like if attached along a 2cm edge, then they share one interface of 2 cm, but in terms of the grid, when two 2x2 blocks share a side, they share 2 small sides? In this case, D and E share one vertical side, which is 2 cm long, but in terms of the small squares, it's one shared side between D and E.

In our case, the left 2x2 and right 2x2 are attached at D and E, which are adjacent, so they share one small side (between D and E).

So for the combined shape, compared to two separate 2x2 blocks, when we attach them, we lose the two sides that are now internal: the right side of D and the left side of E, each 2 cm, so total 4 cm reduction.

Each 2x2 block has perimeter 16 cm, so two separate: 32 cm.

After attaching, perimeter = 32 - 4 = 28 cm, since we remove 2 cm from each block's perimeter for the shared side.

Yes, matches our calculation.

So Problem 6: 28 cm

Now summarize:

1. 32 cm
2. 24 cm
3. 24 cm
4. 24 cm
5. 24 cm
6. 28 cm

But let's double-check problem 1, since it was tricky.

Problem 1: we had 8 squares, with the center of middle row missing, so compared to 3x3=9 squares perimeter 24 cm, removing the center square adds 4 sides of 2 cm each, so +8 cm, total 32 cm, yes.

Problem 6: 28 cm.

So final answers.

Final Answer:
1. 32 cm
2. 24 cm
3. 24 cm
4. 24 cm
5. 24 cm
6. 28 cm