Perimeter with Polly Parakeet - ArgoPrep - Free Printable

Let’s solve each problem step by step. Remember, perimeter is the total distance around the outside of a shape. We add up all the side lengths.

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Problem 1:
The small square has sides of 4 cm each. The big shape is made of 3 rows and 3 columns of these small squares → so it’s a 3x3 grid.

That means:
- Width = 3 × 4 cm = 12 cm
- Height = 3 × 4 cm = 12 cm

It’s a big square! Perimeter = 4 × side = 4 × 12 = 48 cm

Wait — let’s double-check by counting outer edges:

Top: 3 segments × 4 cm = 12 cm
Right: 3 segments × 4 cm = 12 cm
Bottom: 3 segments × 4 cm = 12 cm
Left: 3 segments × 4 cm = 12 cm
Total = 12 + 12 + 12 + 12 = 48 cm ✔

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Problem 2:
Small square side = 3 cm. Shape looks like a “U” or gate.

Let’s count the outer edges:

Top: 3 segments × 3 cm = 9 cm
Right side: goes down 3 segments? Wait — look carefully.

Actually, the shape is 3 units wide and 3 units tall, but with a gap in the middle bottom.

Better to trace the outline:

Start at top-left corner:
→ Right 3 segments (9 cm)
↓ Down 1 segment (3 cm)
← Left 1 segment (3 cm)
↓ Down 1 segment (3 cm)
→ Right 1 segment (3 cm)
↓ Down 1 segment (3 cm)
← Left 3 segments (9 cm)
↑ Up 3 segments (9 cm) — back to start?

Wait, that doesn’t close properly. Let me redraw mentally.

Actually, standard way: this shape is like a rectangle missing the middle bottom part.

Full rectangle would be 3 wide × 3 high → perimeter 2×(3+3)=18 units? But we have extra inner edges.

No — for perimeter, we only care about the *outer* boundary.

Looking at the figure: it’s symmetric.

Top: 3 segments → 9 cm
Right side: from top-right, go down 3 segments? No — actually, on the right, it goes down 1, then left 1, then down 1, then right 1, then down 1? That can’t be.

Wait — better approach: count how many unit edges are on the outside.

Each small square side is 3 cm.

The shape has:

- Top row: 3 squares → top edge: 3 segments
- Middle row: 2 squares (left and right, missing center) → so left and right sides exposed
- Bottom row: 2 squares (left and right) → bottom edges exposed

But for perimeter, walk around:

Start at top-left:
→ Move right 3 units (top) → 3 × 3 = 9 cm
↓ Move down 1 unit (right side of top-right square) → 3 cm
← Move left 1 unit (bottom of top-right square) → 3 cm
↓ Move down 1 unit (right side of middle-right square) → 3 cm
→ Move right 0? Wait no.

I think I’m overcomplicating.

Standard trick: for grid shapes, count the number of unit-length edges on the perimeter.

Each small square side is 3 cm.

In Problem 2, the shape is 3 units wide and 3 units tall, with the center-bottom square missing.

So, imagine a 3x3 grid, remove the bottom-center square.

Now, count outer edges:

Top: 3 edges
Right: 3 edges (full height)
Bottom: left square bottom (1), right square bottom (1) → 2 edges
Left: 3 edges
Plus, the "notch" adds two vertical edges inside? No — when you remove the bottom-center, you expose two new vertical edges (the sides of the hole).

Actually, removing a square from the corner adds 2 edges, but from the middle of a side adds 2 edges too? Let's think.

Original full 3x3 square: perimeter = 12 unit edges (since 3x4=12? No — for n x n grid of squares, perimeter in unit edges is 4n if solid? For 3x3 solid, it's 3*4 = 12 unit edges? Each side has 3 units, so 4*3=12 unit edges. Yes.

When you remove the bottom-center square, you remove 1 square, which had 4 edges, but 2 were shared (with neighbors), so you remove 2 internal edges and expose 2 new edges? Actually, net change: you lose the bottom edge of the removed square, but gain the left and right edges of the hole? And the top was already internal.

Better: original perimeter: 12 unit edges.

Remove bottom-center square:
- You remove its bottom edge (was part of perimeter) → -1
- You remove its top edge (was internal, now becomes perimeter) → +1
- You remove its left and right edges (were internal, now become perimeter) → +2
Net change: -1 +1 +2 = +2

So new perimeter = 12 + 2 = 14 unit edges.

Each unit edge is 3 cm, so 14 × 3 = 42 cm

Let me verify by tracing:

Start at top-left:
→ Right 3 units (top)
↓ Down 1 unit (right side)
← Left 1 unit (along bottom of top-right square)
↓ Down 1 unit (right side of middle-right square)
→ Right 0? No.

After going down 1 from top-right, you're at the top of the middle-right square. Then you go left along the bottom of the top row? But there's no square below the center, so you can't go left yet.

Actually, from top-right corner:
- Go down 1 unit (to between top and middle row on right)
- Now, since there's no square to the left at this level? There is a square to the left (middle-left), but not in the center.

Perhaps draw coordinates.

Assume bottom-left is (0,0). Squares occupy cells.

Squares present:
Row y=2 (top): x=0,1,2
Row y=1 (middle): x=0,2 [missing x=1]
Row y=0 (bottom): x=0,2 [missing x=1]

Perimeter path:

Start at (0,3) [top-left corner of top-left square]
→ to (3,3) : 3 units right
↓ to (3,2) : 1 unit down (right side of top-right square)
← to (2,2) : 1 unit left (bottom of top-right square) — but wait, at y=2, x from 2 to 3 is bottom of top-right square, but below it at y=1, x=2 is present, so this edge is internal? No.

I think I need to think of the outer boundary.

The shape has:

- Top: from x=0 to x=3 at y=3 → length 3
- Right: from y=3 to y=0 at x=3 → but at y=1 and y=0, x=3 is the right side of the right-column squares, so yes, full height 3 units down? From y=3 to y=0 is 3 units.

From (3,3) down to (3,0): that's 3 units.

Then at (3,0), go left to (2,0)? But there's a square at (2,0)? x=2,y=0 is present, so bottom edge from x=2 to x=3 at y=0.

Then from (2,0) , since no square at (1,0), we go up? To where?

At (2,0), we are at bottom-right of bottom-right square. Since no square to the left, we go up along the right side of the hole? But the hole is at x=1,y=0.

Actually, from (2,0), we go left to (1,0)? But there's no square, so we can't; instead, the boundary goes up from (2,0) to (2,1)? Let's list vertices.

Vertices of the polygon:

Start at (0,3)
→ (3,3)
↓ (3,0) [because right side is straight down? But at y=1 and y=0, the right column is there, so yes, from (3,3) to (3,0) is continuous? No, because at y=2 to y=1, it's fine, but the shape is connected.

Actually, the right side is uninterrupted: from top to bottom on the right, it's all there, so from (3,3) to (3,0) is 3 units down.

Then at (3,0), we go left to (2,0) — that's the bottom of the bottom-right square.

Then, since no square at (1,0), we go up to (2,1)? But (2,1) is the bottom of the middle-right square, which is present.

From (2,0) to (2,1): up 1 unit — this is the right side of the hole.

Then from (2,1) to (1,1)? But no square at (1,1), so we go left to (1,1)? But (1,1) is empty.

Actually, from (2,1), since no square to the left, we go left to (1,1), but that's not on the boundary; the boundary should go from (2,1) left to (1,1) only if there's a square, but there isn't.

I recall that for such shapes, the perimeter can be calculated as the perimeter of the bounding box plus twice the depth of any indentations.

Bounding box is 3 wide by 3 high, perimeter 2*(3+3)=12 unit edges.

There is one indentation at the bottom center: it's 1 unit deep and 1 unit wide. When you have a rectangular indentation, it adds 2 times the depth to the perimeter.

Here, depth is 1 unit (since we removed one row), so add 2*1 = 2 unit edges.

Total perimeter = 12 + 2 = 14 unit edges.

Yes, that makes sense.

So 14 × 3 cm = 42 cm

Confirm with another method: count all outer edges.

Top: 3 edges
Bottom: left square bottom (1), right square bottom (1) → 2 edges
Left side: 3 edges (full height)
Right side: 3 edges (full height)
Plus, the two vertical edges of the notch: on the left side of the hole and right side of the hole, each 1 unit high → 2 edges
And the top of the notch? No, the top of the notch is internal.

The notch is at the bottom, so when you remove the bottom-center square, you expose:
- The bottom edge of the middle-center square? But there is no middle-center square; in row y=1, x=1 is missing, so the bottom of the top-center square is exposed? Let's see.

Squares:
- Top row: three squares, so their bottom edges are partially exposed.
Specifically, the bottom edge of the top-center square is exposed because there's no square below it.
Similarly, the left and right edges of the missing squares are exposed.

For the missing bottom-center square (x=1,y=0):
- Its top edge is now exposed (was shared with top-center square? No, top-center is at y=1,x=1, which is also missing? In our case, for problem 2, the shape has:
Looking back at the image description: it's a 3x3 grid with the center of the bottom row missing? Or is it the center of the entire grid?

In the user's image, for problem 2, it shows a shape that is 3 units wide and 3 units tall, with the middle of the bottom cut out, but actually, from the drawing, it seems like it's missing the center square of the 3x3, but that would make it a ring, but the drawing shows a U-shape, so likely only the bottom-center is missing, and the middle-center is present? No, in a U-shape, typically the bottom-center is missing, and the middle is full.

Let me assume from common problems: for a 3x3 grid with the bottom-center square removed, the perimeter is 14 unit edges.

Yes, standard result.

So 14 * 3 = 42 cm.

I'll go with that.

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Problem 3:
Small square side = 5 cm. Shape is like a "C" or backward C.

It's 2 units wide and 5 units tall, with the right side indented in the middle.

Specifically, it's a rectangle 2x5, but with the middle three units on the right side missing? Looking at the drawing: it has 5 rows, and in each row, there are 2 squares except possibly some.

From the image: it's a vertical strip of 5 squares high, and 2 squares wide, but the right column is missing for the middle three rows? No.

Actually, it's shaped like a C: so left column full 5 squares, right column only top and bottom squares, missing the middle three.

So, squares present:
- Left column: y=0 to 4 (5 squares)
- Right column: y=0 and y=4 only (top and bottom), missing y=1,2,3

So, it's like a frame open on the right in the middle.

Perimeter: let's calculate.

Bounding box: 2 wide x 5 high, perimeter 2*(2+5)=14 unit edges.

But there are indentations: on the right side, from y=1 to y=3, it's indented inward by 1 unit.

Each "step" adds to the perimeter.

Specifically, for the right side, instead of a straight line, it goes in and out.

From top to bottom on right:
- At y=4 to y=3: down 1 unit (right side of top-right square)
- Then left 1 unit (bottom of top-right square) — but since no square below, this is exposed
- Then down 3 units? No.

Trace the boundary:

Start at top-left (0,5) assuming y from 0 at bottom.

Set coordinates: let bottom-left be (0,0).

Squares:
- Left column: x=0, y=0,1,2,3,4
- Right column: x=1, y=0 and y=4

So, the shape occupies:
(0,0),(0,1),(0,2),(0,3),(0,4), (1,0), (1,4)

Now, perimeter path:

Start at (0,5) [top-left of top-left square]
→ to (1,5) : right 1 unit (top of top-left square) — but there's a square at (1,4), so top of it is at y=5, x=1 to 2? Let's define.

Each square is 1x1 unit for counting, then multiply by 5 cm later.

Square at (i,j) occupies [i,i+1] x [j,j+1].

So:
- Square (0,4): [0,1]x[4,5]
- Square (0,3): [0,1]x[3,4]
- ...
- Square (0,0): [0,1]x[0,1]
- Square (1,4): [1,2]x[4,5]
- Square (1,0): [1,2]x[0,1]

Now, outer boundary:

Start at (0,5)
→ to (2,5) : because both top-left and top-right squares have top at y=5, from x=0 to x=2? Square (0,4) covers x=0 to1, y=4 to5; square (1,4) covers x=1 to2, y=4 to5, so together, top edge from x=0 to x=2 at y=5 → length 2 units.

Then ↓ to (2,4) : down 1 unit (right side of top-right square)

Then ← to (1,4) : left 1 unit (bottom of top-right square) — but at y=4, x=1 to2 is bottom of square (1,4), and below it, at y=3, there is no square at x=1, so this edge is exposed.

Then ↓ to (1,1) ? From (1,4) down to (1,1)? But there are no squares in between on the right, so yes, down 3 units to (1,1) — this is the left side of the hole.

Then → to (2,1) : right 1 unit? But at y=1, x=1 to2, is there a square? No, square (1,0) is at y=0 to1, so at y=1, it's the top of square (1,0).

From (1,1), we go down to (1,0)? Let's see.

After reaching (1,1), since no square to the right, and below there is square (1,0), so we go down to (1,0) — down 1 unit.

Then → to (2,0) : right 1 unit (top of bottom-right square? Square (1,0) is [1,2]x[0,1], so its top is at y=1, but we are at (1,0), which is bottom-left of it.

From (1,1) [which is top-left of square (1,0)? Square (1,0) has corners at (1,0),(2,0),(2,1),(1,1).

So at (1,1), we are at top-left of square (1,0).

Then we can go right to (2,1) : along the top of square (1,0) — but is this exposed? Below it is nothing, but above it is nothing, so yes, it's the top edge.

Then ↓ to (2,0) : down 1 unit (right side of bottom-right square)

Then ← to (0,0) : left 2 units (bottom of both bottom squares) — square (0,0) and (1,0) cover x=0 to2 at y=0, so bottom edge from x=0 to x=2 at y=0 → length 2 units.

Then ↑ to (0,5) : up 5 units (left side of left column) — since left column is full from y=0 to y=5, so from (0,0) to (0,5) is 5 units up.

Now, let's list all segments:

1. (0,5) to (2,5) : 2 units right
2. (2,5) to (2,4) : 1 unit down
3. (2,4) to (1,4) : 1 unit left
4. (1,4) to (1,1) : 3 units down [from y=4 to y=1]
5. (1,1) to (2,1) : 1 unit right
6. (2,1) to (2,0) : 1 unit down
7. (2,0) to (0,0) : 2 units left
8. (0,0) to (0,5) : 5 units up

Sum: 2+1+1+3+1+1+2+5 = let's add: 2+1=3, +1=4, +3=7, +1=8, +1=9, +2=11, +5=16 units.

Is that correct? Let me check if we missed anything.

From (1,4) to (1,1): that's down 3 units, but is there a square at (0,3), etc.? The left side is covered by the last segment.

Total 16 unit edges.

Each unit edge is 5 cm, so 16 × 5 = 80 cm

Another way: the shape has a certain number of exposed edges.

Left side: 5 units (full height)
Top: 2 units (both top squares)
Right side: only the top and bottom parts: from y=4 to5 and y=0 to1, so 2 units, but also the inner parts.

In the indentation, we have additional edges.

Total perimeter should be: for a solid 2x5 rectangle, perimeter = 2*(2+5)=14 unit edges.

But we removed three squares from the right column: at y=1,2,3.

Each removed square: if it was on the edge, removing it adds edges.

Specifically, for each removed square that was on the boundary, but in this case, the squares at (1,1),(1,2),(1,3) were on the right boundary.

When you remove a square from the side, you remove one edge (the outer one) but add three new edges (the other three sides), so net +2 per removed square.

Here, we removed three squares, each was on the right edge, so for each, net +2 unit edges to perimeter.

Original perimeter 14, plus 3*2 = 6, total 20? But that can't be because we got 16.

Mistake: when you remove multiple adjacent squares, the added edges may share.

In this case, removing three consecutive squares from the right side.

Originally, the right side was a straight line of 5 units.

After removing the middle three, the right side now has: top 1 unit, then a "bay" that goes left 1 unit, down 3 units, right 1 unit, then bottom 1 unit.

So compared to original straight right side of 5 units, now we have: top 1, then instead of down 3, we have left 1, down 3, right 1, then bottom 1.

So the path on the right is: down 1 (top), then left 1, down 3, right 1, down 1 (bottom) — but the down 1 at bottom is separate.

From the tracing, we had on the right: from (2,5) to (2,4) : down 1
then to (1,4) : left 1
then to (1,1) : down 3
then to (2,1) : right 1
then to (2,0) : down 1

So for the right part, we have segments: down1, left1, down3, right1, down1 — total length 1+1+3+1+1=7 units for what was originally 5 units down on the right.

Originally, right side was 5 units down.

Now it's 7 units, so increase of 2 units.

Additionally, the top and bottom are unchanged? Top is still 2 units, bottom is 2 units, left is 5 units.

Original perimeter: left 5, right 5, top 2, bottom 2, total 14.

New: left 5, top 2, bottom 2, and right side modified to 7 units, so total 5+2+2+7=16, same as before.

Yes.

So 16 unit edges × 5 cm = 80 cm

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Problem 4:
Small square side = 2 cm. Shape is 4 units wide and 3 units tall? From the grid: it's 4 columns and 3 rows of small squares.

So width = 4 × 2 = 8 cm
Height = 3 × 2 = 6 cm

It's a rectangle, so perimeter = 2 × (width + height) = 2 × (8 + 6) = 2 × 14 = 28 cm

Count edges: top 4 segments, bottom 4, left 3, right 3, total 4+4+3+3=14 segments, each 2 cm, 14×2=28 cm. ✔

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Problem 5:
Small square side = 4 cm. Shape is like a bottle or podium: base is 3 units wide and 3 units tall, with a 1-unit wide tower on top center.

So, the base: 3x3 squares.
On top, centered, a 1x1 square on top of the middle of the base.

So overall, width is 3 units, height is 4 units (3 for base +1 for tower).

But the tower is only in the middle, so the top is not flat.

Perimeter: let's trace.

Start at bottom-left.

Assume bottom-left corner of base.

Base: from x=0 to3, y=0 to3.
Tower: on top of base, centered, so since base is 3 wide, center is at x=1.5, so tower from x=1 to2, y=3 to4.

Squares:
- Base: all (i,j) for i=0,1,2; j=0,1,2 [3x3]
- Tower: (1,3) [since y=3 to4]

Now, perimeter path:

Start at (0,0)
→ to (3,0) : right 3 units (bottom)
↑ to (3,3) : up 3 units (right side of base)
← to (2,3) : left 1 unit (bottom of tower? At y=3, x=2 to3 is top of base-right square, but tower is at x=1 to2, so from (3,3) to (2,3) is left along the top of the base-right square.

Then ↑ to (2,4) : up 1 unit (right side of tower)
← to (1,4) : left 1 unit (top of tower)
↓ to (1,3) : down 1 unit (left side of tower)
← to (0,3) : left 1 unit (top of base-left square? From (1,3) to (0,3) is left along the top of the base-left square, but at y=3, x=0 to1 is top of base-left square.

Then ↓ to (0,0) : down 3 units (left side of base)

List segments:

1. (0,0) to (3,0) : 3 right
2. (3,0) to (3,3) : 3 up
3. (3,3) to (2,3) : 1 left
4. (2,3) to (2,4) : 1 up
5. (2,4) to (1,4) : 1 left
6. (1,4) to (1,3) : 1 down
7. (1,3) to (0,3) : 1 left
8. (0,3) to (0,0) : 3 down

Sum: 3+3+1+1+1+1+1+3 = let's add: 3+3=6, +1=7, +1=8, +1=9, +1=10, +1=11, +3=14 units.

Each unit is 4 cm, so 14 × 4 = 56 cm

Check: without the tower, base 3x3 square, perimeter 12 units.

Adding a 1x1 square on top center: it covers the top-middle of the base, so we remove the top edge of the base-middle square (1 unit), but add three new edges for the tower: left, top, right (3 units), so net +2 units.

Original 12, plus 2 = 14 units, yes.

So 14 × 4 = 56 cm.

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Now, summarize answers:

1. 48 cm
2. 42 cm
3. 80 cm
4. 28 cm
5. 56 cm

Final Answer:
1. 48 cm
2. 42 cm
3. 80 cm
4. 28 cm
5. 56 cm