Perimeter of a Shape worksheet with nine figures made of unit squares for calculating perimeter.

Let's solve the perimeter of each shape in the worksheet step by step. The shapes are made up of 1 cm × 1 cm squares, so each side of a square is 1 cm long.

We will calculate the perimeter — the total length around the outside edge of each shape — by counting the number of unit-length sides that form the outer boundary.

---

a)

This is a 3 cm × 4 cm rectangle.

- Length = 4 cm
- Width = 3 cm

Perimeter = $ 2 \times (length + width) = 2 \times (4 + 3) = 2 \times 7 = 14 $ cm

✔ Answer: 14 cm

---

b)

This shape is an L-shaped figure made of 5 squares.

Let’s trace the outer edges:

- Top row: 3 squares → top edge = 3 cm
- Right side: 1 vertical segment down from top right → 1 cm
- Then bottom right: 1 cm left → 1 cm
- Then down to bottom-left corner: 1 cm
- Then left along bottom: 3 cm
- Then up on left side: 2 cm
- But wait — let’s count each exposed side.

Alternatively, count all outer edges:

Each small square has 4 sides, but shared sides are internal and not counted.

Total squares: 5
Each contributes 4 sides → $ 5 \times 4 = 20 $ sides
But every internal edge (shared between two squares) is counted twice — subtract 2 for each shared side.

Count shared sides:
- Horizontal shared: 2 (top row has 3 squares → 2 shared)
- Vertical shared: 2 (right column has 2 squares → 1 shared), and middle column has 2 squares → 1 shared? Wait — better to sketch.

Actually, visualizing:

```
□ □ □
□
□
```

Wait — no, it's like:

```
□ □ □
□
□
```

So:
- Top row: 3 squares
- Below the middle one: 2 more squares

Shared edges:
- Between top-left and top-middle: 1
- Between top-middle and top-right: 1
- Between top-middle and bottom-middle: 1
- Between bottom-middle and bottom-right: 1

Total shared edges: 4 → remove 4 × 2 = 8 sides (since each shared edge removes 2 from total)

Total outer edges: $ 20 - 8 = 12 $ → 12 cm?

Wait — this method is error-prone.

Better: Trace the perimeter.

Start at top-left corner:

- Go right: 3 cm
- Down: 1 cm
- Right: 1 cm
- Down: 1 cm
- Left: 1 cm
- Down: 1 cm
- Left: 1 cm
- Up: 1 cm
- Left: 1 cm
- Up: 1 cm
- Right: 1 cm
- Up: 1 cm

Wait — better to go clockwise.

Start at top-left of top-left square:

1. Right → 3 cm (top edge)
2. Down → 1 cm (right edge of first square)
3. Right → 1 cm (to next square)
4. Down → 1 cm
5. Right → 1 cm
6. Down → 1 cm
7. Left → 1 cm
8. Down → 1 cm
9. Left → 1 cm
10. Up → 1 cm
11. Left → 1 cm
12. Up → 1 cm
13. Right → 1 cm
14. Up → 1 cm

Wait — too many steps.

Better: Count all outer edges.

Look at the shape:

```
■ ■ ■
■
■
```

Top edge: 3 cm
Right edge: 3 cm (from top-right to bottom-right)
Bottom edge: 1 cm (bottom-left to bottom-right)
Left edge: 2 cm (from bottom-left to top-left of leftmost square)
Then the "indent" on the left: we have a missing square on the left side.

Actually, let's label the grid.

Assume:
- Row 1: columns 1,2,3 → filled
- Row 2: column 2 → filled
- Row 3: column 2 → filled

So the outline:

- Top: from (1,1) to (1,3): 3 cm
- Right: from (1,3) down to (3,3): but (2,3) and (3,3) are empty → only (1,3) to (2,3)? No.

Wait — coordinates:

Let’s define:
- Each square is 1 cm
- Coordinates: (row, col)

Squares present:
- (1,1), (1,2), (1,3)
- (2,2)
- (3,2)

Now trace the outer path:

Start at (1,1) top-left:

1. Right → (1,3): 3 cm
2. Down → (2,3): but (2,3) is empty → so go down from (1,3) to (2,3) → 1 cm
3. Left → (2,2): but (2,2) is occupied → so can’t go through
Instead, go down from (1,3) → (2,3) → then left to (2,2)? No — (2,2) is there, but the side is shared.

Better: Walk the outer boundary.

Start at top-left corner of (1,1):

- Move right along top of (1,1), (1,2), (1,3): 3 cm
- Down right side of (1,3): 1 cm to (2,3)
- Now, (2,3) is empty, so continue down? No — (2,3) is air, so we must turn left?
No — the shape ends here.

Wait — the shape has no square at (2,3), so the right side stops at (1,3).

From (1,3) go down: only 1 cm down to (2,3), but that’s not part of shape.

So the right edge is only 1 cm high.

Then from (2,3), we need to go left to (2,2), but (2,2) is occupied, so we go down from (1,3) → (2,3) → then left to (2,2)? No — that would be inside.

Actually, the outer boundary:

- Start at (1,1) top-left
- Right → (1,3): 3 cm
- Down → (2,3): 1 cm
- Left → (2,2): 1 cm (but this is internal? No — (2,2) is a square, so its right side is exposed?)

Wait — (2,2) has:
- Left: adjacent to nothing → exposed
- Right: adjacent to nothing → exposed
- Top: adjacent to (1,2) → shared
- Bottom: adjacent to (3,2) → shared

So (2,2) has two exposed sides: left and right

Similarly, (3,2) has:
- Left: exposed
- Right: exposed
- Top: shared with (2,2)
- Bottom: exposed

Let’s list all exposed edges:

For each square, count how many sides are on the boundary.

Square (1,1):
- Top: exposed → 1
- Left: exposed → 1
- Bottom: adjacent to (2,1)? No → (2,1) is empty → exposed → 1
- Right: adjacent to (1,2) → shared → not exposed

→ 3 exposed sides

Square (1,2):
- Top: exposed → 1
- Left: adjacent to (1,1) → shared → no
- Right: adjacent to (1,3) → shared → no
- Bottom: adjacent to (2,2) → shared → no

→ Only top exposed → 1 side

Wait — no! (1,2) has:
- Top: yes
- Left: shared with (1,1)
- Right: shared with (1,3)
- Bottom: shared with (2,2)

All except top are shared → only top exposed → 1 side

Square (1,3):
- Top: exposed → 1
- Right: exposed → 1
- Bottom: adjacent to (2,3)? No → (2,3) is empty → exposed → 1
- Left: shared with (1,2) → no

→ 3 exposed sides

Square (2,2):
- Top: shared with (1,2) → no
- Bottom: shared with (3,2) → no
- Left: exposed → 1
- Right: exposed → 1

→ 2 exposed sides

Square (3,2):
- Top: shared with (2,2) → no
- Bottom: exposed → 1
- Left: exposed → 1
- Right: exposed → 1

→ 3 exposed sides

Now sum exposed sides:
- (1,1): 3
- (1,2): 1
- (1,3): 3
- (2,2): 2
- (3,2): 3

Total: $ 3+1+3+2+3 = 12 $ cm

✔ Answer: 12 cm

---

c)

Shape:
```
■
■ ■
■
```

Squares:
- (1,2)
- (2,1), (2,2)
- (3,1)

So:
- (1,2): top, right, bottom (to (2,2)), left (no) → exposed: top, right, left? Left is empty → yes

(1,2):
- Top: exposed → 1
- Right: exposed → 1
- Bottom: shared with (2,2) → no
- Left: exposed → 1 → 3

(2,1):
- Top: exposed → 1
- Left: exposed → 1
- Bottom: exposed → 1
- Right: shared with (2,2) → no → 3

(2,2):
- Top: shared with (1,2) → no
- Bottom: shared with (3,1)? No → (3,1) is below (2,1), not (2,2) → so bottom exposed → 1
- Left: shared with (2,1) → no
- Right: exposed → 1 → 2

(3,1):
- Top: shared with (2,1) → no
- Bottom: exposed → 1
- Left: exposed → 1
- Right: exposed → 1 → 3

Total exposed sides: 3+3+2+3 = 11 cm

Wait — check (2,2): right is exposed → yes, bottom is exposed → yes → 2

(3,1): right exposed → yes → 3

Yes.

Total: 11 cm

But let’s trace:

Start at (1,2) top-left:
- Right → 1 cm
- Down → 1 cm to (2,2)
- Right → 1 cm
- Down → 1 cm to (3,2)? But no square → so stop

Wait — better to walk perimeter.

Start at top-left of (1,2):

1. Right → 1 cm
2. Down → 1 cm (to (2,2))
3. Right → 1 cm
4. Down → 1 cm → now at (3,2) — but no square, so go down? No — instead, go left to (3,1)? But not connected.

Actually, from (2,2) right, then down — but (3,2) is empty, so we go down from (2,2) → but that’s not possible.

Instead, after (2,2), go down to (3,2) — but no square, so the boundary goes down from (2,2) to (3,2) → 1 cm, then left to (3,1)? No.

Wait — the shape has (3,1) below (2,1), not (2,2).

So from (2,2), go down → no square → so boundary continues down from (2,2) to (3,2) → 1 cm, then left to (3,1) — but (3,1) is separate.

Actually, the shape is:

Row 1: [ ] [■] [ ]
Row 2: [■] [■] [ ]
Row 3: [■] [ ] [ ]

So the boundary:

Start at (1,2) top-left:
- Right → 1 cm
- Down → 1 cm (to (2,2))
- Left → 1 cm (to (2,1)) — but (2,1) is a square, so we go along its top?
No — we’re tracing the outside.

After (2,2), go down → 1 cm to (3,2) — but no square → so the boundary goes down 1 cm from (2,2) to (3,2), then left to (3,1)? No — (3,1) is at same level.

Wait — better to use the edge-counting method.

List each square:

(1,2):
- Top: exposed → 1
- Right: exposed → 1
- Bottom: shared with (2,2) → no
- Left: exposed → 1 → 3

(2,1):
- Top: exposed → 1
- Left: exposed → 1
- Bottom: exposed → 1
- Right: shared with (2,2) → no → 3

(2,2):
- Top: shared → no
- Bottom: exposed → 1 (since (3,2) is empty)
- Left: shared → no
- Right: exposed → 1 → 2

(3,1):
- Top: shared with (2,1) → no
- Bottom: exposed → 1
- Left: exposed → 1
- Right: exposed → 1 → 3

Total: 3+3+2+3 = 11 cm

✔ Answer: 11 cm

---

d)

Shape:
```
■ ■
■
■
```

Squares:
- (1,1), (1,2)
- (2,2)
- (3,2)

Same as b, but mirrored.

(1,1): top, left, bottom (to (2,1)? No → (2,1) is empty → bottom exposed), right (to (1,2)) → shared

So (1,1):
- Top: exposed → 1
- Left: exposed → 1
- Bottom: exposed → 1
- Right: shared → no → 3

(1,2):
- Top: exposed → 1
- Right: exposed → 1
- Bottom: shared with (2,2) → no
- Left: shared → no → 2

(2,2):
- Top: shared → no
- Bottom: shared with (3,2) → no
- Left: exposed → 1
- Right: exposed → 1 → 2

(3,2):
- Top: shared → no
- Bottom: exposed → 1
- Left: exposed → 1
- Right: exposed → 1 → 3

Total: 3+2+2+3 = 10 cm

Wait — (1,2) has top and right exposed → 2

(2,2): left and right → 2

(3,2): bottom, left, right → 3

(1,1): top, left, bottom → 3

Sum: 3+2+2+3 = 10 cm

But earlier b was 12, this is different.

Wait — in b, we had 3 across top, then two below middle.

Here: two across top, then two below right.

So different.

Yes — this is symmetric to b, but rotated.

So perimeter should be same? Let’s see.

In b: top row 3 squares, then two below middle → total exposed: 12

Here: top row 2 squares, then two below right → should be similar.

But let’s count:

(1,1): top, left, bottom → 3
(1,2): top, right, bottom → 3 (bottom shared with (2,2)) → no, bottom shared → so only top and right → 2
(2,2): top shared, bottom shared, left exposed, right exposed → 2
(3,2): top shared, bottom exposed, left exposed, right exposed → 3

Total: 3+2+2+3 = 10 cm

Wait — (1,2) bottom is shared with (2,2) → yes → so only top and right → 2

Yes.

But in b, (1,1) had bottom exposed because (2,1) was empty → so 3

Here, (1,1) has bottom exposed → 3

(1,2) has bottom shared → 2

So total: 3+2+2+3 = 10 cm

✔ Answer: 10 cm

---

e)

Shape:
```
■ ■ ■
■
■ ■ ■
```

Squares:
- (1,1), (1,2), (1,3)
- (2,2)
- (3,1), (3,2), (3,3)

So like a cross.

Count exposed sides.

(1,1):
- Top: exposed → 1
- Left: exposed → 1
- Bottom: shared with (2,1)? No → (2,1) empty → exposed → 1
- Right: shared with (1,2) → no → 3

(1,2):
- Top: exposed → 1
- Left: shared → no
- Right: shared → no
- Bottom: shared with (2,2) → no → only top → 1

(1,3):
- Top: exposed → 1
- Right: exposed → 1
- Bottom: shared with (2,3)? No → (2,3) empty → exposed → 1
- Left: shared → no → 3

(2,2):
- Top: shared → no
- Bottom: shared with (3,2) → no
- Left: exposed → 1
- Right: exposed → 1 → 2

(3,1):
- Top: shared with (2,1)? No → (2,1) empty → exposed → 1
- Bottom: exposed → 1
- Left: exposed → 1
- Right: shared with (3,2) → no → 3

(3,2):
- Top: shared → no
- Bottom: exposed → 1
- Left: shared → no
- Right: shared → no → only bottom → 1

(3,3):
- Top: shared with (2,3)? No → (2,3) empty → exposed → 1
- Right: exposed → 1
- Bottom: exposed → 1
- Left: shared → no → 3

Now sum:

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

Total: 3+1+3+2+3+1+3 = 16 cm

✔ Answer: 16 cm

---

f)

Cross shape:
```
■
■ ■ ■
■
```

Squares:
- (1,2)
- (2,1), (2,2), (2,3)
- (3,2)

Symmetric cross.

Count:

(1,2):
- Top: exposed → 1
- Right: exposed → 1
- Bottom: shared → no
- Left: exposed → 1 → 3

(2,1):
- Top: exposed → 1
- Left: exposed → 1
- Bottom: exposed → 1
- Right: shared → no → 3

(2,2):
- Top: shared → no
- Bottom: shared → no
- Left: shared → no
- Right: shared → no → 0

(2,3):
- Top: exposed → 1
- Right: exposed → 1
- Bottom: exposed → 1
- Left: shared → no → 3

(3,2):
- Top: shared → no
- Bottom: exposed → 1
- Left: exposed → 1
- Right: exposed → 1 → 3

Total: 3+3+0+3+3 = 12 cm

✔ Answer: 12 cm

---

g)

Shape:
```
■ ■
■ ■
■
```

Squares:
- (1,1), (1,2)
- (2,1), (2,2)
- (3,2)

So:
(1,1): top, left, bottom, right → right shared with (1,2), bottom shared with (2,1) → so top and left exposed → 2

(1,2): top, right, bottom → bottom shared with (2,2), left shared → so top and right → 2

(2,1): top shared, left exposed, bottom exposed, right shared → left and bottom → 2

(2,2): top shared, right exposed, bottom shared, left shared → only right → 1

(3,2): top shared, bottom exposed, left exposed, right exposed → 3

Total: 2+2+2+1+3 = 10 cm

✔ Answer: 10 cm

---

h)

Shape:
```
■ ■ ■
■
■ ■ ■
```

Wait — this is the same as e? Yes!

It's identical to e.

So perimeter is also 16 cm

✔ Answer: 16 cm

---

i)

Shape:
```
■ ■ ■
■
■
```

Same as d? Let’s see:

(1,1), (1,2), (1,3) — top row
(2,2)
(3,2)

Wait — this is same as b

So perimeter = 12 cm

✔ Answer: 12 cm

---

Final Answers:

| Shape | Perimeter |
|-------|----------|
| a | 14 cm |
| b | 12 cm |
| c | 11 cm |
| d | 10 cm |
| e | 16 cm |
| f | 12 cm |
| g | 10 cm |
| h | 16 cm |
| i | 12 cm |

---

Summary:

- a: 14 cm
- b: 12 cm
- c: 11 cm
- d: 10 cm
- e: 16 cm
- f: 12 cm
- g: 10 cm
- h: 16 cm
- i: 12 cm

You can fill these in the blanks on the worksheet.