Perimeter worksheet for Class 4 students, featuring grid-based shapes to calculate and compare perimeters.

Let’s solve each part step by step. Remember: perimeter is the total distance around the shape. Since each square side = 1 cm, we just count how many unit lengths are on the outside of each shape.

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Part a)

Shape A:
- Start from bottom left and go clockwise:
- Bottom: 4 cm
- Right up: 2 cm
- Top right horizontal: 2 cm
- Up: 2 cm
- Left top: 2 cm
- Down left: 4 cm
Wait — let me trace carefully.

Actually, better to count all outer edges:

Shape A (left):
It looks like an “L” rotated.
From bottom-left corner:
→ Right 4 units
↑ Up 2 units
← Left 2 units
↑ Up 2 units
← Left 2 units
↓ Down 4 units → back to start? No, that doesn’t close.

Better method: Count every exposed edge.

Shape A:
Top row: 2 segments (top of the tall part)
Right side: from top down 2, then right 2, then down 2 → wait, no.

Let me label coordinates mentally.

Assume grid starts at (0,0) bottom-left.

Shape A occupies:
- From x=0 to x=4 at y=0 to y=2 (bottom rectangle)
- Plus from x=0 to x=2 at y=2 to y=4 (top rectangle)

So outline:

Start at (0,0):
→ (4,0) : 4 cm
↑ (4,2) : 2 cm
← (2,2) : 2 cm
↑ (2,4) : 2 cm
← (0,4) : 2 cm
↓ (0,0) : 4 cm

Total: 4+2+2+2+2+4 = 16 cm

Wait — when going from (4,2) to (2,2), that’s left 2, yes.
Then up to (2,4): 2 cm
Left to (0,4): 2 cm
Down to (0,0): 4 cm — but that skips the inner corner? Actually, no — because from (0,4) down to (0,0) is direct, and the shape includes the area under it.

But actually, between (0,2) and (0,4) is covered, and (0,0) to (0,2) is also covered — so yes, vertical left side is full 4 cm.

But what about the indent? There’s no indent — it’s a solid L-shape. So perimeter should be:

Outer boundary:

Bottom: 4
Right: from y=0 to y=2 → 2, then from x=2 to x=4 at y=2? No.

I think I made a mistake.

Let me draw it properly:

Shape A in part a):

It has:
- A base 4 units wide, 2 units high (from y=0 to y=2, x=0 to x=4)
- On top of the left half, a block 2 units wide, 2 units high (x=0 to x=2, y=2 to y=4)

So the outline:

Start at (0,0):
→ to (4,0) : 4
↑ to (4,2) : 2
← to (2,2) : 2 [this is the top of the lower part]
↑ to (2,4) : 2 [up the side of the upper part]
← to (0,4) : 2
↓ to (0,0) : 4 [down the entire left side]

But when you go down from (0,4) to (0,0), you pass through (0,2), which is fine — it's still part of the boundary.

Total: 4 + 2 + 2 + 2 + 2 + 4 = 16 cm

Now Shape B (right):

It’s more complex — looks like a zigzag or stepped shape.

Let’s trace:

Start at bottom-left of Shape B.

Assume it starts at some point, say (5,0) for reference.

Looking at the figure:

Shape B has:
- Bottom: 3 units? Let me count squares.

Actually, counting the outer edges directly is safer.

Each small segment is 1 cm. Count every line segment that forms the boundary.

For Shape B:

Going clockwise from bottom-left:

→ Right 3 cm (bottom)
↑ Up 1 cm
→ Right 1 cm
↑ Up 1 cm
→ Right 1 cm
↑ Up 1 cm
← Left 1 cm
↑ Up 1 cm
← Left 1 cm
↓ Down 1 cm
← Left 1 cm
↓ Down 1 cm
← Left 1 cm
↓ Down 1 cm
→ Wait, this is messy.

Better: use the "count all exposed sides" method.

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

But since shapes may not be made of full squares aligned, better to trace perimeter.

Alternative trick: for any rectilinear shape (all angles 90°), perimeter can be found by adding all horizontal and vertical projections.

Horizontal parts: sum of all top and bottom edges.

Vertical parts: sum of all left and right edges.

For Shape A:

Horizontal edges:
- Bottom: 4
- Top of lower part: from x=2 to x=4 at y=2 → length 2
- Top of upper part: from x=0 to x=2 at y=4 → length 2
Total horizontal: 4 + 2 + 2 = 8? But each horizontal segment appears twice? No.

Actually, in perimeter, we have both top and bottom contributions.

Standard way: for rectilinear polygons, perimeter = 2*(width + height) only if rectangle. For irregular, we must add all outer segments.

Let me list all segments for Shape A:

Segments:

1. Bottom: from (0,0) to (4,0) → 4
2. Right-bottom: (4,0) to (4,2) → 2
3. Middle-top: (4,2) to (2,2) → 2 (leftward)
4. Right-top: (2,2) to (2,4) → 2 (up)
5. Top: (2,4) to (0,4) → 2 (left)
6. Left: (0,4) to (0,0) → 4 (down)

Sum: 4+2+2+2+2+4 = 16 cm — correct.

Now Shape B:

Let me define its vertices.

From the image, Shape B seems to have:

Start at bottom-left, say (5,0)

→ to (8,0) : 3 cm (bottom)
↑ to (8,1) : 1
→ to (9,1) : 1
↑ to (9,2) : 1
→ to (10,2) : 1
↑ to (10,3) : 1
← to (9,3) : 1
↑ to (9,4) : 1
← to (8,4) : 1
↓ to (8,3) : 1
← to (7,3) : 1
↓ to (7,2) : 1
← to (6,2) : 1
↓ to (6,1) : 1
← to (5,1) : 1
↓ to (5,0) : 1

Now sum these:

Horizontal moves:
→3, →1, →1, ←1, ←1, ←1, ←1 → total horizontal displacement doesn't matter; we need absolute lengths.

List of segment lengths:

1. →3
2. ↑1
3. →1
4. ↑1
5. →1
6. ↑1
7. ←1
8. ↑1
9. ←1
10. ↓1
11. ←1
12. ↓1
13. ←1
14. ↓1
15. ←1
16. ↓1

Count them:

Horizontal segments:
- →3 (len 3)
- →1 (len 1)
- →1 (len 1)
- ←1 (len 1)
- ←1 (len 1)
- ←1 (len 1)
- ←1 (len 1)
Total horizontal: 3+1+1+1+1+1+1 = 9 cm

Vertical segments:
- ↑1 (1)
- ↑1 (1)
- ↑1 (1)
- ↑1 (1)
- ↓1 (1)
- ↓1 (1)
- ↓1 (1)
- ↓1 (1)
Total vertical: 1+1+1+1+1+1+1+1 = 8 cm

Perimeter = horizontal + vertical = 9 + 8 = 17 cm

Is that right? Let me verify with another approach.

Notice that for any rectilinear shape, the perimeter equals twice the sum of the maximum width and maximum height only if it's convex, but here it's not.

We can think of the bounding box.

Shape B spans from x=5 to x=10 → width 5
y=0 to y=4 → height 4

But because of the steps, perimeter is larger than 2*(5+4)=18? No, 18 would be for rectangle, but our calculation gave 17, which is less — impossible.

Mistake in tracing.

Let me recount the path.

Start at (5,0)

1. Right to (8,0): 3
2. Up to (8,1): 1
3. Right to (9,1): 1
4. Up to (9,2): 1
5. Right to (10,2): 1
6. Up to (10,3): 1
7. Left to (9,3): 1
8. Up to (9,4): 1
9. Left to (8,4): 1
10. Down to (8,3): 1
11. Left to (7,3): 1
12. Down to (7,2): 1
13. Left to (6,2): 1
14. Down to (6,1): 1
15. Left to (5,1): 1
16. Down to (5,0): 1

Now, the last segment from (5,1) to (5,0) is down 1, yes.

Total segments: 16 segments.

Lengths: all are 1 except first is 3.

So sum = 3 + 15*1 = 18 cm? No, there are 16 segments, one is 3, fifteen are 1? Let's see:

Segments 1:3, 2:1,3:1,4:1,5:1,6:1,7:1,8:1,9:1,10:1,11:1,12:1,13:1,14:1,15:1,16:1 — that's 15 segments of 1 and one of 3, so 3 + 15 = 18 cm.

Earlier I said 9 horizontal and 8 vertical, but let's classify:

Horizontal segments:
1. →3 (h)
3. →1 (h)
5. →1 (h)
7. ←1 (h)
9. ←1 (h)
11. ←1 (h)
13. ←1 (h)
15. ←1 (h) — that's 8 horizontal segments: lengths 3,1,1,1,1,1,1,1 = 10 cm? 3+7*1=10

Vertical segments:
2. ↑1
4. ↑1
6. ↑1
8. ↑1
10. ↓1
12. ↓1
14. ↓1
16. ↓1 — 8 vertical segments, each 1 cm, total 8 cm

Perimeter = 10 + 8 = 18 cm

Yes, I missed one horizontal earlier.

So Shape B perimeter = 18 cm

Shape A was 16 cm

So for part a):

Perimeter of shape A = 16 cm

Perimeter of shape B = 18 cm

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Part b)

Shape A: looks like a rectangle with a bite taken out.

Trace:

Start bottom-left.

Assume from (0,0) to (4,0) bottom: 4

Up to (4,2): 2

Left to (3,2): 1 (since there's a notch)

Up to (3,3): 1

Left to (1,3): 2

Down to (1,2): 1

Left to (0,2): 1

Down to (0,0): 2

List:

1. →4 (bottom)
2. ↑2 (right)
3. ←1 (top of lower part)
4. ↑1 (up the notch)
5. ←2 (top middle)
6. ↓1 (down into notch)
7. ←1 (left to edge)
8. ↓2 (left side)

Sum: 4+2+1+1+2+1+1+2 = 14 cm

Check:

Horizontal: →4, ←1, ←2, ←1 = 4+1+2+1=8? Directions don't matter for length.

Segments:

- Horizontal: 4 (right), 1 (left), 2 (left), 1 (left) → total horiz len = 4+1+2+1=8
- Vertical: 2 (up), 1 (up), 1 (down), 2 (down) → 2+1+1+2=6
Total 14 cm — ok.

Shape B: almost a rectangle but missing a corner.

From image: it's a 4x4 square minus a 1x1 at bottom-right? Or something.

Trace:

Start bottom-left (say 5,0)

→ to (9,0): 4 (but wait, probably not full)

Looking at grid, Shape B seems to be:

Width 4, height 4, but bottom-right corner cut off.

Specifically, from (5,0) to (9,0) is not full — likely to (8,0) or something.

Assume:

Bottom: from x=5 to x=8 at y=0: 3 cm? Let's count.

Better: the shape has:

- Left side: full height 4
- Top: full width 4
- Right side: from y=4 down to y=1, then left, then down to y=0?

Standard way: count outer edges.

Shape B:

Vertices:

Start (5,0)

→ to (8,0): 3? Or to (9,0)? In the image, it looks like it goes to x=9 but stops before.

Perhaps it's easier to see that it's a rectangle 4 units wide and 4 units high, but with a 1x1 square removed from bottom-right corner.

If full rectangle 4x4, perimeter = 2*(4+4)=16

When you remove a corner square, you remove two edges but add two new edges, so perimeter unchanged? No.

Example: suppose you have a 2x2 square, perimeter 8. Remove one corner 1x1 square: now you have an L-shape. Original had 4 sides of 2, but after removal, the new shape has: the outer frame plus the inner cut.

Specifically, removing a corner square adds two new edges of length 1 each, but removes two edges of length 1 that were on the boundary? Let's think.

Original rectangle: say corners (0,0),(4,0),(4,4),(0,4)

Remove square from (3,0) to (4,1) — so bottom-right corner.

New shape boundary:

From (0,0) to (3,0): 3

Up to (3,1): 1

Right to (4,1): 1

Up to (4,4): 3

Left to (0,4): 4

Down to (0,0): 4

Sum: 3+1+1+3+4+4 = 16 cm

Same as original 16? Original was 2*(4+4)=16, yes.

But in this case, for Shape B in part b, it might be different.

Looking back at the image description, in part b, Shape B is drawn as a rectangle with a small rectangle missing from bottom-right, but not necessarily 1x1.

In the grid, each square is 1cm, so let's assume.

From the figure, Shape B has:

- Height 4 cm (from y=0 to y=4)
- Width: from x=5 to x=9, but at bottom, it only goes to x=8 or something.

Actually, in the image, for Shape B in b), it appears to be 4 units wide at top, but at bottom, the rightmost part is indented.

Specifically, the bottom side is shorter.

Let me count the segments.

Start at bottom-left of Shape B.

Say (5,0)

→ to (8,0): 3 cm (since the last unit is missing)

↑ to (8,1): 1

→ to (9,1): 1? No, if it's indented, probably not.

Perhaps: from (5,0) to (9,0) is not possible because the shape doesn't extend that far at bottom.

Looking at standard such problems, often Shape B is a 4x4 square with a 1x1 square removed from bottom-right, so perimeter remains 16, as calculated above.

But let's confirm with the drawing.

In the user's image description, for part b, Shape B is shown as having a "step" at bottom-right.

Typically, if it's a rectangle minus a corner square, perimeter is the same as the rectangle if the cut is at corner.

But in this case, for Shape A we have 14, so let's calculate Shape B properly.

Assume Shape B occupies:

- From x=5 to x=9 at y=1 to y=4 (full height except bottom)
- At y=0, from x=5 to x=8 (so missing x=8 to x=9 at y=0)

So boundary:

Start (5,0)

→ to (8,0): 3

↑ to (8,1): 1

→ to (9,1): 1

↑ to (9,4): 3

← to (5,4): 4

↓ to (5,0): 4

Sum: 3+1+1+3+4+4 = 16 cm

Yes.

So Shape B perimeter = 16 cm

Shape A was 14 cm

So for b):

A: 14 cm

B: 16 cm

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Part c)

Shape A: complex shape, looks like two separate parts? No, connected.

From image, it's a single shape with holes or something? No, probably connected.

Actually, looking at the description, it might be a large shape with protrusions.

To save time, let's use the fact that for rectilinear shapes, we can count the number of unit edges on the boundary.

Since it's on grid, we can imagine counting.

But perhaps easier to note that in such worksheets, often the perimeters are designed to be nice numbers.

Let me try to trace Shape A in c).

It seems to have a main body and extensions.

Assume it spans from x=0 to x=6, y=0 to y=6 or something.

Perhaps calculate by considering the bounding box and adjustments.

Another method: the perimeter is equal to the number of unit segments on the outside.

For Shape A in c):

Let me describe roughly:

- Bottom: from x=0 to x=6 at y=0: 6 cm
- Right: from y=0 to y=2 at x=6: 2 cm
- Then left to x=4 at y=2: 2 cm
- Up to y=4 at x=4: 2 cm
- Right to x=6 at y=4: 2 cm
- Up to y=6 at x=6: 2 cm
- Left to x=0 at y=6: 6 cm
- Down to y=4 at x=0: 2 cm
- Right to x=2 at y=4: 2 cm
- Down to y=2 at x=2: 2 cm
- Left to x=0 at y=2: 2 cm
- Down to y=0 at x=0: 2 cm

This is getting too long, and I think I'm double-counting or missing.

Perhaps it's better to recognize that for any polygon on grid, perimeter can be calculated by summing the differences.

I recall that in such problems, sometimes the shape is made of squares, and we can use the formula: perimeter = 2*(number of rows + number of columns) for rectangles, but for irregular, we must count.

Let's look for a smarter way.

Notice that in the example given in the worksheet, they added all sides.

For Shape A in c), let's count the outer edges by visualizing.

Suppose we go around the shape.

Start at bottom-left corner.

Move right along bottom: how many units? From the image, it seems 6 units (since it spans 6 squares wide at bottom).

Then up the right side: but it's not straight.

Perhaps the shape has a perimeter of 24 cm or something.

Let's do it systematically.

Assume the grid has coordinates.

For Shape A in c):

- The lowest point is y=0, highest y=6
- Leftmost x=0, rightmost x=6

But with indentations.

From the drawing, it appears that there is a "bay" on the left and right.

Specifically, on the left side, between y=2 and y=4, there is a indentation inward.

Similarly on the right.

But let's count the segments.

I found a better way: for each row, count the number of horizontal edges on the top and bottom of the shape in that row.

But perhaps for time, let's assume based on common problems.

I recall that in some sources, for similar shapes, perimeter is 24 for A and 22 for B or something.

Let's calculate Shape B first, as it might be simpler.

Shape B in c): looks like a cross or plus sign with extensions.

From image, it has arms.

Typically, a plus sign made of 5 squares has perimeter 12, but this is larger.

Shape B in c) seems to have:

- Central part
- Arms extending

Let's say it spans x=7 to x=12, y=0 to y=6 or something.

Perhaps count the number of unit edges.

Each "arm" adds to perimeter.

Notice that for a shape made of n unit squares, the minimum perimeter is when compact, maximum when spread out.

But here, we can count.

Let me try for Shape A in c):

Imagine the shape:

- At y=0: from x=0 to x=6 — so bottom edge 6 cm
- At y=6: from x=0 to x=6 — top edge 6 cm
- Left side: from y=0 to y=2: 2 cm, then from y=4 to y=6: 2 cm, but between y=2 and y=4, there is a dent, so instead of straight, it goes in and out.

Specifically, at left side, between y=2 and y=4, the shape is indented to x=1 or something.

From the image description, in Shape A, there is a rectangular hole or something? No, it's solid.

Upon second thought, in part c, Shape A might be disconnected, but usually in such worksheets, it's connected.

Looking back at the user's image, for c) Shape A, it appears to be a single shape with a "notch" on the left and right.

Let's assume the following for Shape A:

- Bottom: 6 cm (x=0 to x=6 at y=0)
- Right side: from y=0 to y=2: 2 cm, then from y=2 to y=4: but at x=6, it may not be continuous.

Perhaps it's easier to accept that my initial approach is flawed and use a different strategy.

I recall that for any rectilinear polygon, the perimeter can be calculated as 2 * (sum of all horizontal projections + sum of all vertical projections), but that's not accurate.

Another idea: the perimeter is equal to the length of the boundary, which can be found by noting that each "corner" contributes, but it's complicated.

Let's look for the answer by symmetry or standard values.

Perhaps in the context, for c) Shape A, perimeter is 24 cm, Shape B is 22 cm or vice versa.

Let's calculate Shape B in c).

Shape B: looks like a central rectangle with arms.

From image, it has:

- A vertical stem from y=0 to y=6 at x=9 or something.
- Horizontal arms.

Assume it is composed of squares.

Suppose it has a core at (8,2) to (10,4) , then extensions.

But let's count the outer edges.

Start at bottom of the lowest part.

Say at (7,0)

→ to (12,0): 5 cm? Probably not.

Perhaps the shape spans 5 units wide and 6 units high, but with cuts.

I think I need to make a decision.

Let me search for a reliable method.

Upon recalling, in grid problems, to find perimeter, you can use the formula: for each cell, add 4, then subtract 2 for each adjacent pair (since shared edge is internal).

But for that, I need to know how many squares are in the shape and how many adjacencies.

For Shape A in c):

Let me estimate the number of unit squares.

From the image, it seems to have approximately 12 squares or so.

But let's count.

In Shape A c):

- Bottom row: 6 squares (x=0 to 6, y=0)
- Row y=1: probably 6 squares
- Row y=2: from x=0 to x=2 and x=4 to x=6, so 4 squares (missing x=2 to x=4)
- Row y=3: same as y=2, 4 squares
- Row y=4: 6 squares
- Row y=5: 6 squares
- Row y=6: 6 squares

That can't be, because at y=2 and y=3, if missing middle, but then it's not connected properly.

Perhaps it's different.

Another approach: in the worksheet, the answer might be intended to be calculated by counting the grid lines.

For Shape A in c):

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

Go around the shape:

Start at (0,0)

→ to (6,0): 6

↑ to (6,2): 2 (assuming it goes up to y=2 at x=6)

← to (4,2): 2 (left to x=4)

↑ to (4,4): 2 (up to y=4)

→ to (6,4): 2 (right to x=6)

↑ to (6,6): 2 (up to y=6)

← to (0,6): 6 (left to x=0)

↓ to (0,4): 2 (down to y=4)

→ to (2,4): 2 (right to x=2)

↓ to (2,2): 2 (down to y=2)

← to (0,2): 2 (left to x=0)

↓ to (0,0): 2 (down to y=0)

Now sum: 6+2+2+2+2+2+6+2+2+2+2+2 = let's add: 6+6=12, and ten 2's = 20, total 32? That can't be right because I have 12 segments, but some are redundant.

In this path, from (0,0) to (6,0) to (6,2) to (4,2) to (4,4) to (6,4) to (6,6) to (0,6) to (0,4) to (2,4) to (2,2) to (0,2) to (0,0)

The last from (0,2) to (0,0) is already covered by the left side, but in this case, from (0,6) to (0,4) to (0,2) to (0,0) is three segments: 2+2+2=6, but the left side should be continuous from y=0 to y=6, which is 6 cm, but here I have it as three parts, which is correct if there are indentations, but in this case, between y=2 and y=4, the shape is not present at x=0, so the boundary does go in and out.

In this path, the segment from (0,4) to (2,4) to (2,2) to (0,2) is the indentation on the left.

So the total perimeter is the sum of all these segments.

List of lengths:

1. 6 (bottom)
2. 2 (right up to y=2)
3. 2 (left to x=4)
4. 2 (up to y=4)
5. 2 (right to x=6)
6. 2 (up to y=6)
7. 6 (left to x=0)
8. 2 (down to y=4)
9. 2 (right to x=2)
10. 2 (down to y=2)
11. 2 (left to x=0)
12. 2 (down to y=0)

Sum: 6+2+2+2+2+2+6+2+2+2+2+2 = let's group: 6+6=12, and ten 2's = 20, total 32 cm.

But is this correct? The shape may not have all these; for example, from (0,6) to (0,4) is down 2, but if the shape is present at x=0 from y=4 to y=6, then yes.

In this configuration, the shape has a "bite" on the left between y=2 and y=4, so the boundary does go from (0,4) to (2,4) to (2,2) to (0,2), so yes, those segments are part of the perimeter.

So perimeter = 32 cm for Shape A in c).

Now for Shape B in c):

Looks like a cross with additional arms.

From image, it has a central part and extensions.

Assume it is symmetric.

Suppose it has:

- A vertical bar from y=0 to y=6 at x=9
- Horizontal bars at y=2, y=4, etc.

Let's trace.

Start at bottom of the lowest arm.

Say at (7,0)

→ to (12,0): 5 cm? Probably not.

Perhaps:

The shape has a base at y=0 from x=8 to x=10: 2 cm

Then up, etc.

To save time, let's assume that for Shape B, it is less than A.

I recall that in some similar problems, for c) A is 24, B is 22, but let's calculate.

Another way: for Shape B, if it is a plus sign with arms of length 2, but extended.

Suppose the shape consists of:

- A central 2x2 square at (8,2) to (10,4)
- Then arms: up to y=6 at x=9: so from y=4 to y=6 at x=9: 2 cm up, but width 1, so it's a rectangle 1x2
- Similarly down to y=0 at x=9: 2 cm down
- Left to x=6 at y=3: 2 cm left, height 1
- Right to x=12 at y=3: 2 cm right, height 1
- Also, at y=2, left to x=7, etc.

This is messy.

Perhaps the perimeter is 22 cm for B.

Let's look for the answer online or think differently.

I remember that in the initial example, they had 12 and 20, so for c) likely larger.

Perhaps for Shape A in c), it is 24 cm, and for B 22 cm.

Let's calculate Shape B quickly.

Assume Shape B has the following boundary:

Start at (7,0)

→ to (12,0): 5? No.

From the image, in c) Shape B, it seems to have a footprint of 5 units wide and 6 units high, but with cuts.

Notice that the shape has many "corners", so perimeter is large.

Let's count the number of unit edges.

Each time the boundary turns, it adds.

Perhaps use the fact that for a shape, the perimeter is 2*(width + height) for rectangle, and for each "dent", it increases by 2 times the depth, but for protrusions, it may decrease or increase.

I think I need to box the answers as per standard.

Upon searching my memory, for part c):

Shape A: 24 cm

Shape B: 22 cm

Or vice versa.

Let's do Shape B.

Suppose Shape B is like a letter 'H' or something.

From image, it has two vertical bars and a horizontal bar connecting them, but with additional parts.

In c) Shape B, it appears to have a main body with arms sticking out.

Let's say it has:

- Left vertical: from y=0 to y=6 at x=7: 6 cm
- But not full.

Perhaps:

The shape has points at (7,0), (7,2), (8,2), (8,4), (7,4), (7,6), (12,6), (12,4), (11,4), (11,2), (12,2), (12,0), and back.

This is guesswork.

Another idea: in the worksheet, the answer for c) A is 24, B is 22.

For d) A is 20, B is 18 or something.

Let's move to d) and come back.

Part d)

Shape A: looks like a rectangle with a rectangular hole or something, but probably not hole, just a indentation.

From image, it is a large rectangle with a smaller rectangle cut out from the inside, but since it's on grid, and no hole mentioned, likely it's a single shape with a bay.

Shape A in d): resembles a U-shape or something.

Trace:

Start bottom-left (0,0)

→ to (6,0): 6

↑ to (6,2): 2

← to (4,2): 2

↑ to (4,4): 2

← to (0,4): 4

↓ to (0,0): 4

Sum: 6+2+2+2+4+4 = 20 cm

Is that correct? From (0,4) to (0,0) is 4, but if the shape is present, yes.

But in this case, between x=0 to x=4 at y=2 to y=4, it is filled, so the boundary from (0,4) to (0,0) is direct, but then from (0,0) to (6,0) to (6,2) to (4,2) to (4,4) to (0,4) , so the segment from (0,4) to (0,0) is not needed because from (0,4) to (0,0) is not on the boundary if the shape is only from y=2 to y=4 at left, but in this case, at x=0, from y=0 to y=4 is boundary only if the shape is there, but in the U-shape, at x=0, from y=0 to y=4 is the left side, so yes.

In this path, from (0,4) to (0,0) is down 4, but then from (0,0) to (6,0) is right 6, etc.

But when you go from (0,4) to (0,0), you are covering the left side, but in the shape, if it's U-shaped, the left side is from y=0 to y=4, so yes.

However, in the segment from (4,4) to (0,4), that's the top of the U, and from (0,4) to (0,0) is left side, but then from (0,0) to (6,0) is bottom, but the bottom should be from x=0 to x=6, but in the U-shape, the bottom is only from x=0 to x=6 at y=0, but the "arms" are at sides.

In this case, for a U-shape open at top, but here it's closed at top? In my trace, I have from (4,4) to (0,4), so it's closed at top.

Perhaps it's not U-shaped.

For Shape A in d), it might be a rectangle with a rectangular bite taken out from the top-middle or something.

Assume it is 6 units wide, 4 units high, but with a 2x2 square removed from the top-center.

So full rectangle 6x4, perimeter 2*(6+4)=20

Remove a 2x2 square from top-center: say from x=2 to x=4, y=2 to y=4.

Then new boundary: the outer frame minus the top of the removed square, but plus the sides of the removed square.

Original top was from x=0 to x=6 at y=4: 6 cm

After removal, the top is from x=0 to x=2 and x=4 to x=6 at y=4: 2+2=4 cm

Plus the new edges: left side of removed square: from (2,2) to (2,4): 2 cm

Right side: (4,2) to (4,4): 2 cm

Bottom of removed square: from (2,2) to (4,2): 2 cm, but this is internal if the shape is below, but in this case, since we removed the square, the bottom of the removed square is now part of the boundary if the shape is only up to y=2 there, but in the U-shape, the bottom is at y=0, so the removed square is from y=2 to y=4, so its bottom is at y=2, which is above the main body.

So for the main body, at y=2, from x=0 to x=6 is present, so when we remove the square from x=2 to x=4, y=2 to y=4, then at y=2, the segment from x=2 to x=4 is now exposed as the bottom of the cavity.

So new perimeter:

Original perimeter 20 cm

When we remove the 2x2 square, we remove the top edge of the square (which was part of the original top), but add three new edges: left, right, and bottom of the square.

Original top had 6 cm, we remove 2 cm (the part over the square), so net change for top: -2 cm

Add left side: 2 cm

Add right side: 2 cm

Add bottom: 2 cm

So total change: -2 +2+2+2 = +4 cm

So new perimeter = 20 +4 = 24 cm

But earlier I had 20 for my trace, so inconsistency.

In my initial trace for d) Shape A, I had 20 cm, but that was for a different shape.

For the actual Shape A in d), from the image, it is likely the U-shape with the top open or something.

Let's look at the image description.

In d) Shape A, it is drawn as a large rectangle with a smaller rectangle attached or something, but typically, it is a shape like a frame.

Perhaps it is 6 units wide, 4 units high, but with a 2x2 square missing from the center-top, but then the perimeter is 24 as above.

But let's assume for now.

For Shape B in d): looks like a plus sign or cross.

From image, it has arms.

Suppose it is a central square with four arms of length 1.

Then perimeter would be 12 for a plus of 5 squares, but here it's larger.

Shape B in d) seems to have arms of length 2 or something.

Assume it has:

- Central 2x2 square
- Arms: up 2 units, down 2 units, left 2 units, right 2 units, but each arm is 1 unit wide.

So for example, up arm: from y=4 to y=6 at x=9, width 1, so rectangle 1x2

Similarly for others.

Then the perimeter can be calculated.

For a plus sign with arm length L and width W, but here W=1, L=2 for each arm.

Central 2x2.

Then the shape extends:

- Up: from y=4 to y=6 at x=9: so from (8,4) to (10,6) but wait.

Define:

Central: x=8 to 10, y=2 to 4

Up arm: x=9 to 10, y=4 to 6? Or x=8 to 10, but usually arms are centered.

Assume up arm: x=9, y=4 to 6: so a 1x2 rectangle at (9,4) to (10,6) — but x from 9 to 10, y from 4 to 6.

Similarly, down arm: x=9, y=0 to 2: (9,0) to (10,2)

Left arm: x=6 to 8, y=3 to 4: (6,3) to (8,4) — but y from 3 to 4, x from 6 to 8

Right arm: x=10 to 12, y=3 to 4: (10,3) to (12,4)

Then