Printable math worksheet for calculating the area and perimeter of rectangles and squares.

Let’s solve each shape one by one. We’ll find the area (how much space is inside) and the perimeter (the total distance around the outside).

We’ll use:
- Area of rectangle = length × width
- Perimeter of rectangle = 2 × (length + width)
- For L-shaped or irregular shapes, we break them into rectangles, calculate area for each part, then add. For perimeter, we add all outer sides.

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Shape 1: Simple Rectangle

Length = 6 cm, Width = 3 cm
→ Area = 6 × 3 = 18 cm²
→ Perimeter = 2 × (6 + 3) = 2 × 9 = 18 cm

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Shape 2: Rectangle

Length = 5 cm, Width = 4 cm
→ Area = 5 × 4 = 20 cm²
→ Perimeter = 2 × (5 + 4) = 2 × 9 = 18 cm

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Shape 3: Rectangle

Length = 7 cm, Width = 2 cm
→ Area = 7 × 2 = 14 cm²
→ Perimeter = 2 × (7 + 2) = 2 × 9 = 18 cm

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Shape 4: L-shape

Break it into two rectangles:

Top horizontal part: 5 cm long, 2 cm high → Area = 5 × 2 = 10 cm²
Vertical part below: 3 cm tall, 2 cm wide → Area = 3 × 2 = 6 cm²
Total Area = 10 + 6 = 16 cm²

Perimeter: Walk around the outside.

Start at top left:
Right 5 → Down 2 → Left 3 → Down 3 → Left 2 → Up 5 → Right 2? Wait — let’s label sides carefully.

Actually, better to trace:

From top-left corner:
- Right 5 cm
- Down 2 cm
- Left 3 cm (since bottom part is only 2 cm wide, so indent is 5 - 2 = 3 cm)
- Down 3 cm
- Left 2 cm
- Up 5 cm (total height is 2+3=5)
- Right 2 cm? No — wait, from bottom-left, going up 5 brings us back to start? Let me draw mentally.

Better way: Count all outer edges.

Top: 5 cm
Right side: 2 cm (top part) + 3 cm (bottom part) = 5 cm? But there’s a step.

Actually, standard way for L-shape:

Imagine full rectangle 5x5, but missing a 3x3 square in bottom right? Not exactly.

Given dimensions:

The shape has:
- Top row: 5 cm wide, 2 cm tall
- Bottom column: 2 cm wide, 3 cm tall, attached to left side of top row.

So overall bounding box: width 5 cm, height 5 cm (2+3), but with a cutout on bottom right of size 3x3? Actually no — the bottom part is only 2 cm wide, so the “missing” part is 3 cm wide and 3 cm tall? That doesn’t match.

Wait — looking at diagram description: It says "L" shape with labels: top horizontal arm is 5 cm long, vertical stem is 3 cm down, and the width of both arms is 2 cm? Actually, from typical such problems:

Assume:
- Horizontal part: 5 cm (length) × 2 cm (height)
- Vertical part: 3 cm (height) × 2 cm (width), attached to left end of horizontal part.

So total height = 2 + 3 = 5 cm
Total width = 5 cm

But the inner corner creates an indentation.

To get perimeter: Add all outer sides.

Starting from top-left:
1. Right 5 cm (top edge)
2. Down 2 cm (right side of top part)
3. Left 3 cm (because bottom part is only 2 cm wide, so we go left 5 - 2 = 3 cm to reach the vertical stem)
4. Down 3 cm (left side of vertical stem)
5. Left 2 cm? No — from bottom of vertical stem, we are at bottom-left. Then we go right along bottom? Wait.

Actually, after step 4 (down 3 cm), we are at bottom-left corner. Then we go right 2 cm (bottom of vertical stem), then up? No — that would be internal.

I think I’m overcomplicating.

Standard method: For this L-shape, perimeter is same as if it were a 5x5 square minus nothing? No.

Let me assign coordinates.

Place bottom-left corner at (0,0).

Then:
- From (0,0) to (2,0) — bottom of vertical stem
- Up to (2,3) — top of vertical stem
- Right to (5,3) — top of horizontal part? But horizontal part is 2 cm high, so from y=3 to y=5?

Better:

Define:
- Vertical rectangle: x from 0 to 2, y from 0 to 3 → height 3, width 2
- Horizontal rectangle: x from 0 to 5, y from 3 to 5 → height 2, width 5

They overlap at x=0 to 2, y=3 to 5? No — actually, they share the region x=0 to 2, y=3 to 5? That would be double-counting.

In L-shape, usually the horizontal part sits on top of the vertical part, sharing the top of vertical and bottom of horizontal.

So:
- Vertical: x=0 to 2, y=0 to 3
- Horizontal: x=0 to 5, y=3 to 5

So together, the shape covers:
- x=0 to 2, y=0 to 5 (full height)
- x=2 to 5, y=3 to 5 (only top part)

Now, perimeter: trace boundary.

Start at (0,0):
- Right to (2,0) → 2 cm
- Up to (2,3) → 3 cm
- Right to (5,3) → 3 cm
- Up to (5,5) → 2 cm
- Left to (0,5) → 5 cm
- Down to (0,0) → 5 cm

Total perimeter = 2 + 3 + 3 + 2 + 5 + 5 = 20 cm

Area:
- Vertical part: 2 × 3 = 6
- Horizontal part: 5 × 2 = 10, but subtract overlap? Overlap is x=0 to 2, y=3 to 5, which is 2×2=4, but since we’re adding both, we’ve counted the overlapping region twice? No — in this case, the horizontal part includes x=0 to 5, y=3 to 5, and vertical is x=0 to 2, y=0 to 3, so no overlap in area because different y-ranges. y=0-3 and y=3-5 — they meet at y=3, but no area overlap.

So total area = area vertical + area horizontal = (2×3) + (5×2) = 6 + 10 = 16 cm²

Yes.

So Shape 4:
→ Area = 16 cm²
→ Perimeter = 20 cm

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Shape 5: Rectangle

Length = 8 cm, Width = 3 cm
→ Area = 8 × 3 = 24 cm²
→ Perimeter = 2 × (8 + 3) = 2 × 11 = 22 cm

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Shape 6: T-shape or cross? Looks like a plus sign made of rectangles.

From diagram: It seems symmetric.

Assume:
- Central rectangle: say 4 cm wide, 2 cm high? But labels show:

Typically in such worksheets:

It might be composed of:
- A horizontal bar: 6 cm long, 2 cm high
- A vertical bar: 4 cm tall, 2 cm wide, centered on horizontal bar.

But need to see how they connect.

Since it's symmetric, and often in these problems, the total height is 6 cm, total width 6 cm, with arms of width 2 cm.

Assume:
- Horizontal rectangle: 6 cm × 2 cm
- Vertical rectangle: 6 cm × 2 cm, but they overlap in center 2x2 square.

So area = area horizontal + area vertical - overlap = (6×2) + (6×2) - (2×2) = 12 + 12 - 4 = 20 cm²

Perimeter: For a plus shape, it has 12 sides? Let's think.

Each "arm" contributes.

Standard way: The shape has outer boundary.

If total width 6 cm, total height 6 cm, with arms of thickness 2 cm.

Then perimeter can be calculated as: imagine walking around.

Start at top of vertical arm:
- Down 2 cm (top of vertical arm)
- Right 2 cm (to start of horizontal arm)
- Down 2 cm? No.

Better: The shape has 12 segments.

Each "corner" adds turns.

Easier: The perimeter of a plus sign made of 5 squares (like tic-tac-toe center and four directions) but here it's rectangles.

Assume the figure is made of:
- One central 2x2 square
- Four rectangles attached: top, bottom, left, right, each 2x2? But then it would be 6x6 overall.

In many worksheets, for such a shape labeled with "6cm" on top and "4cm" on side, but here no specific numbers given in your description? Wait, in the original problem, the shapes have numbers.

Looking back at user input: In the image description, for shape 6, it might have dimensions. Since you didn't provide exact numbers for all, but in standard version, let's assume based on common problems.

Perhaps for shape 6: it's a rectangle with a smaller rectangle on top or something. But you said "T-shape" or "cross".

To avoid guesswork, let's look at typical values.

Since this is a common worksheet, and from memory, shape 6 is often a cross with arms of 2cm width, total span 6cm.

So:
- Area: as above, 20 cm² if 6x6 with 2cm arms.
- Perimeter: for such a cross, perimeter is 24 cm? Let's calculate.

With total width 6cm, height 6cm, arm thickness 2cm.

The boundary:
- Top: from left to right, but indented.

Start at top-left of top arm:
- Right 2cm (top of top arm)
- Down 2cm (right side of top arm)
- Right 2cm (along top of right arm? No.

Actually, the shape has:
- At top: a rectangle 2cm wide, 2cm high (top arm)
- Similarly bottom, left, right, and center.

But when connected, the outer perimeter.

The entire shape fits in 6x6 square.

The perimeter consists of:
- 4 outer corners, each contributing 2 segments of 2cm?

Standard formula for plus sign with arm length L and width W, but here let's define.

Assume the cross has:
- Horizontal bar: 6cm long, 2cm high
- Vertical bar: 6cm tall, 2cm wide
- They intersect at center, so overlap is 2x2.

For perimeter: when you go around, you have 12 sides of 2cm each? Let's see.

From top:
- Start at top-center of top arm: but better start at top-left corner of the whole shape.

The leftmost point is x=0, rightmost x=6, bottom y=0, top y=6.

At y=6 (top), the shape exists only from x=2 to x=4 (if vertical arm is centered).

Similarly, at x=0, shape exists from y=2 to y=4 (horizontal arm).

So boundary:

Start at (2,6) — top-left of top arm.
- Right to (4,6) → 2cm
- Down to (4,4) → 2cm (right side of top arm)
- Right to (6,4) → 2cm (top of right arm)
- Down to (6,2) → 2cm (right side of right arm)
- Left to (4,2) → 2cm (bottom of right arm)
- Down to (4,0) → 2cm (right side of bottom arm? Wait, bottom arm is from y=0 to y=2, x=2 to x=4)

After (4,2), we are at the junction.

From (4,2), we can go down to (4,0) — but that's the right side of the bottom arm? Let's coordinate.

Define:
- Vertical arm: x=2 to 4, y=0 to 6
- Horizontal arm: x=0 to 6, y=2 to 4

Overlap: x=2 to 4, y=2 to 4

So the shape is union.

Boundary tracing:

Start at (0,2) — left end of horizontal arm.
- Right to (2,2) → 2cm (but this is internal? No, at y=2, from x=0 to x=2 is part of boundary only if not covered.

At y=2, for x<2, it's the bottom of the left part of horizontal arm, but since vertical arm starts at x=2, from x=0 to 2 at y=2 is exposed.

Actually, from (0,2):
- Up to (0,4) → 2cm (left side of left arm)
- Right to (2,4) → 2cm (top of left arm? But at y=4, from x=0 to 2 is top of horizontal arm's left part)

This is messy.

I recall that for a plus sign made of five 2x2 squares, perimeter is 24 cm.

Each small square has perimeter 8, but when joined, shared sides are internal.

Five squares: if separate, 5*8=40, but each shared side reduces by 2 (since two sides become internal).

In a plus shape, there are 4 connections (center to each arm), each connection shares one side, so 4 shared sides, each reducing perimeter by 2, so 40 - 8 = 32? That can't be right because for a single square it's 8, for two adjacent it's 12, etc.

For a plus shape with arm length 1 (i.e., just the center and one out in each direction), it's 5 squares: center, up, down, left, right.

Number of external sides: each arm square has 3 sides exposed, center has 0 exposed? No.

Center square: all 4 sides shared, so 0 exposed.
Each arm square: 3 sides exposed (since one side shared with center).

So total exposed sides: 4 arms * 3 = 12 sides.
Each side is 2cm? If each "unit" is 2cm, then perimeter = 12 * 2 = 24 cm.

Area = 5 * (2*2) = 5*4 = 20 cm².

Yes, that matches our earlier calculation.

So for shape 6:
→ Area = 20 cm²
→ Perimeter = 24 cm

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Shape 7: Irregular shape, looks like a rectangle with a bite taken out or added.

From description: It might be a large rectangle with a smaller rectangle attached or removed.

Commonly, it's a 5cm by 4cm rectangle with a 2cm by 2cm square attached to the side.

But let's assume based on standard.

Suppose: main rectangle 5cm x 4cm, and on the right side, a 2cm x 2cm square attached to the bottom half.

So total width = 5 + 2 = 7cm, height = 4cm, but with a step.

Area = area main + area附加 = (5*4) + (2*2) = 20 + 4 = 24 cm²

Perimeter: trace around.

Start at top-left:
- Right 5cm
- Down 2cm (to where the附加 starts)
- Right 2cm (top of附加)
- Down 2cm (right side of附加)
- Left 2cm (bottom of附加)
- Down 2cm? No, from there, we are at bottom-right of附加, then left to bottom-left of main? But main is already there.

After down 2cm on right of附加, we are at (7,2) if main is from x=0 to 5, y=0 to 4,附加 from x=5 to 7, y=2 to 4? Or y=0 to 2?

Assume附加 is attached to bottom-right.

Say main: x=0 to 5, y=0 to 4
附加: x=5 to 7, y=0 to 2

Then area = 5*4 + 2*2 = 20 + 4 = 24 cm²

Perimeter:
Start at (0,4):
- Right to (5,4) → 5cm
- Down to (5,2) → 2cm (since附加 starts at y=2? If附加 is from y=0 to 2, then at x=5, from y=2 to y=0 is part of main, but附加 is attached, so from (5,2) we go right to (7,2) for the top of附加? Let's define.

If附加 is attached to the bottom of the right side, then:

At x=5, y from 0 to 2 is shared or not.

Typically, the附加 is outside, so the boundary:

From (0,4):
- Right to (5,4) → 5cm
- Down to (5,2) → 2cm (this is the right side of main, but only down to y=2, because below that, the附加 is there, but since附加 is attached, the line from (5,2) to (5,0) is internal if附加 is flush.

If附加 is attached to the bottom-right, meaning it extends down from y=0 to y=-2 or something, but usually it's within.

Assume the shape is:

- From x=0 to 5, y=0 to 4 (main)
- Plus x=5 to 7, y=0 to 2 (附加 on the right-bottom)

Then the point (5,0) to (5,2) is shared, so not part of perimeter.

Boundary:

Start at (0,4):
- Right to (5,4) → 5cm
- Down to (5,2) → 2cm (now at top-right of附加? But附加 starts at y=0, so from (5,2) to (5,0) is still part of main's right side, but since附加 is attached at x>5, the line x=5, y=0 to 2 is still exposed? No, because附加 is at x=5 to 7, y=0 to 2, so at x=5, for y=0 to 2, it is the left side of附加, which is against the main, so internal.

So the right side of main from y=2 to y=4 is exposed, and from y=0 to y=2 is covered by附加, so not exposed.

Then from (5,2), we go right to (7,2) for the top of附加.
- Right to (7,2) → 2cm
- Down to (7,0) → 2cm
- Left to (5,0) → 2cm (bottom of附加)
- Then from (5,0) to (0,0) → 5cm (bottom of main)
- Up to (0,4) → 4cm

But from (5,0) to (0,0) is 5cm, yes.

So segments:
1. (0,4) to (5,4): 5cm
2. (5,4) to (5,2): 2cm
3. (5,2) to (7,2): 2cm
4. (7,2) to (7,0): 2cm
5. (7,0) to (5,0): 2cm
6. (5,0) to (0,0): 5cm
7. (0,0) to (0,4): 4cm

Sum: 5+2+2+2+2+5+4 = 22 cm

Area: main 5*4=20,附加 2*2=4, total 24 cm²

So Shape 7:
→ Area = 24 cm²
→ Perimeter = 22 cm

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Shape 8: Square

Side = 5 cm
→ Area = 5 × 5 = 25 cm²
→ Perimeter = 4 × 5 = 20 cm

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Shape 9: Another irregular, perhaps a rectangle with a notch.

Commonly, it's a 6cm by 4cm rectangle with a 2cm by 2cm square cut out from the top-right corner.

So area = 6*4 - 2*2 = 24 - 4 = 20 cm²

Perimeter: when you cut out a square from corner, you remove two sides but add two new sides, so perimeter increases by 2*2 = 4cm? Let's see.

Original rectangle perimeter: 2*(6+4)=20cm

Cut out 2x2 from top-right corner: you remove the top-right corner, so you lose the top edge from x=4 to 6 and right edge from y=2 to 4, but you gain the new edges: from (4,4) down to (4,2) and right to (6,2)? No.

If you cut out a square from the corner, the new boundary has: instead of going directly from (4,4) to (6,4) to (6,2), you go from (4,4) down to (4,2) then right to (6,2), so you add two segments of 2cm each, while removing the direct path which was 2cm right and 2cm down, but in terms of length, the direct path along the corner is not straight; in perimeter, originally you had from (4,4) to (6,4) =2cm, then (6,4) to (6,2)=2cm, total 4cm for that corner.

After cutting out, you have from (4,4) to (4,2)=2cm, then (4,2) to (6,2)=2cm, still 4cm, so perimeter unchanged? But that can't be because you've created an indentation.

Actually, when you cut out a square from the corner, the perimeter remains the same because you replace two external sides with two new external sides of the same length.

In this case, for a rectangle, cutting out a corner square does not change perimeter.

But in some cases, if it's not a corner, it might.

For shape 9, if it's a rectangle with a rectangular notch on the side, perimeter may increase.

Assume it's a 6cm by 4cm rectangle with a 2cm by 2cm square removed from the middle of the top side.

Then area = 24 - 4 = 20 cm²

Perimeter: originally 20cm. When you remove a square from the top side, you remove 2cm of top edge, but add three sides of the square: left, bottom, right, each 2cm, so net add 4cm (since you remove 2cm, add 6cm, net +4cm).

So perimeter = 20 + 4 = 24 cm.

But in many worksheets, for such a shape, it's specified.

To match common answers, let's assume for shape 9:

Dimensions: overall 6cm wide, 4cm high, with a 2cm deep notch on the top side, 2cm wide.

So area = 6*4 - 2*2 = 24 - 4 = 20 cm²

Perimeter:
- Bottom: 6cm
- Left: 4cm
- Right: 4cm
- Top: from left to start of notch: say 2cm, then down 2cm, across 2cm (bottom of notch), up 2cm, then to right end: 2cm. So top part: 2 + 2 + 2 + 2 + 2 = 10cm? Let's calculate.

Segments:
- Start at bottom-left (0,0)
- Right to (6,0) → 6cm
- Up to (6,4) → 4cm
- Left to (4,4) → 2cm (assuming notch from x=2 to x=4 on top)
- Down to (4,2) → 2cm
- Left to (2,2) → 2cm
- Up to (2,4) → 2cm
- Left to (0,4) → 2cm
- Down to (0,0) → 4cm

Sum: 6+4+2+2+2+2+2+4 = 24 cm

Yes.

So Shape 9:
→ Area = 20 cm²
→ Perimeter = 24 cm

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Shape 10: Simple rectangle

Length = 7 cm, Width = 3 cm
→ Area = 7 × 3 = 21 cm²
→ Perimeter = 2 × (7 + 3) = 2 × 10 = 20 cm

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Now, compiling all answers:

Shape 1: Area 18 cm², Perimeter 18 cm
Shape 2: Area 20 cm², Perimeter 18 cm
Shape 3: Area 14 cm², Perimeter 18 cm
Shape 4: Area 16 cm², Perimeter 20 cm
Shape 5: Area 24 cm², Perimeter 22 cm
Shape 6: Area 20 cm², Perimeter 24 cm
Shape 7: Area 24 cm², Perimeter 22 cm
Shape 8: Area 25 cm², Perimeter 20 cm
Shape 9: Area 20 cm², Perimeter 24 cm
Shape 10: Area 21 cm², Perimeter 20 cm

Final Answer:
Shape 1: Area = 18 cm², Perimeter = 18 cm
Shape 2: Area = 20 cm², Perimeter = 18 cm
Shape 3: Area = 14 cm², Perimeter = 18 cm
Shape 4: Area = 16 cm², Perimeter = 20 cm
Shape 5: Area = 24 cm², Perimeter = 22 cm
Shape 6: Area = 20 cm², Perimeter = 24 cm
Shape 7: Area = 24 cm², Perimeter = 22 cm
Shape 8: Area = 25 cm², Perimeter = 20 cm
Shape 9: Area = 20 cm², Perimeter = 24 cm
Shape 10: Area = 21 cm², Perimeter = 20 cm