Compound Shapes (B) | Cazoom Maths Worksheets - Free Printable

Let’s solve each problem one by one. Remember, perimeter means adding up all the outer sides of the shape — even if some lengths aren’t labeled, we can figure them out using what’s given.

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Problem 1)
Shape is like an L with a diagonal side.

Labeled sides:
- Top = 10 cm
- Left top vertical = 1 cm
- Horizontal step inward = 4 cm
- Diagonal = 10 cm
- Right vertical = 9 cm

We need to find the missing bottom horizontal part.

Look at the top: total width is 10 cm. The inner horizontal part is 4 cm, so the remaining bottom horizontal must be 10 - 4 = 6 cm.

Now add all outer sides:

Top (10) + left small vertical (1) + inner horizontal (4) + diagonal (10) + right vertical (9) + bottom horizontal (6)

Wait — that counts the inner parts? No! Perimeter is only the *outer* edge.

Actually, let’s trace the outside:

Start from top-left corner:

→ Go right 10 cm (top)
→ Go down 9 cm (right side)
→ Go left along bottom — but how long? We know the top is 10 cm, and there's a 4 cm indent on the left, so the bottom straight part is 10 - 4 = 6 cm
→ Then go up the diagonal? Wait — no, the diagonal is part of the outer edge.

Actually, looking again: the shape has these outer edges:

- Top: 10 cm
- Right side: 9 cm
- Bottom-right slant: 10 cm (diagonal)
- Then from bottom of diagonal, go left 4 cm (the “step”)
- Then up 1 cm (small vertical)
- Then... wait, that brings us back to start? But we’re missing the connection between the 1 cm up and the top.

Actually, better way: list all outer segments without double-counting.

From diagram:

Outer path:

1. Top horizontal: 10 cm
2. Right vertical: 9 cm
3. Diagonal down-left: 10 cm
4. Bottom horizontal (leftward): ?
5. Small vertical up: 1 cm
6. Inner horizontal right: 4 cm → but this is NOT outer! It’s inside.

Wait — I think I’m confusing myself.

Let me label points mentally.

Imagine starting at top-left corner:

- Move right 10 cm → top edge
- Move down 9 cm → right edge
- Now, instead of going left, you go diagonally down-left for 10 cm → that’s the hypotenuse
- From end of diagonal, move left horizontally — how far? Since the top was 10 cm, and the “cut-in” is 4 cm from left, then the bottom horizontal segment should be 10 - 4 = 6 cm? But wait, the diagonal connects to the point that is 4 cm from left on the bottom level? Not necessarily.

Alternative approach: use the fact that in such shapes, sometimes opposite sides help.

Notice: the total height on the right is 9 cm. On the left, we have a 1 cm vertical segment. So the vertical drop from the end of the 1 cm segment to the bottom must be 9 - 1 = 8 cm? But that’s not directly helpful.

Wait — perhaps the diagonal is connecting two points: one is 4 cm right from left edge at some height, and the other is at bottom right.

Actually, let’s calculate the missing horizontal bottom part.

The full width at top is 10 cm. The shape indents 4 cm from the left on the lower part. So the bottom horizontal part (from the end of the diagonal to the leftmost point) should be 4 cm? No.

I think I need to redraw mentally.

Assume coordinates:

Set top-left as (0,9) — since right side is 9 cm tall.

Then:

- From (0,9) to (10,9) → top, 10 cm
- From (10,9) to (10,0) → right side, 9 cm
- From (10,0) to some point via diagonal 10 cm — but where?

The diagonal is labeled 10 cm, and it goes to a point that is connected to a 4 cm horizontal and 1 cm vertical.

From the diagram: after the diagonal, you go left 4 cm, then up 1 cm to meet the top.

So, suppose the diagonal ends at point P. From P, go left 4 cm to Q, then up 1 cm to R, which should connect to the top-left.

Since going up 1 cm from Q reaches the top level (y=9), then Q is at y=8.

And since from Q you go left 4 cm to reach the left edge? Or to connect to the start?

Actually, the 1 cm vertical is on the left, so likely:

After going up 1 cm from Q, you are at the same x as the left edge, and y=9.

So if you go up 1 cm to reach y=9, then Q is at y=8.

And you went left 4 cm to get to Q from P, so P is 4 cm right of Q.

Also, the diagonal is from (10,0) to P, length 10 cm.

Let P be at (x,8), since from P you go left 4 cm to Q(x-4,8), then up 1 cm to (x-4,9), which should be the top-left corner? But top-left is (0,9), so x-4 = 0 → x=4.

So P is at (4,8).

Now, distance from (10,0) to (4,8):

Δx = 6, Δy = 8, so distance = √(6² + 8²) = √(36+64) = √100 = 10 cm — matches!

Perfect.

So now, the outer perimeter path:

Start at (0,9):

→ to (10,9): 10 cm
→ to (10,0): 9 cm
→ to (4,8): 10 cm (diagonal)
→ to (0,8): 4 cm (left horizontal)
→ to (0,9): 1 cm (up vertical)

That’s all.

Add them: 10 + 9 + 10 + 4 + 1 = 34 cm

But is (0,8) to (0,9) correct? Yes, and it closes the shape.

So perimeter = 34 cm.

Wait — but in the diagram, the 4 cm is shown as going left from the diagonal end, and 1 cm up, so yes.

So answer for 1) is 34 cm.

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Problem 2)

Shape: looks like a house with a roof.

Sides:

- Left vertical: 3 cm
- Roof slope: 5 cm
- Right top vertical: 3 cm
- Right horizontal: 4 cm
- Bottom: 8 cm
- And a right angle at bottom-right.

Missing: the left part of the bottom? Or the connection.

Trace outer edge:

Start at bottom-left.

Go right along bottom: 8 cm
Go up right side: but right side has two parts: first up 3 cm? No.

Diagram shows:

Bottom: 8 cm
Right side: from bottom-right, go up 3 cm? But there’s a horizontal 4 cm on top right.

Actually, from bottom-right corner:

- Up 3 cm? But then left 4 cm, then up to roof.

Label:

Assume bottom-left A, bottom-right B.

B to C: up 3 cm? But diagram says right side has a 3 cm vertical, then 4 cm horizontal left, then diagonal up-left 5 cm to meet left side.

Left side is 3 cm vertical.

So:

Points:

A (bottom-left)
B (bottom-right) — AB = 8 cm
C: up from B, BC = ? Diagram says on right, there’s a 3 cm vertical, but is that from bottom?

Looking: "3 cm" is labeled on the right vertical segment above the bottom.

Actually, from the diagram description:

- Bottom: 8 cm
- Right side: from bottom, go up some amount, but it says "3 cm" on the vertical part before the horizontal.

Perhaps:

From bottom-right, go up 3 cm to D, then left 4 cm to E, then diagonal up-left 5 cm to F, then down 3 cm to A (bottom-left).

Is that closed?

Distance from E to F is 5 cm, F to A is 3 cm down.

What is horizontal distance from E to A?

Total bottom is 8 cm. From B to D is up, D to E is left 4 cm, so E is 4 cm left of B.

A is 8 cm left of B, so E is 4 cm right of A.

F is directly above A? Because F to A is vertical 3 cm.

So F is at same x as A, y higher.

E is at x = A_x + 4, y = ?

From E to F: diagonal 5 cm, and horizontal distance is 4 cm (since E is 4 cm right of F), so vertical distance must be √(5² - 4²) = √(25-16) = √9 = 3 cm.

So F is 3 cm above E.

But F to A is down 3 cm, so A is at same height as E? Contradiction.

If F is 3 cm above E, and F to A is down 3 cm, then A is at same height as E.

But E is reached by going up from B, so if B is bottom, E is higher, A should be bottom, so same height as B.

Inconsistency.

Perhaps the 3 cm on left is from top to bottom, but let's read labels.

Diagram says:

- Left side: 3 cm (vertical)
- Roof: 5 cm (diagonal)
- Right top vertical: 3 cm
- Right horizontal: 4 cm
- Bottom: 8 cm

And right angle at bottom-right.

So likely:

Start at bottom-left.

Go right 8 cm to bottom-right.

Go up — but how much? The right side has a vertical segment of 3 cm, but is that from bottom or from the horizontal?

Typically in such diagrams, the 3 cm on right is the vertical part above the bottom.

Assume:

From bottom-right, go up 3 cm to point P.

Then go left 4 cm to point Q.

Then go diagonal up-left 5 cm to point R.

Then go down 3 cm to bottom-left.

Now, check if it closes.

Horizontal: from bottom-right to P is up, no horizontal change.

P to Q: left 4 cm.

Q to R: diagonal 5 cm.

R to bottom-left: down 3 cm.

For R to be directly above bottom-left, the horizontal distance from Q to R must be such that when you go left from Q, you reach above bottom-left.

Total bottom is 8 cm. Q is 4 cm left of bottom-right, so 4 cm right of bottom-left.

So if R is directly above bottom-left, then from Q to R, you go left 4 cm and up some amount.

Distance is 5 cm, so up amount = √(5² - 4²) = 3 cm, as before.

Then from R down 3 cm to bottom-left — perfect, matches.

So the sides are:

- Bottom: 8 cm
- Right vertical: 3 cm (from bottom-right to P)
- Top-right horizontal: 4 cm (P to Q)
- Roof diagonal: 5 cm (Q to R)
- Left vertical: 3 cm (R to bottom-left)

All outer edges.

Perimeter = 8 + 3 + 4 + 5 + 3 = 23 cm

Yes.

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Problem 3)

Shape: similar to previous but different dimensions.

Sides:

- Bottom: 10 cm
- Right side: 4 cm (vertical)
- Then left 2 cm (horizontal)
- Then up 6 cm (vertical)
- Then diagonal down-left 10 cm to bottom-left?
- And left side is implied.

Trace:

Start at bottom-left A.

Go right 10 cm to B (bottom-right).

Go up 4 cm to C.

Go left 2 cm to D.

Go up 6 cm to E.

Go diagonal down-left 10 cm to A.

Check if closes.

From E to A: diagonal 10 cm.

Horizontal distance: from D to A: D is 2 cm left of C, C is above B, B is 10 cm right of A, so D is 10 - 2 = 8 cm right of A.

E is directly above D, so E is 8 cm right of A.

Vertical distance: from E to A: E is at height 4 + 6 = 10 cm above bottom? C is 4 cm up, D same height, E is 6 cm up from D, so E is 10 cm above bottom.

A is at bottom, so vertical distance 10 cm.

Horizontal distance 8 cm.

So distance E to A = √(8² + 10²) = √(64 + 100) = √164 ≈ 12.8, but diagram says 10 cm — contradiction.

Mistake.

Perhaps the diagonal is from E to a point not A.

Diagram shows diagonal labeled 10 cm, and it connects to the left side.

Left side is not labeled, but probably vertical.

Assume from E, diagonal down-left to F on left side, then down to A.

But diagram doesn't show that; it seems direct to bottom-left.

Perhaps the 10 cm diagonal is the hypotenuse, and we can find missing sides.

Another way: the total height on left should match.

From bottom to E: the right part has up 4 cm, then up 6 cm, so total height 10 cm.

On left, if it's vertical, it should be 10 cm, but not labeled.

The diagonal is 10 cm, from E to bottom-left A.

As calculated, if horizontal distance is 8 cm (since from A to B is 10 cm, D is 2 cm left of B, so 8 cm from A), vertical distance is 10 cm (from bottom to E), then distance is √(8^2 + 10^2) = √164 > 10, but should be 10, so impossible.

Unless the 6 cm is not up from D, but something else.

Read diagram: "6 cm" is on the vertical segment after the 2 cm horizontal.

And "10 cm" on diagonal.

Perhaps the diagonal is not to bottom-left, but to a point on the left side.

But in standard interpretation, it should close.

Perhaps the left side is not full height.

Let's calculate the missing left vertical part.

Suppose from bottom-left A, go up h cm to G, then diagonal to E.

But diagram shows diagonal from E directly to A, I think.

Perhaps the 10 cm diagonal is correct, and we need to find the horizontal or vertical.

Another idea: the shape might have the diagonal connecting E to a point that is not A, but the diagram suggests it does.

Let's look at the numbers.

Suppose the horizontal distance from E to A is x, vertical is y, then x^2 + y^2 = 10^2 = 100.

From the right side, the total width is 10 cm. The indent is 2 cm, so the horizontal distance from E to the left edge is 10 - 2 = 8 cm? Only if E is aligned with the right part.

E is above D, D is 2 cm left of C, C is above B, B is bottom-right, so if A is bottom-left, then distance from A to D is 10 - 2 = 8 cm horizontally.

Vertically, E is at height equal to the sum of the verticals on right: 4 cm + 6 cm = 10 cm.

So if A is at (0,0), B at (10,0), C at (10,4), D at (8,4), E at (8,10).

Then distance from E(8,10) to A(0,0) is √(8^2 + 10^2) = √(64+100) = √164 ≈ 12.8, but diagram says 10 cm for that diagonal — contradiction.

So my assumption must be wrong.

Perhaps the 6 cm is not the height from D, but the length of the vertical segment, but in a different orientation.

Or perhaps the diagonal is not from E to A, but from E to a point on the left side at a different height.

But the diagram likely intends that the diagonal connects to the bottom-left corner.

Perhaps the "6 cm" is the length of the vertical segment, but it's not added to the 4 cm; maybe it's separate.

Another possibility: the 4 cm on right is not from bottom, but from the horizontal.

Let's read the diagram carefully.

In problem 3, it shows:

- Bottom: 10 cm
- Right side: 4 cm (vertical) — probably from bottom up
- Then left 2 cm (horizontal)
- Then up 6 cm (vertical)
- Then diagonal 10 cm down-left to bottom-left
- And the left side is not labeled, but should be the remaining.

But as calculated, distance is not 10 cm.

Unless the 6 cm is downward or something, but that doesn't make sense.

Perhaps the diagonal is 10 cm, and we can use Pythagoras to find the missing side.

Suppose from the end of the 6 cm up, we go diagonal to bottom-left.

Let the horizontal distance be a, vertical distance be b, a^2 + b^2 = 100.

From the geometry, the horizontal distance from the point to bottom-left is the total width minus the indent.

Total width 10 cm, indent 2 cm, so if the point is 2 cm left of the right edge, then horizontal distance to left edge is 8 cm.

Vertical distance: the point is at height h above bottom.

On the right, we have up 4 cm, then up 6 cm, so h = 4 + 6 = 10 cm.

Same as before.

But 8^2 + 10^2 = 64 + 100 = 164 > 100, so impossible.

Unless the 6 cm is not additional height, but the length of the segment, and it's not vertical? But it's drawn vertical.

Perhaps the "6 cm" is the length of the vertical segment, but it's measured differently.

Another idea: perhaps the 4 cm on right is the height from the bottom to the horizontal, but the horizontal is not at the top of that; no.

Let's look at the diagram description again.

In the user's image, for problem 3, it shows:

- Bottom: 10 cm
- Right vertical: 4 cm
- Then left horizontal: 2 cm
- Then up vertical: 6 cm
- Then diagonal: 10 cm to bottom-left
- And the left side is implied to be vertical, but not labeled.

But mathematically, it doesn't work with those numbers.

Perhaps the diagonal is not to bottom-left, but to a point on the left side, and the left side has a length that we need to find.

But the diagram likely assumes it closes to bottom-left.

Perhaps the 10 cm diagonal is correct, and the vertical on left is not 10 cm.

Let's calculate what the vertical distance should be.

Suppose horizontal distance is 8 cm (as before), diagonal 10 cm, then vertical distance v = √(10^2 - 8^2) = √(100-64) = √36 = 6 cm.

Oh! So if the vertical distance from E to A is 6 cm, then it works.

But earlier I said E is at height 10 cm, but if vertical distance to A is 6 cm, then E is at height 6 cm.

But on the right, we have up 4 cm to C, then up 6 cm to E, so E should be at 10 cm, but if it's at 6 cm, contradiction.

Unless the "6 cm" is not the height gain, but the length, and it's not vertical? But it's drawn vertical.

Perhaps the 4 cm and 6 cm are not both upward from bottom.

Another possibility: the 4 cm on right is from bottom to the horizontal, but the horizontal is at the top, and the 6 cm is down or something.

Let's think differently.

Perhaps the shape is such that from bottom-right, go up 4 cm, then left 2 cm, then the diagonal is from there to bottom-left, and the 6 cm is the left vertical side.

But the diagram shows "6 cm" on the vertical segment after the 2 cm horizontal, which would be on the right part.

In the text: "6 cm" is labeled on the vertical segment that is after the 2 cm horizontal, so on the right side of the shape.

Perhaps for the diagonal to be 10 cm, and horizontal distance 8 cm, vertical must be 6 cm, so the point E is at height 6 cm above bottom.

Then, on the right, from bottom to C is 4 cm up, then from C to D is left 2 cm, then from D to E is up 2 cm? But diagram says 6 cm.

Unless the "6 cm" is a mistake, or I misread.

Perhaps the 6 cm is the length of the left vertical side.

Let's assume that.

Suppose the left vertical side is 6 cm.

Then, from bottom-left A, go up 6 cm to F.

Then diagonal to E.

E is at (8, h) , A at (0,0), F at (0,6).

Diagonal from E to F or to A?

Diagram shows diagonal from E to A, I think.

If to A, and A at (0,0), E at (8,h), distance 10 cm, so 8^2 + h^2 = 100, h^2 = 36, h=6.

So E is at (8,6).

On the right, from B(10,0) to C(10,4) — up 4 cm.

Then to D(8,4) — left 2 cm.

Then to E(8,6) — up 2 cm.

But the diagram labels "6 cm" on that last vertical segment, but it should be 2 cm, not 6 cm.

Contradiction.

Unless the "6 cm" is for the left side.

In many such problems, the unlabeled side can be found.

Perhaps the "6 cm" is the length of the left vertical side, and the vertical segment on the right is not 6 cm.

Let's read the diagram as per common practice.

In problem 3, likely:

- Bottom: 10 cm
- Right vertical: 4 cm
- Top-right horizontal: 2 cm
- Then the diagonal: 10 cm to bottom-left
- And the left vertical side is missing, but can be found.

From earlier, if E is at (8,6) for diagonal to be 10 cm to (0,0), then the vertical from D(8,4) to E(8,6) is 2 cm, but diagram says "6 cm" on that segment — so probably not.

Perhaps the 6 cm is the left side.

Assume that the vertical segment after the 2 cm horizontal is of length v, and the left side is 6 cm.

But diagram labels "6 cm" on the right-side vertical segment.

Perhaps it's a typo, or I need to interpret.

Another idea: perhaps the "6 cm" is the length of the diagonal, but no, it's labeled on the vertical.

Let's look at the numbers given: bottom 10, right vertical 4, horizontal 2, then vertical 6, diagonal 10.

Perhaps the diagonal is not to bottom-left, but to a point, and the left side is vertical with length to be determined.

But in that case, the perimeter would include that left side.

Let's calculate the position.

Set A(0,0) bottom-left.

B(10,0) bottom-right.

C(10,4) — up 4 cm.

D(8,4) — left 2 cm.

E(8,10) — up 6 cm from D, so y=4+6=10.

Then diagonal from E(8,10) to F, and F is on the left side, say (0,f), and distance 10 cm.

So distance from (8,10) to (0,f) = 10.

So √(8^2 + (10-f)^2) = 10

64 + (10-f)^2 = 100

(10-f)^2 = 36

10-f = 6 or 10-f = -6

f = 4 or f = 16

f=16 is too high, probably f=4.

So F is at (0,4).

Then from F(0,4) to A(0,0) is down 4 cm.

So the left side has two parts: from A to F: 4 cm up, then from F to E via diagonal, but the diagonal is from E to F, so the outer path is:

A to B: 10 cm
B to C: 4 cm
C to D: 2 cm
D to E: 6 cm
E to F: 10 cm (diagonal)
F to A: 4 cm (down)

Perimeter = 10 + 4 + 2 + 6 + 10 + 4 = 36 cm

And the left side is not a single segment; it's from A to F and then to E, but F to E is diagonal, so the vertical part on left is only from A to F, 4 cm, but not labeled, but we have it.

In the diagram, the left side might be shown as vertical from A to some point, but in this case, from A to F is 4 cm, then diagonal to E.

But typically, the perimeter includes all outer edges, so yes.

And the "6 cm" is used for D to E.

So perimeter = 10 (bottom) + 4 (right up) + 2 (left horizontal) + 6 (up vertical) + 10 (diagonal) + 4 (left down) = 36 cm

Yes, and it makes sense.

So for problem 3, perimeter is 36 cm.

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Problem 4)

T-shaped or arrowhead.

Sides:

- Top: 10 cm
- Right top vertical: 2 cm
- Then left 4 cm (horizontal)
- Then down diagonal 6 cm to tip
- Then up diagonal 6 cm to left side
- Then right 4 cm (horizontal)
- Then up 2 cm to top-left? But top is already there.

Trace:

Start at top-left.

Go right 10 cm to top-right.

Go down 2 cm to P.

Go left 4 cm to Q.

Go down-left diagonal 6 cm to R (tip).

Go up-left diagonal 6 cm to S.

Go right 4 cm to T.

Go up 2 cm to top-left.

Now, is T directly below top-left? Probably.

Horizontal: from top-left to top-right: 10 cm.

From P to Q: left 4 cm, so Q is 4 cm left of P, P is below top-right, so if top-right is at x=10, P at (10,-2) if y-down, but let's set coordinates.

Set top-left A(0,0), top-right B(10,0).

Then down 2 cm to C(10,-2).

Left 4 cm to D(6,-2).

Then diagonal down-left 6 cm to E.

Then diagonal up-left 6 cm to F.

Then right 4 cm to G.

Then up 2 cm to A(0,0).

From G to A: up 2 cm, so G is at (0,-2).

From F to G: right 4 cm, so F is at (-4,-2)? But then from E to F is up-left 6 cm, etc.

Distance from D(6,-2) to E: 6 cm down-left.

Similarly, E to F: 6 cm up-left.

And F to G: right 4 cm to (0,-2).

So if G is (0,-2), and F to G is right 4 cm, then F is at (-4,-2).

Then from E to F: up-left 6 cm.

From D to E: down-left 6 cm.

Let E be at (x,y).

From D(6,-2) to E(x,y): distance 6, and direction down-left, so dx<0, dy<0.

Similarly, from E to F(-4,-2): distance 6, up-left, so dx<0, dy>0.

Vector from D to E: (dx1, dy1), with dx1^2 + dy1^2 = 36, dx1<0, dy1<0.

Vector from E to F: (dx2, dy2), dx2^2 + dy2^2 = 36, dx2<0, dy2>0.

Also, F is at (-4,-2), D at (6,-2), so the vector from D to F is (-10,0).

But D to E to F, so vector D to F = D to E + E to F = (dx1+dx2, dy1+dy2) = (-10,0)

So dx1 + dx2 = -10

dy1 + dy2 = 0

Also, dx1^2 + dy1^2 = 36

dx2^2 + dy2^2 = 36

From dy1 + dy2 = 0, dy2 = -dy1

Then dx1^2 + dy1^2 = 36

dx2^2 + (-dy1)^2 = dx2^2 + dy1^2 = 36

So dx1^2 = dx2^2, so |dx1| = |dx2|

Since both dx1 and dx2 are negative (leftward), dx1 = dx2

Then dx1 + dx2 = 2 dx1 = -10, so dx1 = -5, dx2 = -5

Then from dx1^2 + dy1^2 = 25 + dy1^2 = 36, so dy1^2 = 11, dy1 = -√11 (since down)

dy2 = -dy1 = √11

But then the diagonals are 6 cm, but in the diagram, it's symmetric, and likely integer, but here not.

Perhaps the tip is directly below the center.

Top is 10 cm, so center at x=5.

The two diagonals are equal, 6 cm each, and the horizontal parts are 4 cm each on the sides.

From D(6,-2) to E, and E to F(-4,-2), but F is at (-4,-2), D at (6,-2), distance 10 cm, and E is such that DE=6, EF=6, so triangle DEF with DE=EF=6, DF=10.

Then the height from E to DF can be found.

DF is from (6,-2) to (-4,-2), length 10 cm, horizontal.

E is below, distance 6 from D and 6 from F.

So it's isosceles triangle with sides 6,6,10.

Height h from E to DF: by Pythagoras, half of DF is 5, so h = √(6^2 - 5^2) = √(36-25) = √11 ≈3.316, not nice.

But in the diagram, it's probably intended to be symmetric, and the tip is at (5, y), and distance to D(6,-2) is 6 cm.

So distance from (5,y) to (6,-2) = √((5-6)^2 + (y+2)^2) = √(1 + (y+2)^2) = 6

So 1 + (y+2)^2 = 36

(y+2)^2 = 35

y+2 = ±√35, not nice.

Perhaps the 6 cm is the length, and we don't need coordinates.

For perimeter, we can just add the given sides, as long as we have all.

From the path:

- Top: 10 cm
- Right down: 2 cm
- Left horizontal: 4 cm
- Down-left diagonal: 6 cm
- Up-left diagonal: 6 cm
- Right horizontal: 4 cm
- Up: 2 cm

That's seven segments.

Sum: 10 + 2 + 4 + 6 + 6 + 4 + 2 = let's calculate: 10+2=12, +4=16, +6=22, +6=28, +4=32, +2=34 cm.

And it closes, as per the path.

So perimeter = 34 cm.

Even though the geometry might not be exact with integers, for perimeter calculation, we add the given lengths, assuming they are correct as labeled.

So for problem 4, 34 cm.

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Problem 5)

Shape like a rectangle with a bite taken out of the bottom.

Sides:

- Top: 12 cm
- Right side: 9 cm
- Bottom-right diagonal: 5 cm
- Bottom horizontal: 6 cm
- Bottom-left diagonal: 5 cm
- Left side: ?

And right angle at top-right.

So, likely, the left side is vertical, same as right side, 9 cm, but let's see.

Trace:

Start at top-left A.

Go right 12 cm to B (top-right).

Go down 9 cm to C (bottom-right).

Go up-left diagonal 5 cm to D.

Go left 6 cm to E.

Go up-right diagonal 5 cm to F.

Then to A? But F should be on left side.

From C to D: diagonal 5 cm up-left.

D to E: left 6 cm.

E to F: diagonal 5 cm up-right.

F to A: should be vertical or what.

Since the shape is symmetric probably, and top is 12 cm, bottom horizontal is 6 cm, so the overhang on each side is (12-6)/2 = 3 cm.

So from C, go up-left 5 cm to D, which should be 3 cm left and some up.

Distance 5 cm, horizontal 3 cm, so vertical = √(5^2 - 3^2) = √(25-9) = √16 = 4 cm.

So D is 4 cm up from C.

Similarly, from E, go up-right 5 cm to F, 3 cm right and 4 cm up.

Then F should be at the same height as D, and since D is 4 cm up from bottom, and C is at bottom, so D is at height 4 cm.

Then F is also at height 4 cm.

Then from F to A: A is top-left, at height 9 cm (since right side is 9 cm), so from F at height 4 cm to A at 9 cm, difference 5 cm up.

Horizontally, if F is 3 cm right of left edge, then to A is 3 cm left, but since it's vertical, probably F is directly below A, so horizontal distance 0, but then distance would be 5 cm up, but we have diagonal 5 cm from E to F, which is already used.

In this case, from F to A should be vertical 5 cm, but is it labeled? No, but in the perimeter, we need to include it.

The outer path:

A to B: 12 cm
B to C: 9 cm
C to D: 5 cm (diagonal)
D to E: 6 cm (left)
E to F: 5 cm (diagonal)
F to A: ?

If F is at (3,4) if A is (0,9), C is (12,0), then D is at (12-3,0+4)=(9,4)? Let's set coordinates.

Set A(0,9), B(12,9), C(12,0).

From C(12,0) to D: up-left 5 cm, and since overhang 3 cm, so D is at (12-3, 0+4) = (9,4) — because horizontal change 3 cm, vertical 4 cm, distance 5 cm.

Then D to E: left 6 cm, so E(3,4).

Then E to F: up-right 5 cm, to F(3+3,4+4)=(6,8)? But should be to left side.

If symmetric, F should be at (0,4) or something.

From E(3,4) to F: up-right 5 cm, and to reach the left side, probably to (0,8) or (0,4).

Distance to (0,4): from (3,4) to (0,4) is 3 cm left, not 5 cm.

To (0,8): dx=3, dy=4, distance 5 cm, yes.

So F(0,8).

Then from F(0,8) to A(0,9): up 1 cm.

So the left side has from A to F: 1 cm down? But A is (0,9), F is (0,8), so down 1 cm, but usually we go up, but in perimeter, direction doesn't matter.

So segments:

A to B: 12 cm
B to C: 9 cm
C to D: 5 cm
D to E: 6 cm
E to F: 5 cm
F to A: 1 cm (since from y=8 to y=9)

Perimeter = 12 + 9 + 5 + 6 + 5 + 1 = 38 cm

And the left side is not full; only 1 cm from F to A, and from bottom to F is not direct.

So yes.

Perimeter = 38 cm.

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Problem 6)

Arrowhead or dart shape.

Sides:

- Two sides with tick marks: one is 15 cm, one is 7 cm, and they are equal to their counterparts? Tick marks indicate equal length.

Diagram shows:

- One side: 15 cm, with a tick mark.

- Another side: 7 cm, with a tick mark.

- And the other two sides have the same tick marks, so probably the opposite sides are equal.

Specifically, it looks like a kite or dart.

Typically, in such shapes, the two sides with single tick are equal, and the two with double tick are equal, but here only one type of tick? In the image, for problem 6, it shows:

- The long diagonal side: 15 cm, with a tick mark.

- The short side: 7 cm, with a tick mark.

- And the other two sides have the same tick marks, so likely, there are two sides of 15 cm and two of 7 cm.

But in a quadrilateral, if it's symmetric.

From the shape, it seems like a dart: two pairs of adjacent equal sides.

So probably, the perimeter is 15 + 7 + 15 + 7 = 44 cm.

But let's confirm.

The shape has four sides:

- Side 1: 15 cm (labeled)

- Side 2: 7 cm (labeled)

- Side 3: should be equal to side 1, 15 cm, by tick mark

- Side 4: equal to side 2, 7 cm, by tick mark

Yes, so perimeter = 15 + 7 + 15 + 7 = 44 cm.

No missing sides; all are given or implied equal.

So 44 cm.

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Now, summarize:

1) 34 cm

2) 23 cm

3) 36 cm

4) 34 cm

5) 38 cm

6) 44 cm

Final Answer:

1) 34 cm
2) 23 cm
3) 36 cm
4) 34 cm
5) 38 cm
6) 44 cm