Geometric shapes and their 2D representations for educational purposes.

Looking at the worksheet, we need to find the surface area of each 3D shape shown. The instruction says: “Calculate the area of each shape. Use cm² for area (small shapes).”

We’ll go one by one.

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Shape 1: Rectangular prism (box)
Dimensions: 4 cm × 3 cm × 2 cm
Surface area = 2(lw + lh + wh)
= 2[(4×3) + (4×2) + (3×2)]
= 2[12 + 8 + 6] = 2×26 = 52 cm²

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Shape 2: Cylinder
Radius = 1 cm, Height = 3 cm
Surface area = 2πr² + 2πrh
= 2π(1)² + 2π(1)(3)
= 2π + 6π = 8π ≈ 8 × 3.14 = 25.12 cm²

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Shape 3: Cylinder
Radius = 3 cm, Height = 5 cm
Surface area = 2πr² + 2πrh
= 2π(9) + 2π(3)(5)
= 18π + 30π = 48π ≈ 48 × 3.14 = 150.72 cm²

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Shape 4: Rectangular prism
Dimensions: 5 cm × 3 cm × 2 cm
Surface area = 2(lw + lh + wh)
= 2[(5×3) + (5×2) + (3×2)]
= 2[15 + 10 + 6] = 2×31 = 62 cm²

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Shape 5: Triangular pyramid (tetrahedron)
This is tricky — it looks like a triangular pyramid with equilateral triangle base and three triangular faces. But from the diagram, it seems to be a regular tetrahedron with all edges 3 cm? Wait — actually, looking again, it’s labeled with “3 cm” on one edge and “3 cm” on height? Hmm.

Actually, let’s re-express: It appears to be a pyramid with a triangular base. If it’s a regular tetrahedron (all faces are equilateral triangles), then:

Area of one equilateral triangle = (√3/4) × side²
= (√3/4) × 9 ≈ (1.732/4) × 9 ≈ 0.433 × 9 ≈ 3.897 cm²
Total surface area = 4 × 3.897 ≈ 15.59 cm²

But wait — the diagram shows “3 cm” as the slant height? Or edge? Let me check again.

Actually, in many school worksheets, if it’s drawn as a pyramid with a triangular base and labeled “3 cm” on the base edge and “3 cm” as the height of the triangle face, then:

Each face is a triangle with base 3 cm and height 3 cm → area = (1/2)×3×3 = 4.5 cm²
If there are 4 faces → 4 × 4.5 = 18 cm²

But that would only be true if all faces are identical — which they might not be.

Alternatively, perhaps it’s a pyramid with a triangular base where the base is an equilateral triangle of side 3 cm, and the lateral faces are also triangles with base 3 cm and height 3 cm.

Let’s assume that for simplicity (common in middle school):

Base area = (√3/4) × 3² ≈ 3.897 cm²
Each lateral face = (1/2) × 3 × 3 = 4.5 cm² → 3 faces = 13.5 cm²
Total = 3.897 + 13.5 ≈ 17.4 cm²

But this is messy. Maybe the problem expects us to treat it as having 4 triangular faces each 3 cm base and 3 cm height? Then 4 × 4.5 = 18 cm².

I think for level consistency, and since other problems use simple numbers, maybe it’s intended to be 4 faces × (1/2 × 3 × 3) = 18 cm².

Wait — actually, looking at the drawing again: it has a square base? No, it’s labeled with 3 cm on two sides and a diagonal? Actually, no — it’s a triangular pyramid with all edges 3 cm? That would make it a regular tetrahedron.

Standard formula for regular tetrahedron surface area: √3 × a²
a = 3 → √3 × 9 ≈ 1.732 × 9 = 15.588 cm²

I’ll go with 15.59 cm² (rounded to two decimals).

But let’s hold off — maybe the worksheet expects integer answers? Let’s see others.

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Shape 6: L-shaped prism (or composite shape)
It’s a 3D shape made of two rectangular prisms.

From the diagram:
Top part: 3 cm long, 1 cm wide, 1 cm high
Bottom part: 3 cm long, 1 cm wide, 2 cm high? Wait — total height is 3 cm, and top part is 1 cm, so bottom is 2 cm? But width is same?

Actually, it’s better to think of it as a single prism with L-shaped cross-section.

Cross-section area:
L-shape: outer rectangle 3×3 minus inner 2×2? No.

Looking:
The L-shape has:
- Vertical leg: 3 cm tall, 1 cm wide
- Horizontal leg: 3 cm long, 1 cm wide
But they overlap 1×1, so total cross-section area = (3×1) + (3×1) - (1×1) = 3+3-1 = 5 cm²

Then the depth (into page) is 1 cm? From the diagram, it says “1 cm” on the depth.

So volume would be 5 cm³, but we need surface area.

Surface area of a prism = 2 × base area + perimeter of base × height (depth)

Base area = 5 cm²
Perimeter of L-shape: let’s trace it.

Starting from bottom left:
→ right 3 cm
↑ up 1 cm
← left 2 cm
↑ up 2 cm
→ right 1 cm
↓ down 3 cm
← left 3 cm
↓ down 0? Wait, better to list all outer edges.

Actually, standard way: for an L-shape made of two rectangles:

Imagine:
Rectangle A: 3 cm (height) × 1 cm (width)
Rectangle B: 2 cm (width) × 1 cm (height) attached to the right of A, but only 1 cm high? No.

From the diagram:
The L-shape has:
- Left vertical: 3 cm high, 1 cm wide
- Bottom horizontal: 3 cm long, 1 cm wide
But the corner is shared.

So the outline:
Start at bottom left:
Go right 3 cm (bottom of horizontal part)
Go up 1 cm (right end of horizontal part)
Go left 2 cm (top of horizontal part, but only 2 cm because vertical part is 1 cm wide)
Go up 2 cm (left side of vertical part, above the horizontal)
Go right 1 cm (top of vertical part)
Go down 3 cm (left side)
Back to start.

So perimeter = 3 + 1 + 2 + 2 + 1 + 3 = 12 cm

Then surface area = 2 × base area + perimeter × depth
Base area = 5 cm² (as calculated)
Depth = 1 cm
So SA = 2×5 + 12×1 = 10 + 12 = 22 cm²

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Shape 7: Triangular prism
Base is a triangle with base 3 cm and height 2 cm? From diagram: “3 cm” on base, “2 cm” as height of triangle, and length of prism is 4 cm.

First, area of triangular base = (1/2) × 3 × 2 = 3 cm²
Two bases: 2 × 3 = 6 cm²

Now lateral faces: three rectangles.

The triangle has sides: base 3 cm, and two other sides. Since height is 2 cm, and assuming it’s isosceles? Or right triangle?

Diagram shows “2 cm” as height, and “3 cm” as base, and the prism length is 4 cm.

The lateral faces:
- One rectangle: 3 cm × 4 cm = 12 cm² (for the base)
- Two rectangles: each is the slant side × 4 cm.

What is the slant side? If the triangle is isosceles with base 3 cm and height 2 cm, then each half is 1.5 cm, so slant side = √(1.5² + 2²) = √(2.25 + 4) = √6.25 = 2.5 cm

So each of the two side rectangles: 2.5 × 4 = 10 cm² → total 20 cm²

Total surface area = 6 + 12 + 20 = 38 cm²

If it’s a right triangle with legs 2 cm and 3 cm, then hypotenuse = √(4+9)=√13≈3.606, but diagram doesn’t specify.

Given the numbers, likely isosceles with height 2 cm, so slant 2.5 cm is nice number.

So I’ll go with 38 cm²

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Now, summarizing:

1. 52 cm²
2. 25.12 cm² (or 8π)
3. 150.72 cm² (or 48π)
4. 62 cm²
5. 15.59 cm² (regular tetrahedron) or 18 cm²? Let's decide.

For shape 5, if it's a pyramid with 4 triangular faces, each with base 3 cm and height 3 cm, then area per face = 4.5 cm², total 18 cm². This is simpler and matches the level.

In many textbooks, they simplify. So I'll use 18 cm² for shape 5.

Shape 6: 22 cm²
Shape 7: 38 cm²

But let's confirm shape 5.

Upon second thought, the diagram for shape 5 shows a pyramid with a triangular base, and labels "3 cm" on the base edge and "3 cm" as the height from base to apex? But in 3D, it's ambiguous.

To match the simplicity of other problems, and since 3 cm is given, perhaps it's intended to be a tetrahedron with all faces equilateral, but then area is irrational.

Maybe it's a pyramid with a square base? No, it's drawn as triangular.

Another idea: perhaps "3 cm" is the edge length, and it's a regular tetrahedron, so SA = √3 * a^2 = 1.732 * 9 = 15.588, round to 15.6 cm².

But let's see the answer format. The user didn't specify, but in school, sometimes they want exact or decimal.

For consistency, I'll use decimals rounded to two places.

So:

1. 52.00 cm²
2. 25.12 cm²
3. 150.72 cm²
4. 62.00 cm²
5. 15.59 cm² (if regular tetrahedron) or 18.00 cm²? I think 15.59 is more accurate.

Wait, let's calculate regular tetrahedron surface area exactly.

Formula: SA = √3 * a^2
a=3, so SA = 9√3
√3 ≈ 1.7320508, so 9*1.7320508 = 15.5884572 → 15.59 cm²

I think that's correct.

Shape 6: 22.00 cm²
Shape 7: 38.00 cm²

But for shape 7, if the triangle is right-angled with legs 2 cm and 3 cm, then area = 3 cm², same as before.

Hypotenuse = √(2^2 + 3^2) = √13 ≈ 3.606 cm

Then lateral faces:
- Base rectangle: 3 * 4 = 12 cm²
- Side1: 2 * 4 = 8 cm²
- Side2: 3.606 * 4 ≈ 14.424 cm²
Total lateral = 12 + 8 + 14.424 = 34.424 cm²
Plus two bases: 2*3 = 6 cm²
Total = 40.424 cm² — not nice number.

Whereas if it's isosceles with height 2 cm and base 3 cm, then equal sides are 2.5 cm, as I had, giving 38 cm², which is nice.

And 2.5 is 5/2, so likely intended.

So I'll stick with 38 cm² for shape 7.

Final list:

1. 52 cm²
2. 25.12 cm²
3. 150.72 cm²
4. 62 cm²
5. 15.59 cm²
6. 22 cm²
7. 38 cm²

But for shape 2 and 3, should we leave in terms of π? The instruction says "use cm²", and for small shapes, so probably decimal is fine.

In many curricula, they use π ≈ 3.14.

So yes.

Now, for the final answer, since the user didn't specify which shape, but the task is to calculate for each, I think we need to provide all.

But the response should be the final answer for the task. Since it's a worksheet with multiple shapes, perhaps list them.

But the instruction says "provide only the final correct answer", and "after all explanations".

So I'll list the surface areas in order.

Final Answer:
52 cm², 25.12 cm², 150.72 cm², 62 cm², 15.59 cm², 22 cm², 38 cm²

But let's write it clearly.