Composite Surface Area and Volume worksheet with diagrams and space for answers.

Let’s solve each problem one by one. We’ll calculate surface area and volume for composite shapes — that means shapes made of two or more basic 3D figures stuck together.

We’ll use these formulas:

Cone:
- Surface Area (lateral only, since base is attached) = π × r × l
(r = radius, l = slant height)
- Volume = (1/3) × π × r² × h

Cylinder:
- Surface Area (if one end is open or attached, we adjust) → usually lateral + one base if needed
Lateral SA = 2πrh, Base area = πr²
- Volume = πr²h

Pyramid (square base):
- Surface Area = base area + 4 triangular faces
Each triangle area = (1/2) × base × slant height
- Volume = (1/3) × base area × height

Hemisphere:
- Surface Area (curved part only, since flat side is attached) = 2πr²
- Volume = (2/3)πr³

We’ll assume all measurements are in cm unless stated otherwise. Answers rounded to 2 decimal places where needed. Use π ≈ 3.1416.

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Problem 1: Cone on top of a cylinder? Wait — looking at diagram description: It says “5 ft” height total, “3 in” radius? That can’t be right — units mixed? Probably typo. Let me re-read.

Actually, from the image description (even though I can’t see it), standard problems like this usually have consistent units. Since problem 1 says “5 ft” and “3 in”, that must be an error — likely both in same unit. But let’s check common versions.

Wait — perhaps it's a cone with height 5 ft and radius 3 ft? Or maybe 3 inches? That doesn't make sense. Alternatively, maybe it’s a cone with slant height 5 ft and radius 3 ft? The label says “5 ft” along the side (slant height?) and “3 in” as radius? That would be inconsistent.

But since this is a textbook-style worksheet, and other problems use cm, probably problem 1 has a typo. Let me assume it’s cone with radius 3 cm and slant height 5 cm, and no cylinder — just a single cone? But the shape looks like a cone attached to something? Actually, looking at typical worksheets, problem 1 is often a cone alone.

Wait — user said “composite”, so likely multiple parts. Let me think differently.

Perhaps problem 1 is a cone on top of a cylinder? But dimensions given: “5 ft” and “3 in” — still messy.

Alternatively, maybe “5 ft” is height of cone, “3 in” is diameter? No.

I think there’s a miscommunication. Since I can’t see the image, I’ll rely on standard problems.

Looking at problem 2: pyramid on cube — clear.

Problem 3: hemisphere on cylinder — clear.

Problem 4: cone on hemisphere — clear.

Problem 5: pyramid on cube — similar to 2.

So problem 1 might be a single cone? But title says “composite”. Hmm.

Wait — perhaps problem 1 is a cone with radius 3 cm and height 4 cm, making slant height 5 cm (3-4-5 triangle). That makes sense! And “5 ft” might be a typo — should be 5 cm slant height.

Yes, that’s common. So let’s assume:

Problem 1: Cone with radius 3 cm, slant height 5 cm (so height = 4 cm by Pythagoras)

But is it composite? If it’s just a cone, then not composite. Unless... maybe it’s a cone attached to nothing? But the worksheet says “composite”.

Another possibility: problem 1 is a cylinder with a cone on top, but only cone dimensions given? Not likely.

Perhaps “5 ft” is total height, “3 in” is radius — but units don’t match. This is problematic.

To move forward, I’ll assume that in problem 1, it’s a cone with radius 3 cm and slant height 5 cm, and we’re to find its surface area and volume — even though it’s not composite. Maybe the worksheet includes some non-composite too? Or perhaps it’s a mistake.

But let’s look at the answer format — it asks for Formula, Numbers, Answer — so for each.

I recall that in many such worksheets, problem 1 is a cone with r=3, l=5, h=4.

So I’ll go with that.

Problem 1: Single Cone (assuming r=3 cm, l=5 cm, h=4 cm)

Surface Area (lateral, since no base mentioned or it’s open? But usually for solid, we include base? Wait — in composite, if attached, we exclude the base. But here, if it’s standalone, we include base.

The problem says “composite”, so likely it’s part of a larger shape, but in this case, perhaps it’s just the cone shown, and we treat it as is.

To avoid confusion, let’s define:

For a cone standing alone:
- Total Surface Area = πr(l + r) [lateral + base]
- Volume = (1/3)πr²h

But if it’s attached to another shape, we might exclude the base.

Since the diagram isn't visible, and to match common problems, I'll assume for problem 1 it's a cone with r=3 cm, h=4 cm, l=5 cm, and we calculate full surface area and volume.

So:

Formula for SA: πr(r + l)
Numbers: 3.1416 * 3 * (3 + 5) = 3.1416 * 3 * 8 = 75.3984 ≈ 75.40 cm²

Volume: (1/3)*π*r²*h = (1/3)*3.1416*9*4 = (1/3)*113.0976 = 37.6992 ≈ 37.70 cm³

But is this composite? Probably not. Perhaps problem 1 is different.

Another thought: maybe "5 ft" is the height of a cylinder, and "3 in" is radius — but that would be unusual. Perhaps it's 3 cm and 5 cm.

I think I need to proceed with assumptions.

Let me try to infer from the sequence.

Problem 2: pyramid on cube — composite.

Problem 3: hemisphere on cylinder — composite.

Problem 4: cone on hemisphere — composite.

Problem 5: pyramid on cube — composite.

So problem 1 must also be composite. What could it be? Perhaps a cone on a cylinder? But dimensions given are "5 ft" and "3 in" — still issue.

Maybe "5 ft" is slant height of cone, "3 in" is radius of cone, and the cylinder below has same radius, but height not given? Not specified.

This is tricky. Perhaps in the diagram, it's a cone with height 5 ft and radius 3 ft, and no other part — but then why "composite"?

Another idea: perhaps "composite" refers to the fact that we're calculating both SA and V, but that doesn't make sense.

I recall that in some worksheets, problem 1 is a cone with r=3, h=4, l=5, and it's considered as is.

To resolve, I'll assume that for problem 1, it's a cone with radius 3 cm, slant height 5 cm, and we calculate lateral surface area only (as if it's attached or open), but typically for such problems, if it's a solid, we include base.

Let's look for standard answers.

Perhaps the "5 ft" is a typo and should be "5 cm", and "3 in" should be "3 cm", and it's a cone with r=3, l=5, h=4.

And for surface area, since it's a composite shape worksheet, but this might be the first part, we'll calculate full SA.

I think I have to make a decision.

Let me set:

Problem 1: Cone with r = 3 cm, l = 5 cm, h = 4 cm (since 3-4-5 triangle)

Surface Area (total): πr(r + l) = π*3*(3+5) = 24π ≈ 75.40 cm²

Volume: (1/3)πr²h = (1/3)π*9*4 = 12π ≈ 37.70 cm³

But let's confirm if it's composite. Perhaps it's not, and the worksheet has a mix.

Maybe problem 1 is a cylinder with a cone on top, but only cone dimensions given? Unlikely.

Another possibility: "5 ft" is the height of the cone, "3 in" is the diameter, so radius 1.5 in — but then units mixed.

I think for the sake of progressing, I'll assume all units are cm, and for problem 1, it's a cone with r=3 cm, h=4 cm, l=5 cm, and we calculate SA and V as above.

So for Problem 1:

Formula: SA = πr(r + l), V = (1/3)πr²h
Numbers: SA = 3.1416*3*(3+5) = 3.1416*24 = 75.3984, V = (1/3)*3.1416*9*4 = 37.6992
Answer: SA = 75.40 cm², V = 37.70 cm³

But let's move to problem 2, which is clearer.

Problem 2: Square pyramid on top of a cube

From description: pyramid height 6 cm, base edge 8 cm (same as cube), cube side 8 cm.

So composite shape: cube with pyramid on top.

For surface area: we need to exclude the base of the pyramid and the top face of the cube, since they are attached.

So SA = SA of cube minus top face + lateral SA of pyramid

Cube SA normally 6*s², but remove top face, so 5*s² for cube part.

Pyramid lateral SA: 4 triangles, each with base 8 cm, and slant height needs to be calculated.

Given pyramid height 6 cm, base 8 cm, so for each triangular face, the slant height l can be found from the right triangle: half-base = 4 cm, height = 6 cm, so l = sqrt(4² + 6²) = sqrt(16+36) = sqrt(52) = 2√13 ≈ 7.211 cm

Then area of one triangle = (1/2)*base*slant height = (1/2)*8*7.211 = 4*7.211 = 28.844 cm²

Four triangles: 4*28.844 = 115.376 cm²

Cube part: 5 faces * (8*8) = 5*64 = 320 cm²

Total SA = 320 + 115.376 = 435.376 ≈ 435.38 cm²

Volume: volume of cube + volume of pyramid

Cube: s³ = 8^3 = 512 cm³

Pyramid: (1/3)*base area*height = (1/3)*64*6 = (1/3)*384 = 128 cm³

Total V = 512 + 128 = 640 cm³

So for Problem 2:

Formula: SA = 5s² + 4*(1/2)*b*l, V = s³ + (1/3)b²h_pyr
Numbers: s=8, b=8, l=sqrt(4^2+6^2)=sqrt(52)≈7.211, h_pyr=6
SA = 5*64 + 4*(0.5)*8*7.211 = 320 + 2*8*7.211 = 320 + 16*7.211 = 320 + 115.376 = 435.376
V = 512 + (1/3)*64*6 = 512 + 128 = 640
Answer: SA = 435.38 cm², V = 640.00 cm³

Problem 3: Hemisphere on top of a cylinder

Dimensions: cylinder height 12 cm, radius 4 cm; hemisphere radius 4 cm (same as cylinder).

Composite: so the flat face of hemisphere is attached to top of cylinder, so we exclude that circle from both.

Surface Area:
- Cylinder: lateral SA + bottom base (top is covered)
Lateral SA = 2πrh = 2*π*4*12 = 96π
Bottom base = πr² = 16π
So cylinder part: 96π + 16π = 112π
- Hemisphere: curved surface only = 2πr² = 2*π*16 = 32π
- Total SA = 112π + 32π = 144π ≈ 452.39 cm²

Volume:
- Cylinder: πr²h = π*16*12 = 192π
- Hemisphere: (2/3)πr³ = (2/3)π*64 = 128/3 π ≈ 42.6667π
- Total V = 192π + 42.6667π = 234.6667π ≈ 737.00 cm³ (let's calculate: 234.6667 * 3.1416 ≈ ?)

Better to keep symbolic.

V = πr²h + (2/3)πr³ = π*16*12 + (2/3)π*64 = 192π + 128/3 π = (576/3 + 128/3)π = 704/3 π ≈ 234.6667 * 3.1416 ≈ 737.00 cm³

Calculate: 704/3 = 234.666..., times π ≈ 3.1415926535 = 234.6667 * 3.1416 ≈ let's compute: 234.6667 * 3 = 704.0001, 234.6667 * 0.1416 ≈ 33.23, total approx 737.23, but better exact.

704/3 * π = (704π)/3

Numerically: 704 * 3.1416 / 3 = 2211.6864 / 3 = 737.2288 ≈ 737.23 cm³

Earlier I said 737.00, but let's use precise.

π ≈ 3.1415926535, so 704/3 * π = (704 * π)/3

704 * 3.1415926535 = 2211.681228, divided by 3 = 737.227076 ≈ 737.23 cm³

SA: 144 * π = 144 * 3.1415926535 = 452.3893421 ≈ 452.39 cm²

So for Problem 3:

Formula: SA = 2πrh + πr² + 2πr² = 2πrh + 3πr²? No.

Cylinder: lateral + one base = 2πrh + πr²

Hemisphere: 2πr²

Total SA = 2πrh + πr² + 2πr² = 2πrh + 3πr²

With r=4, h=12: 2π*4*12 + 3π*16 = 96π + 48π = 144π, yes.

V = πr²h + (2/3)πr³ = πr²(h + 2r/3) = π*16*(12 + 8/3) = 16π*(36/3 + 8/3) = 16π*(44/3) = 704π/3, same as before.

Numbers: SA = 144 * 3.1416 = 452.3904 ≈ 452.39 cm²

V = 704 * 3.1416 / 3 = 2211.6864 / 3 = 737.2288 ≈ 737.23 cm³

Answer: SA = 452.39 cm², V = 737.23 cm³

Problem 4: Cone on top of a hemisphere

Dimensions: cone height 8 cm, radius 5 cm; hemisphere radius 5 cm (same).

Composite: so the base of the cone is attached to the flat face of the hemisphere, so we exclude those circles.

Surface Area:
- Cone: lateral SA only = πrl, where l = slant height = sqrt(r² + h²) = sqrt(25 + 64) = sqrt(89) ≈ 9.43398 cm
So SA_cone_lateral = π*5*9.43398 ≈ 5*9.43398*π = 47.1699π
- Hemisphere: curved surface only = 2πr² = 2π*25 = 50π
- Total SA = 47.1699π + 50π = 97.1699π ≈ 305.25 cm² (calculate later)

Volume:
- Cone: (1/3)πr²h = (1/3)π*25*8 = (200/3)π ≈ 66.6667π
- Hemisphere: (2/3)πr³ = (2/3)π*125 = 250/3 π ≈ 83.3333π
- Total V = (200/3 + 250/3)π = 450/3 π = 150π ≈ 471.24 cm³

Now numerically:

Slant height l = sqrt(5² + 8²) = sqrt(25+64) = sqrt(89) ≈ 9.433981132

SA_cone_lateral = π * 5 * 9.433981132 = 47.16990566π

SA_hemisphere = 2 * π * 25 = 50π

Total SA = 97.16990566π ≈ 97.1699 * 3.1415926535 ≈ let's compute: 97.1699 * 3.1416 ≈ 305.25 (more precisely: 97.16990566 * 3.1415926535 = calculate step by step)

First, 97.16990566 * 3 = 291.50971698

97.16990566 * 0.1415926535 ≈ 97.17 * 0.1416 ≈ 13.76, more accurately: 97.1699 * 0.1415926535

Approx: 97.17 * 0.1416 = 97.17*0.14 = 13.6038, 97.17*0.0016=0.155472, total 13.759272

So total SA ≈ 291.5097 + 13.7593 = 305.269, but this is rough.

Better: use calculator if possible, but since text, assume π=3.1416

SA = 97.1699 * 3.1416 ≈ let's multiply: 97.1699 * 3 = 291.5097, 97.1699 * 0.1416 = approximately 13.759, total 305.2687 ≈ 305.27 cm²

But earlier I had 97.1699π, and π≈3.1415926535, so 97.16990566 * 3.1415926535 = let's say approximately 305.25? I think I miscalculated.

Standard way: l = sqrt(89) = √89

SA = πr l + 2πr² = π*5*√89 + 2π*25 = 5π√89 + 50π

√89 ≈ 9.433981132056603

5*9.433981132056603 = 47.169905660283015

Times π ≈ 47.169905660283015 * 3.141592653589793 = 148.185 (wait no)

47.1699 * 3.1416 ≈ 148.185? That can't be right because 50π is about 157, so total should be around 305.

47.1699 * π = 47.1699 * 3.1415926535 ≈ let's compute:

47.1699 * 3 = 141.5097

47.1699 * 0.1415926535 ≈ 47.17 * 0.1416 ≈ 6.68, more accurately 47.1699 * 0.14 = 6.603786, 47.1699 * 0.0015926535 ≈ 0.0751, total 6.678886

So 141.5097 + 6.6789 = 148.1886 for cone part

Hemisphere: 50 * π = 50 * 3.1415926535 = 157.079632675

Total SA = 148.1886 + 157.0796 = 305.2682 ≈ 305.27 cm²

Volume: 150 * π = 150 * 3.1415926535 = 471.238898025 ≈ 471.24 cm³

So for Problem 4:

Formula: SA = πr l + 2πr², V = (1/3)πr²h + (2/3)πr³
Numbers: r=5, h_cone=8, l=√(25+64)=√89≈9.434, h_hemi not needed
SA = π*5*9.434 + 2π*25 = 47.17π + 50π = 97.17π ≈ 305.27 cm²
V = (1/3)π*25*8 + (2/3)π*125 = (200/3)π + (250/3)π = 450/3 π = 150π ≈ 471.24 cm³
Answer: SA = 305.27 cm², V = 471.24 cm³

Problem 5: Pyramid on top of a cube

Similar to problem 2, but different dimensions.

From description: pyramid height 10 cm, base edge 6 cm; cube side 6 cm.

So composite: cube with pyramid on top.

SA: exclude top face of cube and base of pyramid.

Cube part: 5 faces * (6*6) = 5*36 = 180 cm²

Pyramid lateral SA: 4 triangles, each with base 6 cm, slant height l = sqrt((half-base)^2 + height^2) = sqrt(3^2 + 10^2) = sqrt(9+100) = sqrt(109) ≈ 10.4403 cm

Area of one triangle = (1/2)*6*10.4403 = 3*10.4403 = 31.3209 cm²

Four triangles: 4*31.3209 = 125.2836 cm²

Total SA = 180 + 125.2836 = 305.2836 ≈ 305.28 cm²

Volume:
Cube: 6^3 = 216 cm³
Pyramid: (1/3)*base area*height = (1/3)*36*10 = 120 cm³
Total V = 216 + 120 = 336 cm³

So for Problem 5:

Formula: SA = 5s² + 4*(1/2)*b*l, V = s³ + (1/3)b²h_pyr
Numbers: s=6, b=6, l=sqrt(3^2+10^2)=sqrt(109)≈10.4403, h_pyr=10
SA = 5*36 + 2*6*10.4403 = 180 + 12*10.4403 = 180 + 125.2836 = 305.2836
V = 216 + (1/3)*36*10 = 216 + 120 = 336
Answer: SA = 305.28 cm², V = 336.00 cm³

Now back to Problem 1. Since it's listed first and likely simpler, and given "5 ft" and "3 in", but probably typo, and in many sources, it's a cone with r=3, h=4, l=5.

Moreover, in the context, perhaps it's not composite, or perhaps it is a cone alone.

Maybe "5 ft" is the slant height, "3 in" is radius, but then units不一致. Assume both in cm.

I think for consistency, I'll assume Problem 1 is a cone with radius 3 cm, slant height 5 cm, height 4 cm, and we calculate full surface area and volume.

So:

SA = πr(r + l) = π*3*(3+5) = 24π ≈ 75.40 cm²

V = (1/3)πr²h = (1/3)π*9*4 = 12π ≈ 37.70 cm³

But to match the composite theme, perhaps it's a different shape. Another common problem is a cylinder with hemispherical ends, but not here.

Perhaps problem 1 is a cone with r=3, h=5, then l=sqrt(9+25)=sqrt(34)≈5.83095, but the "5 ft" might be height.

The label says "5 ft" along the side, so likely slant height.

I think my initial assumption is fine.

So let's summarize all.

For each problem, we'll write Formula, Numbers, Answer.

Also, note that in some cases, for surface area, if the shape is closed, we include all external surfaces, which we did by excluding the attached faces.

Now for the final answer, since the user wants the answers for each, I'll list them.

But the response should have the reasoning first, then final answer.

Since ExplainAnswer is false, I won't explain, but I need to provide the answers.

The instruction is to solve the problem, and provide final answer after reasoning.

But in this case, since it's multiple problems, I'll provide the answers for each.

Perhaps the "final answer" is the set of answers.

Let me compile.

First, for Problem 1:

Assume cone with r=3 cm, l=5 cm, h=4 cm.

Formula: SA = πr(r + l), V = (1/3)πr²h

Numbers: r=3, l=5, h=4

SA = 3.1416 * 3 * (3+5) = 3.1416 * 24 = 75.3984 ≈ 75.40 cm²

V = (1/3) * 3.1416 * 9 * 4 = (1/3)*113.0976 = 37.6992 ≈ 37.70 cm³

But is this correct for composite? Perhaps not, but I think it's acceptable.

Maybe in the diagram, it's a different shape. Another possibility: "5 ft" is height of a cylinder, "3 in" is radius, but then for a cylinder alone, SA = 2πr(h+r), V=πr²h, but units mixed.

I think I have to go with this.

So for Problem 1: SA = 75.40 cm², V = 37.70 cm³

Problem 2: SA = 435.38 cm², V = 640.00 cm³

Problem 3: SA = 452.39 cm², V = 737.23 cm³

Problem 4: SA = 305.27 cm², V = 471.24 cm³

Problem 5: SA = 305.28 cm², V = 336.00 cm³

Now for the final answer, since the user may expect the answers listed, I'll box them.

But the format says "Final Answer: [provide only the final correct answer]", so perhaps list all.

Since it's a worksheet, maybe provide for each.

I think I'll write the answers as per the problems.

So Final Answer:

1. Surface Area: 75.40 cm², Volume: 37.70 cm³
2. Surface Area: 435.38 cm², Volume: 640.00 cm³
3. Surface Area: 452.39 cm², Volume: 737.23 cm³
4. Surface Area: 305.27 cm², Volume: 471.24 cm³
5. Surface Area: 305.28 cm², Volume: 336.00 cm³

But to be precise, let's double-check Problem 1.

Upon second thought, in many online sources, for a similar worksheet, problem 1 is a cone with radius 3 in, height 4 in, slant height 5 in, and they calculate SA and V.

And units are inches, but in the problem it says "5 ft" and "3 in", which is likely a typo, and should be consistent.

Perhaps "5 ft" is a mistake, and it's 5 in or 5 cm.

To match, I'll assume units are cm for all, as other problems use cm.

In problem 2,3,4,5, dimensions are in cm, so probably problem 1 is also in cm.

So r=3 cm, l=5 cm, h=4 cm.

SA = π*3*(3+5) = 24π = 75.39822368615503 cm² ≈ 75.40 cm²

V = (1/3)*π*9*4 = 12π = 37.69911184307752 cm³ ≈ 37.70 cm³

Yes.

For Problem 4, I have SA=305.27, but let's confirm calculation.

l = sqrt(5^2 + 8^2) = sqrt(25+64) = sqrt(89) = 9.433981132056603

SA_cone_lateral = π * 5 * 9.433981132056603 = 47.169905660283015 * π

SA_hemisphere = 2 * π * 25 = 50π

Total SA = (47.169905660283015 + 50) π = 97.169905660283015 π

π = 3.141592653589793

97.169905660283015 * 3.141592653589793 = let's calculate:

97.169905660283015 * 3 = 291.509716980849045

97.169905660283015 * 0.141592653589793 = approximately 13.759 (use calculator if possible, but in text)

97.169905660283015 * 0.14 = 13.6037867924396221

97.169905660283015 * 0.001592653589793 ≈ 0.1548 (since 97.17*0.00159265≈0.1548)

More accurately: 97.169905660283015 * 0.001 = 0.097169905660283015

97.169905660283015 * 0.0005 = 0.0485849528301415075

97.169905660283015 * 0.000092653589793 ≈ 97.17 * 0.00009265 ≈ 0.009005

So total for 0.001592653589793: approximately 0.09717 + 0.04858 + 0.00900 = 0.15475

Then for 0.141592653589793: 13.6037867924396221 + 0.15475 = 13.7585367924396221

Then total SA = 291.509716980849045 + 13.7585367924396221 = 305.2682537732886671 ≈ 305.27 cm²

Yes.

Similarly, others are fine.

So now for the final answer.

Since the user may want the answers filled in, I'll provide them as per the problems.

Final Answer:
1. Surface Area: 75.40 cm², Volume: 37.70 cm³
2. Surface Area: 435.38 cm², Volume: 640.00 cm³
3. Surface Area: 452.39 cm², Volume: 737.23 cm³
4. Surface Area: 305.27 cm², Volume: 471.24 cm³
5. Surface Area: 305.28 cm², Volume: 336.00 cm³

Note: For Problem 1, assumed a cone with radius 3 cm, slant height 5 cm, height 4 cm. If the actual diagram shows a different composite shape, the answer may vary, but based on common problems, this is likely.

To be accurate, perhaps in the diagram, problem 1 is a cylinder with a cone on top, but without dimensions for cylinder, it's hard. Given the information, I'll stick with this.

So the final answer is as above.