Surface Area of Composite Figures | PDF - Free Printable

Let’s solve each composite figure step by step. We’ll calculate surface area for each, rounding to two decimal places.

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Figure 1: Rectangular prism with triangular prism on top

Dimensions:
- Base rectangular prism: length = 15 in, width = 8 in, height = 4 in
- Triangular prism on top: base triangle has sides 6 in, 6 in, and base 8 in (same as width of box), height of triangle = ? Wait — actually, the diagram shows a triangular face with sides labeled 6 in, 6 in, and the base is 8 in? But that doesn’t make sense because 6+6 > 8, so it’s possible. Actually, looking again — the triangle is drawn with two slanted edges of 6 in each, and the base is along the 8-in side of the rectangle. So we can find the height of the triangle using Pythagoras.

Wait — actually, in many such problems, if not given, sometimes they assume you use the given numbers directly. Let me re-express:

Actually, from standard interpretation of this common problem:

The composite shape is:
- A rectangular prism: 15 in (length) × 8 in (width) × 4 in (height)
- On top, centered or aligned, a triangular prism whose triangular face has base = 8 in (same as width), and two equal sides of 6 in each → so it's an isosceles triangle.

We need surface area of entire composite figure — meaning we do NOT include areas where shapes are glued together.

So:

Step 1: Surface area of rectangular prism alone:
SA_rect = 2(lw + lh + wh) = 2(15×8 + 15×4 + 8×4) = 2(120 + 60 + 32) = 2(212) = 424 in²

But wait — the top face (15×8) is partially covered by the triangular prism. The triangular prism sits on top, covering a rectangle of size 15 in (length) × 8 in (width)? No — actually, the triangular prism runs along the length? Or across?

Looking at typical diagrams: usually, the triangular prism is placed so its triangular faces are on the ends (across the width), and it extends the full length.

In Figure 1, the triangle is shown on the front, with base 8 in (matching the width of the box), and the prism extends 15 in long (same as length of box). So yes — the triangular prism covers the entire top face of the rectangular prism? Not exactly — the top face of the rectangular prism is 15×8, and the triangular prism sits on it, but since the triangular prism has a rectangular base of 15×8? No — the base of the triangular prism is the triangle itself? I think I’m confusing myself.

Better approach: Think of the composite solid as made of:

- Bottom: rectangular prism (15x8x4)
- Top: triangular prism sitting on top, with triangular cross-section (base 8 in, legs 6 in each), and length 15 in.

When joined, the bottom face of the triangular prism (which is a rectangle 15 in x 8 in?) — no, the “base” of the triangular prism is the triangle, and it extrudes 15 in. So the face that touches the rectangular prism is a rectangle of size 15 in (length) × 8 in (base of triangle).

Therefore, when calculating total surface area, we must subtract the overlapping area twice? No — only once from each part? Actually, standard rule: add surface areas of both solids, then subtract 2 × area of overlap (because both had that face included, but now it’s internal).

But easier: calculate exposed surfaces only.

Let’s break down Figure 1:

Exposed surfaces:

1. Bottom of rectangular prism: 15 × 8 = 120 in²
2. Front and back of rectangular prism: each is 15 × 4 = 60 → total 120 in²
3. Left and right sides of rectangular prism: each is 8 × 4 = 32 → total 64 in²
BUT — the top of the rectangular prism is mostly covered, except possibly the parts not under the triangle? Actually, the triangular prism sits on top and covers the entire 15x8 top? No — because the triangular prism has a triangular cross-section, so its "bottom" face is a rectangle 15 in long and 8 in wide? That would mean it covers the whole top.

Wait — perhaps the triangular prism is oriented differently. In many textbooks, for such figures, the triangular prism is placed so that its rectangular lateral faces are vertical, and the triangular faces are on the ends.

Given the drawing shows a triangle on the front with sides 6,6, and base 8, and the prism goes back 15 in, then:

The triangular prism has:
- Two triangular faces: each with base 8 in, height h (we need to find h)
- Three rectangular faces:
- One base rectangle: 8 in × 15 in (this is the one touching the rectangular prism — so hidden)
- Two slanted rectangles: each 6 in × 15 in

First, find height of triangle: isosceles triangle with sides 6,6, base 8.

Height h = √(6² - 4²) = √(36 - 16) = √20 ≈ 4.472 in

Now, surface area of triangular prism alone (if separate):
= 2*(area of triangle) + perimeter_of_triangle * length
Area of one triangle = (1/2)*8*√20 = 4*√20 ≈ 4*4.472 = 17.888 in²
Perimeter of triangle = 6+6+8 = 20 in
Lateral SA = 20 * 15 = 300 in²
Total SA_tri = 2*17.888 + 300 ≈ 35.776 + 300 = 335.776 in²

But when attached, the base rectangle (8x15=120 in²) is hidden.

Similarly, for rectangular prism: SA = 2(15*8 + 15*4 + 8*4) = 2(120+60+32)=2*212=424 in²
But its top face (15*8=120 in²) is also hidden.

So total composite SA = SA_rect + SA_tri - 2*overlap_area
Overlap area = 120 in² (the contact face)

So SA_total = 424 + 335.776 - 2*120 = 759.776 - 240 = 519.776 ≈ 519.78 in²

But let me verify by adding exposed faces only:

Rectangular prism exposed:
- Bottom: 15*8 = 120
- Front/back: 2*(15*4) = 120
- Left/right: 2*(8*4) = 64
→ Total rect exposed: 120+120+64 = 304

Triangular prism exposed:
- Two triangular faces: 2*(1/2*8*√20) = 8*√20 ≈ 8*4.472 = 35.776
- Two slanted rectangular faces: 2*(6*15) = 180
- The base rectangle (8*15) is hidden, so not included.
→ Total tri exposed: 35.776 + 180 = 215.776

Total composite SA = 304 + 215.776 = 519.776 ≈ 519.78 in²

Okay, good.

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Figure 2: Hemisphere on top of cylinder? Wait — looks like a sphere cut in half on top of a cone? Or just a hemisphere?

Diagram shows: a hemisphere (half-sphere) with radius 10 cm, and below it... actually, looking closely, it seems to be a hemisphere sitting on a flat surface, but there's a line indicating diameter 10 cm? Wait, label says "10 cm" pointing to the radius? Or diameter?

In the image, it says "10 cm" next to the curved part, likely radius. And it's a hemisphere — so surface area of hemisphere includes the curved surface plus the circular base? But if it's sitting on a surface, the base might not be exposed. However, in most problems like this, unless specified, we consider the total surface area including the base if it's part of the figure.

Wait — actually, in Figure 2, it appears to be just a hemisphere — no other shape. The diagram shows a dome shape with a flat bottom, labeled 10 cm — probably radius.

Standard surface area of a hemisphere:
Curved surface area = 2πr²
Base area = πr²
Total SA = 3πr² if including base.

But if it's a standalone figure, and we're to find total surface area, we include the base.

However, sometimes in composite figures, if it's attached, we exclude, but here it seems standalone.

Looking at the label: "10 cm" — likely radius.

So r = 10 cm

SA_hemisphere = 2πr² + πr² = 3πr² = 3 * π * 100 = 300π ≈ 300 * 3.1416 = 942.48 cm²

But wait — is it a full hemisphere or something else? The diagram might show a different shape. Another possibility: it could be a cone with a hemisphere on top? But the diagram doesn't show a cone — it shows a smooth curve from base to top, which is characteristic of a hemisphere.

I think it's safe to assume it's a hemisphere with radius 10 cm.

So SA = 3πr² = 3 * π * 10^2 = 300π ≈ 942.48 cm²

But let me confirm: some sources say for a hemisphere, if it's a solid, total surface area is 3πr², yes.

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Figure 3: Cube with hemisphere on top

Cube: side 4 yd

Hemisphere on top: diameter 2 yd, so radius 1 yd

Surface area:

Cube alone: 6 * (side)^2 = 6 * 16 = 96 yd²

But the top face of the cube is partially covered by the hemisphere. The hemisphere sits on the cube, covering a circle of radius 1 yd.

So, we remove the area of the circle from the cube's top face, and add the curved surface area of the hemisphere.

Also, the flat base of the hemisphere is not exposed — it's glued to the cube.

So:

Exposed SA of cube: original SA minus area of circle on top
= 96 - πr² = 96 - π(1)^2 = 96 - π

Exposed SA of hemisphere: only curved surface = 2πr² = 2π(1)^2 = 2π

Total SA = (96 - π) + 2π = 96 + π ≈ 96 + 3.1416 = 99.1416 ≈ 99.14 yd²

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Figure 4: Two rectangular prisms stacked

Bottom prism: 14 m (l) × 5 m (w) × 3 m (h)

Top prism: 6 m (l) × 4 m (w) × 4 m (h) — assuming it's placed on top, centered or aligned.

From diagram, likely the top prism is smaller and sits on the bottom one.

To find total surface area, we add SA of both, then subtract twice the area of overlap (since both had that face, but now internal).

First, SA_bottom = 2(lw + lh + wh) = 2(14*5 + 14*3 + 5*3) = 2(70 + 42 + 15) = 2(127) = 254 m²

SA_top = 2(6*4 + 6*4 + 4*4) = 2(24 + 24 + 16) = 2(64) = 128 m²

Overlap area: the area where they touch. Assuming the top prism sits entirely on the bottom one, and since top is 6x4, and bottom is 14x5, the contact area is 6*4 = 24 m² (if aligned properly).

So total SA = SA_bottom + SA_top - 2*overlap = 254 + 128 - 2*24 = 382 - 48 = 334 m²

But let's verify by exposed faces.

Bottom prism exposed:
- Bottom: 14*5 = 70
- Front/back: 2*(14*3) = 84
- Left/right: 2*(5*3) = 30
- Top: but part is covered. The top face is 14*5=70, minus the covered part 6*4=24, so exposed top = 70 - 24 = 46
→ Total bottom exposed: 70 + 84 + 30 + 46 = 230

Top prism exposed:
- All faces except the bottom one that's glued.
SA_top without bottom = total SA_top - area of bottom = 128 - 24 = 104? No.

Better: top prism has:
- Top: 6*4 = 24
- Front/back: 2*(6*4) = 48? Height is 4 m, so front/back are 6m x 4m each → 2*24=48
- Left/right: 2*(4*4) = 32
- Bottom: hidden, so not included.
→ Exposed SA_top = 24 + 48 + 32 = 104

Total composite SA = 230 + 104 = 334 m² — same as before.

Good.

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Figure 5: Square pyramid on top of cube

Cube: side 10 ft

Pyramid: square base 10 ft x 10 ft (same as cube top), and slant height? Diagram shows 13 ft — likely the slant height of the triangular faces.

For a square pyramid, surface area (lateral) is 4 * (1/2 * base * slant height) = 2 * base * slant height

But total SA of pyramid includes base? When attached to cube, the base is hidden.

So:

Cube SA alone: 6 * 10^2 = 600 ft²

But top face is covered by pyramid base, so subtract 10*10 = 100 ft²

Pyramid lateral SA: 4 triangles, each with base 10 ft, height (slant height) 13 ft

Area of one triangle = (1/2)*10*13 = 65 ft²

Four triangles: 4*65 = 260 ft²

Pyramid base is hidden, so not added.

Total SA = (cube SA - top face) + pyramid lateral SA = (600 - 100) + 260 = 500 + 260 = 760 ft²

Is that correct? Yes.

Note: if the 13 ft was the edge length, not slant height, we'd need to calculate, but diagram likely indicates slant height.

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Figure 6: Cylinder with hemispheres on both ends? Or just a sphere?

Diagram shows a sphere with a cylinder inside? Or a capsule shape? Label says "28 ft" — likely diameter.

Actually, it looks like a sphere with a cylindrical section, but the label "28 ft" is across the middle, suggesting diameter of sphere.

But there's a horizontal line through the center, and labels — wait, it might be a cylinder with two hemispherical ends, making a capsule.

Label: "28 ft" — probably the diameter of the sphere/hemisphere.

And there's a dimension for the cylinder part? In the diagram, it shows a cylinder with two hemispheres, and the total length isn't given, but the diameter is 28 ft.

Wait — looking closely, it might be just a sphere. But the diagram has a rectangle in the middle with semicircles on ends, which is typical for a capsule.

But in the label, it says "28 ft" pointing to the diameter, and no other dimension. Perhaps it's a sphere.

Another possibility: it's a cylinder with height equal to diameter, but that doesn't match.

I recall that in some worksheets, Figure 6 is a sphere with diameter 28 ft.

Let me assume it's a sphere.

Then SA_sphere = 4πr²

Diameter = 28 ft, so radius r = 14 ft

SA = 4 * π * 14^2 = 4 * π * 196 = 784π ≈ 784 * 3.1416 = 2463.0144 ≈ 2463.01 ft²

But if it's a capsule (cylinder with two hemispheres), we need the length of the cylinder. Since not given, likely it's just a sphere.

In the diagram description, it says "a sphere" or similar, but based on common problems, and since no cylinder height is given, I'll go with sphere.

To confirm: if it were a capsule, we'd need the straight part length, which isn't provided, so probably sphere.

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Now, summarizing all answers:

1) 519.78 in²
2) 942.48 cm²
3) 99.14 yd²
4) 334.00 m²
5) 760.00 ft²
6) 2463.01 ft²

But let me double-check Figure 2. Is it really a hemisphere? In some interpretations, if it's a dome on a flat surface, and we're to find external surface area, we might not include the base. But the problem says "surface area of composite figures", and for a single hemisphere, it's ambiguous.

Looking back at the user's image description: for Figure 2, it's labeled with "10 cm" and it's a half-sphere shape. In many math problems, when they show a hemisphere and ask for surface area, they mean total surface area including the base, unless specified otherwise.

But let's see the context: other figures have multiple parts, but Figure 2 seems standalone. To be safe, I'll check both.

If only curved surface: 2πr² = 2*π*100 = 200π ≈ 628.32 cm²

But I think including base is standard for "surface area of the figure".

Upon second thought, in the worksheet title "Surface Area of Composite Figures", and Figure 2 might be intended as a hemisphere alone, so total SA = 3πr².

I'll stick with 942.48 cm².

For Figure 6, if it's a sphere, 2463.01 ft² is correct.

Now, final answers rounded to two decimals.

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Final Answer:
1) 519.78
2) 942.48
3) 99.14
4) 334.00
5) 760.00
6) 2463.01