Practice worksheet for finding the surface area of various solid figures.

Let’s solve each problem step by step. We’re finding the surface area of solid figures made from rectangular prisms (like boxes). Surface area means the total area of all the outside faces.

We’ll use this idea:
For any box, surface area = 2×(length×width + length×height + width×height)
But when shapes are joined together (like L-shapes or steps), some faces are hidden inside — so we can’t just add two separate boxes. Instead, we count only the exposed faces.

A better way for these problems:
→ Break the shape into parts you know (like front, back, top, bottom, left, right).
→ Or imagine “unfolding” it and adding up all visible rectangles.
→ Sometimes it’s easier to think: “What would the surface area be if it were a full big box? Then subtract the areas that are missing because of cuts or overlaps.”

Actually, for these specific problems, since they’re all made by combining or cutting rectangular prisms, we can calculate by counting each face carefully.

Let me go one by one.

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Problem 1: Shape looks like an L made from two blocks. Dimensions given: overall height 5, base 6, depth 3. The cut-out part is 3 units wide and 2 units high on the top right.

Think of it as a big box minus a small box.

Big box: 6 (long) × 5 (high) × 3 (deep)

Surface area of big box = 2×(6×5 + 6×3 + 5×3) = 2×(30 + 18 + 15) = 2×63 = 126

Now, we removed a small block: 3 (wide) × 2 (high) × 3 (deep) — but wait, removing it doesn’t just subtract its surface area — because when you remove it, you expose new inner faces!

Actually, better method: Count all outer faces.

Front view: It’s an L-shape. Area = (6×5) - (3×2) = 30 - 6 = 24? No — that’s not right for surface area. Surface area counts both sides.

Better approach: Use net or count each side.

Alternative standard method for such problems:

For composite solids made of cubes or rectangular prisms glued together, surface area = sum of surface areas of individual pieces MINUS twice the area of each glued face (because those faces are no longer exposed).

Let’s try that.

Shape 1: Can be seen as two rectangular prisms:

- Bottom piece: 6 long × 3 deep × 3 high → SA = 2*(6*3 + 6*3 + 3*3) = 2*(18+18+9)=2*45=90

Wait, no — actually looking at diagram: total height is 5, and the lower part is 3 high, upper part is 2 high, and the upper part is only 3 units long (since 6-3=3).

So:

Piece A (bottom): 6 × 3 × 3 → SA = 2*(6*3 + 6*3 + 3*3) = 2*(18+18+9)=90

Piece B (top right): 3 × 3 × 2 → SA = 2*(3*3 + 3*2 + 3*2) = 2*(9+6+6)=2*21=42

They are glued together on a face of size 3×3 = 9 square units.

So total SA = 90 + 42 - 2*9 = 132 - 18 = 114

Is that correct? Let me verify with another method.

Imagine the whole thing as a 6x5x3 box, which has SA = 2*(6*5 + 6*3 + 5*3) = 2*(30+18+15)=126

But we have a notch cut out: 3 wide, 2 high, 3 deep. When you cut out a rectangular hole from the corner, you remove 3 faces from the original surface, but add 3 new internal faces.

Original surface had: front, top, and right-side faces where the notch is.

Removed area: front: 3×2=6, top: 3×3=9, right side: 2×3=6 → total removed = 21

Added area: the three new inner faces: same sizes: 6, 9, 6 → total added = 21

So net change = 0? That can't be — because in reality, when you cut a notch, you do expose new surfaces.

Wait — actually, for a corner notch, you remove 3 external faces but add 3 internal faces of the same area, so surface area stays the same? But that contradicts our earlier calculation.

I think I’m confusing myself.

Let me look for a reliable method.

Standard trick for these worksheets: For each figure, calculate by considering the projections or by counting faces.

Perhaps the intended method is to treat each as a combination and subtract overlapping areas.

Let me check Problem 2 first — it might be simpler.

Problem 2: Looks like a flat prism, almost like a parallelogram base, but labeled with dimensions: 7, 5, 3, and angle? Wait, no angles given — probably it's a rectangular prism tilted, but dimensions suggest it's a box with length 7, width 5, height 3? But drawn slanted.

Looking at labels: it says 7, 5, 3 — likely length, width, height. And it's a rectangular prism, even if drawn obliquely.

So SA = 2*(7*5 + 7*3 + 5*3) = 2*(35 + 21 + 15) = 2*71 = 142

That seems straightforward.

Similarly, Problem 3: L-shaped, similar to Problem 1.

Dimensions: total length 10, height 8, depth 4. The vertical part is 4 wide, horizontal part is 6 long (10-4=6), and the step is 4 high? Let's see.

From diagram: left part is 4 wide, 8 high; right part is 6 long, 4 high; depth is 4 for both.

So two pieces:

Left: 4×4×8 → SA = 2*(4*4 + 4*8 + 4*8) = 2*(16+32+32)=2*80=160

Right: 6×4×4 → SA = 2*(6*4 + 6*4 + 4*4) = 2*(24+24+16)=2*64=128

Glued on a face of 4×4 = 16

Total SA = 160 + 128 - 2*16 = 288 - 32 = 256

But let's verify with big box minus nothing — it's not a full box.

Another way: think of the entire bounding box: 10×8×4, SA = 2*(10*8 + 10*4 + 8*4) = 2*(80+40+32)=2*152=304

But there is a missing part: a rectangle of 6×4 on the top right? No, the shape is L, so compared to full box, we are missing a block of 6×4×4? No.

Full box 10x8x4 would include the space that is empty in the L-shape. The empty part is 6 (long) × 4 (high) × 4 (deep)? But in the L-shape, the right part is only 4 high, while left is 8 high, so the missing part is above the right section: size 6×4×4.

When you remove that block, you remove 3 faces from the exterior, but add 3 new interior faces.

Removed faces: top: 6×4=24, front: 6×4=24, right side: 4×4=16? Let's define coordinates.

Assume the full box has front face 10x8. We remove a block from the top-right-front corner: size 6 (x) × 4 (y) × 4 (z).

The faces removed from exterior:
- Top face: area 6×4 = 24
- Front face: area 6×4 = 24
- Right face: area 4×4 = 16 (since depth is 4)

Total removed = 24+24+16=64

New faces exposed: the three faces of the cavity:
- Bottom of cavity: 6×4=24 (but this is now internal, not part of surface area? No, in surface area of the solid, we include all external surfaces, including the inside of cavities if they are open.

In this case, since it's a solid with a notch, the cavity is open, so we do include the three new faces.

So added area = 24 (back of cavity) + 24 (left side of cavity) + 16 (bottom of cavity)? Let's think.

When you remove the block, the new surfaces are:
- The face that was against the left part: size 4×4 = 16 (this is now exposed on the left side of the cavity)
- The face that was against the bottom: size 6×4 = 24 (exposed on the bottom of the cavity)
- The face that was against the back: size 6×4 = 24 (exposed on the back of the cavity)

So added area = 16 + 24 + 24 = 64

Same as removed, so net change 0, so SA = 304

But earlier I calculated 256, which is different. So which is correct?

I think I made a mistake in the piece method.

In the piece method, when I said left piece 4x4x8, that's not accurate because the depth is 4, but the width is 4, height 8, so volume is 4*4*8=128, but in the L-shape, the left part is 4 wide, 8 high, 4 deep, yes.

Right part is 6 long, 4 high, 4 deep, so 6*4*4=96.

Total volume 128+96=224, while full box 10*8*4=320, so missing 96, which matches the removed block 6*4*4=96.

Now for surface area, when two pieces are glued, we subtract twice the glued area.

Glued area: where they touch. The left piece and right piece share a face of size 4 (depth) × 4 (height of right piece) = 16, since the right piece is only 4 high, and it's attached to the bottom 4 units of the left piece's side.

So glued area = 4*4 = 16

SA of left piece alone: 2*(4*4 + 4*8 + 4*8) = 2*(16+32+32)=2*80=160

SA of right piece alone: 2*(6*4 + 6*4 + 4*4) = 2*(24+24+16)=2*64=128

Sum 288, minus 2*16 = 32, so 256

But according to the full box method, if we start with full box SA 304, and remove a block, and since the removed block had SA 2*(6*4 + 6*4 + 4*4) = 2*(24+24+16)=128, but when removed, we lose the three faces that were on the exterior, and gain the three faces that were internal, but in this case, the three faces lost are part of the full box's surface, and the three gained are new, but their areas may not be equal if the block was on the corner.

In this case, for the removed block of 6x4x4, the three faces that were on the exterior of the full box are:
- The top face: 6x4 = 24
- The front face: 6x4 = 24
- The right face: 4x4 = 16 (assuming depth is z, height y, length x)

Sum 64

The three new faces exposed are:
- The face that was adjacent to the left piece: size 4 (depth) × 4 (height) = 16 (this is the y-z face)
- The face that was adjacent to the bottom: size 6 (length) × 4 (depth) = 24 (x-z face)
- The face that was adjacent to the back: size 6 (length) × 4 (height) = 24 (x-y face)

Sum 64

So net change 0, so SA should be 304

But 256 ≠ 304, so contradiction.

I see the mistake: in the full box, when we remove the block, the "right face" of the full box is 8 high × 4 deep = 32, but after removal, the right face is only the part below the notch, which is 4 high × 4 deep = 16, and the notch exposes a new face on the right side of the cavity, which is 4 high × 4 deep = 16, so for the right side, originally 32, now we have 16 (lower part) + 16 (cavity wall) = 32, same.

Similarly for front: originally 10x8=80, after removal, front has the L-shape: area 10*8 - 6*4 = 80-24=56, but then we have the back of the cavity, which is 6x4=24, so total front-facing area 56 + 24 = 80, same.

For top: originally 10x4=40, after removal, top has the left part 4x4=16, and the cavity bottom is not on top, so top area is only 16, but we have the bottom of the cavity which is 6x4=24, but that's not on the top; it's on the bottom of the cavity, which is facing down, so for surface area, we include it, but it's not part of the top projection.

I think I need to accept that for these problems, the intended method is to calculate as composite solids with glued faces.

Perhaps for Problem 1, let's calculate manually.

Shape 1: L-shape, dimensions: overall 6 long, 5 high, 3 deep. The vertical part is 3 wide, 5 high; the horizontal part is 3 wide, 3 high, but wait, from diagram, it's like a step: left part 3x5x3, right part 3x3x3, but they overlap in the middle.

Standard way: the shape can be divided into two rectangular prisms:

- Prism A: 3 (w) × 5 (h) × 3 (d) -- left column
- Prism B: 3 (w) × 3 (h) × 3 (d) -- right base, but this would overlap with A in the bottom 3x3x3 region.

So better: Prism A: 3x5x3 for the left part.
Prism B: 3x3x3 for the right part, but positioned so that it is attached to the bottom of A on the right side.

So they share a face of 3x3 = 9 (the interface between them).

SA of A: 2*(3*5 + 3*3 + 5*3) = 2*(15+9+15)=2*39=78

SA of B: 2*(3*3 + 3*3 + 3*3) = 2*27=54

Sum 132, minus 2*9 = 18, so 114

Now, let's list all faces to verify.

Front view: L-shape: area = 3*5 + 3*3 = 15+9=24, but this is for one side; back is the same, so 48 for front and back.

Top view: the top is only the left part: 3x3 = 9, and the right part is lower, so top surface is 3x3 = 9 for the left top, and for the right part, its top is at height 3, so from above, we see the top of the right part: 3x3=9, but since it's lower, in orthographic projection, we still see it, so total top area = 3*3 + 3*3 = 18? No, in terms of actual surface, the top surface consists of two parts: the top of the left prism: 3x3=9, and the top of the right prism: 3x3=9, so 18.

Bottom view: similarly, bottom of both: left 3x3=9, right 3x3=9, so 18.

Left side: the left face of the left prism: 5x3=15

Right side: the right face of the right prism: 3x3=9, and also the right face of the left prism above the right prism: from y=3 to y=5, x=3, so size 2x3=6, so total right side area = 9 + 6 = 15

Now, also, there is the front and back, which we have 24 each, so 48.

Let's list:

- Front: 24
- Back: 24
- Top: 18 (3x3 for left top + 3x3 for right top)
- Bottom: 18 (3x3 for left bottom + 3x3 for right bottom)
- Left: 5x3 = 15
- Right: as above, 3x3 (for right prism) + 2x3 (for the part of left prism above) = 9+6=15

Sum: 24+24+18+18+15+15 = let's calculate: 24+24=48, 18+18=36, 15+15=30, total 48+36=84, +30=114

Yes! So 114 is correct for Problem 1.

For Problem 3, similarly.

Shape 3: L-shape, total length 10, height 8, depth 4. Left part: 4 wide, 8 high, 4 deep. Right part: 6 long, 4 high, 4 deep. Attached at the bottom.

So Prism A: 4x8x4, SA = 2*(4*8 + 4*4 + 8*4) = 2*(32+16+32)=2*80=160

Prism B: 6x4x4, SA = 2*(6*4 + 6*4 + 4*4) = 2*(24+24+16)=2*64=128

Glued area: where they touch: the interface is 4 (depth) × 4 (height of B) = 16, since B is 4 high, and attached to the bottom 4 units of A's right side.

So SA total = 160 + 128 - 2*16 = 288 - 32 = 256

Verify with face counting.

Front view: L-shape: area = 4*8 + 6*4 = 32 + 24 = 56, so front and back: 112

Top view: top of left part: 4x4=16, top of right part: 6x4=24, so top area 40

Bottom view: similarly, bottom of both: left 4x4=16, right 6x4=24, so 40

Left side: left face of A: 8x4=32

Right side: right face of B: 4x4=16, and the part of A above B: from y=4 to y=8, x=4, so size 4x4=16, so total right side 16+16=32

Sum: front/back 112, top/bottom 80, left/right 64, total 112+80=192, +64=256. Yes.

So method is confirmed.

Now let's do all problems.

Problem 1: As above, 114

Problem 2: Rectangular prism, dimensions 7,5,3. SA = 2*(7*5 + 7*3 + 5*3) = 2*(35+21+15)=2*71=142

Problem 3: As above, 256

Problem 4: Another L-shape. Dimensions: total height 10, width 4, depth 3. Vertical part 4x10x3, horizontal part 4x4x3? From diagram: it's like a tall L, with the base extending to the right.

Specifically: left part 4 wide, 10 high, 3 deep. Right part 4 long, 4 high, 3 deep. Attached at bottom.

So Prism A: 4x10x3, SA = 2*(4*10 + 4*3 + 10*3) = 2*(40+12+30)=2*82=164

Prism B: 4x4x3, SA = 2*(4*4 + 4*3 + 4*3) = 2*(16+12+12)=2*40=80

Glued area: 4 (depth) × 4 (height of B) = 16

SA total = 164 + 80 - 2*16 = 244 - 32 = 212

Verify: front: 4*10 + 4*4 = 40+16=56, times 2 for front/back = 112

Top: top of A: 4x3=12, top of B: 4x3=12, so 24

Bottom: similarly 24

Left: 10x3=30

Right: right of B: 4x3=12, and part of A above: from y=4 to 10, size 6x3=18, so 12+18=30

Sum: 112 + 24+24=48, +30+30=60, total 112+48=160, +60=220? Wait, 112+48=160, +60=220, but I got 212. Mistake.

Front/back: 56 each? Front area is 4*10 for left + 4*4 for right = 40+16=56, yes, so 112 for front and back.

Top: the top surface: for the left prism, top is 4x3=12. For the right prism, top is 4x3=12. But are they at different heights? In terms of actual surface area, yes, both are exposed, so 24.

Bottom: similarly, bottom of left: 4x3=12, bottom of right: 4x3=12, so 24.

Left side: the left face of left prism: 10x3=30

Right side: the right face of right prism: 4x3=12 (since it's 4 high, 3 deep). Additionally, the right face of the left prism above the right prism: from y=4 to y=10, which is 6 high, 3 deep, so 6x3=18. So total right side 12+18=30.

Now sum: front 56, back 56, top 24, bottom 24, left 30, right 30.

56+56=112, 24+24=48, 30+30=60, total 112+48=160, 160+60=220.

But earlier calculation gave 212. Inconsistency.

What's wrong? In the glued area, when I said glued area is 4x4=16, but is that correct?

Prism A is 4x10x3, Prism B is 4x4x3. They are attached along the face where x=4 (if A is from x=0 to 4, B from x=4 to 8, but in this case, for Problem 4, the diagram shows the right part extending to the right, so if A is left, B is right, attached at x=4.

The interface is at x=4, for y from 0 to 4 (since B is only 4 high), z from 0 to 3. So area 4 (y) × 3 (z) = 12, not 16.

I see! I forgot the depth. Depth is 3, not 4.

In Problem 4, depth is 3, as per diagram.

So glued area = height of B × depth = 4 × 3 = 12

SA of A: 2*(4*10 + 4*3 + 10*3) = 2*(40+12+30)=2*82=164

SA of B: 2*(4*4 + 4*3 + 4*3) = 2*(16+12+12)=2*40=80

Sum 244, minus 2*12 = 24, so 220

Yes, matches the face counting.

So for Problem 4: 220

Problem 5: Similar to Problem 4, but mirrored. Dimensions: total length 10, height 8, depth 4. Left part 4x8x4, right part 6x4x4? From diagram: it's L-shape with the short leg on the left.

Specifically: the vertical part is on the right: 4 wide, 8 high, 4 deep. The horizontal part is on the left: 6 long, 4 high, 4 deep. Attached at bottom.

So Prism A (right): 4x8x4, SA = 2*(4*8 + 4*4 + 8*4) = 2*(32+16+32)=2*80=160

Prism B (left): 6x4x4, SA = 2*(6*4 + 6*4 + 4*4) = 2*(24+24+16)=2*64=128

Glued area: where they touch: at the interface, y from 0 to 4, z from 0 to 4, so area 4*4=16

SA total = 160 + 128 - 2*16 = 288 - 32 = 256

Same as Problem 3, which makes sense by symmetry.

Problem 6: Another L-shape. Dimensions: total length 12, height 8, depth 4. Vertical part on left: 4 wide, 8 high, 4 deep. Horizontal part on right: 8 long, 4 high, 4 deep. Attached at bottom.

Prism A: 4x8x4, SA = 2*(4*8 + 4*4 + 8*4) = 2*(32+16+32)=160

Prism B: 8x4x4, SA = 2*(8*4 + 8*4 + 4*4) = 2*(32+32+16)=2*80=160

Glued area: 4 (height of B) × 4 (depth) = 16

SA total = 160 + 160 - 2*16 = 320 - 32 = 288

Verify: front: 4*8 + 8*4 = 32+32=64, times 2 = 128

Top: top of A: 4x4=16, top of B: 8x4=32, so 48

Bottom: similarly 48

Left: 8x4=32

Right: right of B: 4x4=16, and part of A above: from y=4 to 8, size 4x4=16, so 32

Sum: 128 + 48+48=96, +32+32=64, total 128+96=224, +64=288. Yes.

Problem 7: Rectangular prism, dimensions 5,4,3. SA = 2*(5*4 + 5*3 + 4*3) = 2*(20+15+12)=2*47=94

Problem 8: Rectangular prism, dimensions 6,3,2. SA = 2*(6*3 + 6*2 + 3*2) = 2*(18+12+6)=2*36=72

Problem 9: L-shape. Dimensions: total length 10, height 6, depth 4. Vertical part on left: 4 wide, 6 high, 4 deep. Horizontal part on right: 6 long, 3 high, 4 deep. Attached at bottom.

Prism A: 4x6x4, SA = 2*(4*6 + 4*4 + 6*4) = 2*(24+16+24)=2*64=128

Prism B: 6x3x4, SA = 2*(6*3 + 6*4 + 3*4) = 2*(18+24+12)=2*54=108

Glued area: height of B × depth = 3 × 4 = 12

SA total = 128 + 108 - 2*12 = 236 - 24 = 212

Verify: front: 4*6 + 6*3 = 24+18=42, times 2 = 84

Top: top of A: 4x4=16, top of B: 6x4=24, so 40

Bottom: similarly 40

Left: 6x4=24

Right: right of B: 3x4=12, and part of A above: from y=3 to 6, size 3x4=12, so 24

Sum: 84 + 40+40=80, +24+24=48, total 84+80=164, +48=212. Yes.

Now summarize:

1. 114

2. 142

3. 256

4. 220

5. 256

6. 288

7. 94

8. 72

9. 212

I think that's all.

Final Answer:
1. 114
2. 142
3. 256
4. 220
5. 256
6. 288
7. 94
8. 72
9. 212