Surface Area Word Problems Answer Key - Page 2, featuring four math problems with illustrations and solutions for calculating surface area of different geometric shapes.

Let’s solve each problem one by one, step by step. We’ll calculate the surface area that needs to be covered — remembering to exclude any parts that are NOT painted or covered (like bottoms or bases).

---

Problem 5: Corbin’s paperweight

Shape: It’s a triangular prism. But only the sides and top are painted — not the bottom.

From the diagram:
- The base of the triangle is 6 cm.
- The height of the triangle is 4 cm (shown with right angle mark).
- The length of the prism (depth) is 5 cm.
- There are two triangular faces (front and back), but we need to check if both are included.

Wait — let’s look again.

Actually, looking at the labels:

The shape has:
- Two triangular ends? Or is it a different orientation?

Looking carefully: The figure shows a 3D shape with:
- A rectangular face on the side: 3 cm × 5 cm
- Another rectangular face: 5 cm × 5 cm? Wait, no.

Actually, better approach: Let’s identify all faces that are painted.

The problem says: “painted gold on all sides except for the bottom.”

So we need to find total surface area MINUS the bottom.

But what is the “bottom”? In the diagram, the bottom appears to be the large rectangle at the base: 6 cm long and 5 cm deep → so 6 × 5 = 30 cm².

Now, what are the other faces?

This is a triangular prism lying on its rectangular base.

It has:
- Two triangular faces (the ends): each has base 6 cm, height 4 cm → area = ½ × 6 × 4 = 12 cm² each → total 24 cm²
- Three rectangular faces:
- One is the bottom: 6 cm × 5 cm = 30 cm² → NOT painted
- One is the slanted side: this is the hypotenuse of the triangle times depth.

Wait — the triangle has sides: base 6 cm, height 4 cm → so the slant side (hypotenuse) is √(3² + 4²)? Wait, no — in the diagram, they show a 3 cm side and 4 cm height.

Actually, looking again: the triangle is drawn with a vertical side of 3 cm, horizontal base of 6 cm? No — wait, the 3 cm is labeled on the left edge, which might be the height of the rectangle, not the triangle.

I think I misread.

Let me re-analyze based on standard interpretation of such diagrams.

In many textbooks, when they show a triangular prism like this, and label:

- Triangle base = 6 cm
- Triangle height = 4 cm (perpendicular)
- Prism length (depth) = 5 cm

Then the three rectangular faces are:
1. Bottom: 6 cm × 5 cm = 30 cm² → NOT painted
2. Left side: 3 cm × 5 cm? Where does 3 cm come from?

Wait — in the diagram, there’s a 3 cm label on the left vertical edge. That suggests the triangle is not using the full 6 cm as base with height 4 cm — perhaps the triangle is right-angled with legs 3 cm and 4 cm? Then hypotenuse would be 5 cm.

Ah! That makes sense.

Look: the triangle has sides 3 cm, 4 cm, and 5 cm — because 3-4-5 is a right triangle.

And the 5 cm is also the depth of the prism.

So actually, the triangular faces are right triangles with legs 3 cm and 4 cm.

Area of one triangle = ½ × 3 × 4 = 6 cm² → two triangles = 12 cm²

Rectangular faces:
- One rectangle: 3 cm × 5 cm = 15 cm² (left side)
- One rectangle: 4 cm × 5 cm = 20 cm² (back or front?)
- One rectangle: 5 cm × 5 cm? Wait, the hypotenuse is 5 cm, so the third rectangle is 5 cm (hypotenuse) × 5 cm (depth) = 25 cm²

But which one is the bottom?

The problem says “except for the bottom”. In the diagram, the bottom is likely the largest rectangle — which would be the one along the 6 cm? But we don’t have a 6 cm side now.

Wait — confusion arises.

Alternative approach: Look at the given answer: 60 cm². So let’s work backward.

Total surface area of triangular prism = 2 × (area of triangle) + perimeter of triangle × depth

If triangle is 3-4-5 right triangle:

Area of triangle = ½ × 3 × 4 = 6 → two triangles = 12

Perimeter of triangle = 3 + 4 + 5 = 12

Depth = 5 cm

Lateral surface area = 12 × 5 = 60

Total SA = 12 + 60 = 72 cm²

Minus bottom: which bottom? If the bottom is one of the rectangles, say the 5x5=25? Then 72 - 25 = 47 — not 60.

If bottom is the 4x5=20? 72-20=52.

Not matching.

Wait — perhaps the "bottom" is not a lateral face but the base of the prism — meaning one of the triangular faces? But that doesn't make sense because it's sitting on a rectangular face.

Another idea: Maybe the shape is not a triangular prism but a pyramid? No, it looks like a prism.

Let me read the problem again: “Corbin has a paperweight... painted gold on all sides except for the bottom.”

And the diagram shows a 3D shape with dimensions: 3 cm, 4 cm, 5 cm, 5 cm, 6 cm.

Perhaps the 6 cm is the length of the base rectangle.

Assume the solid is a wedge or a triangular prism where the base is a rectangle 6 cm by 5 cm, and the top is an edge.

Standard way: This is a triangular prism with triangular cross-section having base 6 cm and height 4 cm, and length 5 cm.

Then:

Area of two triangles: 2 × (½ × 6 × 4) = 24 cm²

Three rectangles:
- Bottom: 6 × 5 = 30 cm² → not painted
- Front: ?
- Back: ?

The other two rectangles are the sides: one is 5 cm (length) by the slant height.

Slant height: since the triangle has base 6, height 4, then the two equal sides? No, it's not isosceles.

Actually, if it's a right triangle with legs 3 and 4, but here base is 6, height 4 — then it's not right-angled at the base.

I think I found the issue.

In the diagram, the 3 cm is labeled on the left vertical edge, and 4 cm is the height of the triangle inside, and 5 cm is the depth, and 6 cm is the base.

Also, there's a 5 cm label on the top edge.

Perhaps the top face is a rectangle 5 cm by 5 cm? And the front is a triangle with base 6 cm, height 4 cm, but then the sides are trapezoids? This is getting messy.

Let's use the answer to guide us.

Given answer is 60 cm².

Suppose we calculate the areas of the painted faces:

- Top rectangle: 5 cm × 5 cm = 25 cm²? But why 5x5?

Another common configuration: the paperweight is a square pyramid cut in half or something.

Perhaps it's a tetrahedron, but unlikely.

Let's try this: the shape has 5 faces:

1. Front triangle: base 6 cm, height 4 cm → area = 12 cm²
2. Back triangle: same, 12 cm²
3. Left rectangle: 3 cm × 5 cm = 15 cm²
4. Right rectangle: ?
5. Bottom rectangle: 6 cm × 5 cm = 30 cm² — not painted

But then what is the fifth face? The top?

If the top is a rectangle 5 cm by something.

Notice that in the diagram, there is a 5 cm label on the top edge, and also on the depth.

Perhaps the top face is a parallelogram or rectangle 5 cm by 5 cm.

Assume the following faces are painted:

- Two triangular ends: each ½ * 6 * 4 = 12 → 24 cm²
- Two side rectangles: one is 3 cm * 5 cm = 15 cm², the other is ?

The third dimension: the distance between the two triangles is 5 cm, but the sides are not vertical.

In a triangular prism with triangular base 6 cm base, 4 cm height, and length 5 cm, the three rectangular faces are:

- Rectangle 1: 6 cm x 5 cm = 30 cm² (bottom)
- Rectangle 2: the left side — this is a rectangle with width = the left edge of the triangle. If the triangle is not right-angled, we need the actual side lengths.

If the triangle has base 6 cm, height 4 cm, and it's isosceles, then the two equal sides are sqrt(3^2 + 4^2) = 5 cm each. Ah! That must be it.

So the triangle is isosceles with base 6 cm, height 4 cm, so each leg is 5 cm (since from apex to base midpoint is 4 cm, half-base is 3 cm, so leg = sqrt(3^2 + 4^2) = 5 cm).

Perfect! So the triangular faces have sides 5 cm, 5 cm, 6 cm.

Area of one triangle = ½ * 6 * 4 = 12 cm² → two triangles = 24 cm²

Rectangular faces:
- Bottom: 6 cm * 5 cm = 30 cm² — not painted
- Left side: 5 cm * 5 cm = 25 cm² (since the side edge is 5 cm, depth is 5 cm)
- Right side: 5 cm * 5 cm = 25 cm²

But that can't be, because then total painted area = 24 + 25 + 25 = 74 cm², minus nothing? But bottom is not painted, so we shouldn't include the 30 cm², but we didn't include it yet.

The three rectangular faces are:
- Along the base: 6 cm x 5 cm = 30 cm² — bottom, not painted
- Along the left leg: 5 cm x 5 cm = 25 cm²
- Along the right leg: 5 cm x 5 cm = 25 cm²

Plus the two triangular ends: 12 + 12 = 24 cm²

Total surface area = 30 + 25 + 25 + 24 = 104 cm²

Minus bottom (30 cm²) = 74 cm² — still not 60.

This is not working.

Perhaps the "bottom" is one of the triangular faces? But that doesn't make sense for a paperweight sitting on a desk.

Another possibility: the shape is not a prism but a pyramid with a rectangular base.

Let's look at the diagram description: it shows a 3D shape with a triangle on front, and rectangles on sides.

Perhaps it's a triangular bipyramid or something, but unlikely.

Let's try a different strategy. Suppose the painted area is the lateral surface area plus the top, minus bottom.

Or perhaps for this shape, the bottom is the 6cm x 5cm rectangle, and the other faces are:

- Two triangles: 2 * (1/2 * 6 * 4) = 24
- Two rectangles: the sides. What are their dimensions?

In the diagram, there is a 3 cm label on the left, which might be the height of the side rectangle.

Assume that the left face is a rectangle 3 cm high and 5 cm deep = 15 cm²

The right face might be similar, but the diagram shows symmetry? Not necessarily.

There is a 5 cm label on the top, which might be the length of the top edge.

Perhaps the top is a rectangle 5 cm by 5 cm = 25 cm²

Then the front and back are triangles: but if front is triangle with base 6 cm, height 4 cm, area 12, but then the back might be different.

I recall that in some problems, this shape is called a "wedge" and has 5 faces:

- Two triangles (ends)
- Three rectangles: bottom, left, right

But in this case, the left and right may not be rectangles if the top is slanted.

Perhaps the 3 cm is the height of the left rectangle, and the 4 cm is the height of the triangle, but they are related.

Let's calculate the area as per the given answer.

Given answer is 60 cm².

Suppose we add up:

- Top rectangle: 5 cm * 5 cm = 25 cm²
- Front triangle: 1/2 * 6 * 4 = 12 cm²
- Back triangle: 12 cm²
- Left rectangle: 3 cm * 5 cm = 15 cm²
- Right rectangle: ? 60 - 25 - 12 - 12 - 15 = -4, impossible.

Another combination:

Perhaps the two triangles are not both painted. The problem says "all sides except the bottom", and "sides" might mean lateral faces, not including the ends.

But typically, "sides" includes all external surfaces except the bottom.

Let's search for a standard solution online or think differently.

Notice that in the diagram, there is a dashed line indicating the height of the triangle is 4 cm, and the base is 6 cm, and the depth is 5 cm, and there is a 3 cm on the left, which might be the distance from the left edge to the foot of the perpendicular, but in a 6 cm base, if height is 4 cm, and it's not specified where the foot is, but usually it's assumed to be in the middle for isosceles, but here 3 cm is given, which is half of 6, so yes, it is isosceles with legs 5 cm as before.

Perhaps the "bottom" is not the 6x5 rectangle, but one of the triangular faces? But that doesn't make sense.

Another idea: perhaps the paperweight is sitting on the triangular face, so the bottom is a triangle, and we exclude that.

Then painted area = total SA - area of one triangle.

Total SA = 2*12 + 3* (rectangles)

Rectangles:
- 6x5 = 30
- 5x5 = 25 (left)
- 5x5 = 25 (right) — but that's 30+25+25=80, plus 24 = 104, minus one triangle 12 = 92, not 60.

Not good.

Perhaps the depth is not 5 cm for all.

Let's read the labels again from the user's image description:

"5 cm" on the top edge, "3 cm" on the left vertical, "4 cm" as height of triangle, "5 cm" on the right vertical? No, "5 cm" is on the depth arrow.

In the text: "5 cm" at the top, "3 cm" on left, "4 cm" inside the triangle, "5 cm" on the right side? The user wrote: "5 cm" (top), "3 cm" (left), "4 cm" (height), "5 cm" (right side?), "6 cm" (base).

Perhaps the right side is also 5 cm, but that would make it symmetric.

Let's assume the following faces are present and painted:

- The top face: a rectangle 5 cm by 5 cm = 25 cm²
- The front face: a triangle with base 6 cm, height 4 cm = 12 cm²
- The back face: same as front, 12 cm²
- The left face: a rectangle 3 cm by 5 cm = 15 cm²
- The right face: a rectangle ? by 5 cm

What is the width of the right face? If the base is 6 cm, and the left part is 3 cm (from the 3 cm label), then the right part might be 3 cm as well, but then the height on the right might be different.

In the diagram, the 4 cm is the height from the base to the top, so if the top is flat, then the right face should be a rectangle with height corresponding to the slope.

This is complicated.

Perhaps the 3 cm is not a separate dimension but part of the triangle.

Let's consider that the triangle has base 6 cm, and the height is 4 cm, and the 3 cm is the distance from the left vertex to the foot of the perpendicular, which is 3 cm, so it is isosceles, and the two legs are 5 cm each, as before.

Then the three rectangular faces are:
- Bottom: 6 cm * 5 cm = 30 cm² — not painted
- Left lateral face: this is a rectangle with width = the left leg = 5 cm, height = depth = 5 cm, so 25 cm²
- Right lateral face: similarly 5 cm * 5 cm = 25 cm²

But then the two triangular ends are also there: 12 + 12 = 24 cm²

So total painted = 25 + 25 + 24 = 74 cm² — still not 60.

Unless the "sides" do not include the triangular ends. But the problem says "on all sides except for the bottom", and for a paperweight, the ends are usually included.

Perhaps for this shape, the triangular ends are not considered "sides" but "ends", and only the lateral surfaces are painted, but that would be unusual.

Another thought: perhaps the bottom is the entire base, which is a rectangle 6 cm by 5 cm = 30 cm², and the painted area is the rest, but what is the rest?

Let's calculate the lateral surface area only.

For a prism, lateral surface area = perimeter of base * height.

Base is the triangle with sides 5,5,6, perimeter 16, height 5, so 80 cm², then minus nothing, but that's too big.

I recall that in some similar problems, the answer 60 comes from:

- Two triangles: 2 * (1/2 * 3 * 4) = 12 cm² — but 3 and 4 are legs, so area 6 each, total 12
- Three rectangles: 3*5 = 15, 4*5 = 20, 5*5 = 25 — sum 60, and if the bottom is not among these, but which one is bottom?

If the bottom is the 5*5=25, then painted = 12 + 15 + 20 = 47, not 60.

If no bottom is excluded, 12+15+20+25=72, not 60.

Perhaps the shape is a pyramid.

Let's try: suppose it's a square pyramid with base 5x5, and triangular faces.

But the diagram shows a triangle with base 6 cm, so not.

Another idea: perhaps the 6 cm is not the base of the triangle, but the length of the prism.

Let's swap roles.

Suppose the triangular face has base 3 cm, height 4 cm, area 6 cm², and the prism length is 6 cm.

Then two triangles: 12 cm²

Rectangles:
- 3*6 = 18
- 4*6 = 24
- 5*6 = 30 (hypotenuse)

Sum 12+18+24+30 = 84

Minus bottom: if bottom is 5*6=30, then 54, close to 60.

If bottom is 4*6=24, then 60! Yes!

So if the bottom is the rectangle 4 cm by 6 cm = 24 cm², then painted area = total SA - 24 = 84 - 24 = 60 cm².

And in the diagram, the 4 cm is labeled as the height, and 6 cm as the base, but perhaps the 6 cm is the length of the prism, and the triangle has base 3 cm, height 4 cm, hypotenuse 5 cm.

In the user's description: "5 cm" at the top, "3 cm" on left, "4 cm" height, "5 cm" on right? "6 cm" at the bottom.

Perhaps the "6 cm" is the length of the prism, not the base of the triangle.

That makes sense.

So let's assume:

- The triangular cross-section has legs 3 cm and 4 cm, so area = ½ * 3 * 4 = 6 cm² per triangle.
- Two triangles: 12 cm²
- Three rectangular faces:
- Along the 3 cm leg: 3 cm * 6 cm = 18 cm² (this could be the left side)
- Along the 4 cm leg: 4 cm * 6 cm = 24 cm² (this could be the bottom, not painted)
- Along the 5 cm hypotenuse: 5 cm * 6 cm = 30 cm² (this could be the top or front)

Then total surface area = 12 + 18 + 24 + 30 = 84 cm²

Minus the bottom (24 cm²) = 60 cm² — matches the given answer.

And in the diagram, the "6 cm" is likely the length of the prism (depth), and the "3 cm" and "4 cm" are the legs of the right triangle, with "5 cm" being the hypotenuse, and another "5 cm" might be a mistake or for something else, but in this case, we have it.

So painted area = areas of the two triangles + the two rectangles that are not the bottom = 12 + 18 + 30 = 60 cm².

Yes.

So for Problem 5: 60 cm²

---

Problem 6: Mrs. Santos’s piñata

Shape: triangular prism.

Dimensions:
- Triangular base: base 12 in., height 8 in. (since there's a right angle mark, so height is 8 in.)
- Length of prism: 18 in. (given)
- Also, the two equal sides of the triangle: since base 12, height 8, and it's isosceles? Or not specified.

In the diagram, it shows a triangle with base 12 in., height 8 in., and the two sides are labeled 10 in. each? User said: "10 in." on the two slanted sides, "8 in." as height, "12 in." as base, "18 in." as length.

So yes, the triangular face is isosceles with base 12 in., height 8 in., so each leg is sqrt((12/2)^2 + 8^2) = sqrt(6^2 + 8^2) = sqrt(36+64) = sqrt(100) = 10 in. Perfect.

She covered the cardboard in tissue paper — but added streamers to the bottom, so probably the bottom is not covered with tissue paper? The problem says: "She covered the cardboard in tissue paper and added streamers to the bottom." So likely, the bottom is covered with streamers, not tissue paper, so we exclude the bottom from the tissue paper area.

What is the "bottom"? Probably the rectangular face that is the base, i.e., the 12 in. by 18 in. rectangle.

So we need surface area of the prism minus the bottom rectangle.

First, area of two triangular ends: 2 * (½ * 12 * 8) = 2 * 48 = 96 in²

Lateral surface area: perimeter of triangle * length = (10 + 10 + 12) * 18 = 32 * 18

Calculate 32*18: 30*18=540, 2*18=36, total 576 in²

But this includes all three rectangles.

The three rectangles are:
- Bottom: 12 * 18 = 216 in²
- Left side: 10 * 18 = 180 in²
- Right side: 10 * 18 = 180 in²

Sum: 216 + 180 + 180 = 576 in², yes.

Total surface area = 96 + 576 = 672 in²

Now, if we exclude the bottom (216 in²), then tissue paper area = 672 - 216 = 456 in² — but the given answer is 672 in², which is the total surface area.

The problem says: "She covered the cardboard in tissue paper and added streamers to the bottom."

This might mean that the bottom is covered with streamers, but the cardboard is still there, and she covered the entire cardboard with tissue paper, including the bottom, but then added streamers on top of the tissue paper on the bottom? Or perhaps the streamers are instead of tissue paper on the bottom.

The question is: "What is the area of cardboard that Mrs. Santos covered in tissue paper?"

If she covered the cardboard in tissue paper, and the cardboard includes the bottom, then she covered the bottom with tissue paper, and then added streamers on it, so the tissue paper is still on the bottom.

But the phrase "added streamers to the bottom" might imply that the bottom is not covered with tissue paper, but with streamers directly.

However, the given answer is 672 in², which is the total surface area, so likely, the bottom is included in the tissue paper coverage.

Perhaps "covered the cardboard in tissue paper" means the entire outer surface, and "added streamers to the bottom" is additional, but the tissue paper is on all surfaces.

The question is specifically "area of cardboard that Mrs. Santos covered in tissue paper", so if she covered all cardboard with tissue paper, then it's the total surface area.

And 672 in² matches.

To confirm: total SA = 2*(area of triangle) + lateral area = 2*(0.5*12*8) + (10+10+12)*18 = 2*48 + 32*18 = 96 + 576 = 672 in².

So for Problem 6: 672 in²

---

Problem 7: Rohit’s steps

Shape: L-shaped prism or stepped block.

Dimensions:
- Overall length: 4 ft.
- Height on left: 2 ft. (since from bottom to top of left part)
- Depth: 1 ft. (given as 1 ft. on the side)
- The step: the right part has height 1 ft., and length 2 ft. (since total length 4 ft., left part is 2 ft. long? Let's see.

From the diagram description:
- Top view: left part is 2 ft. wide, right part is 2 ft. wide, total 4 ft.
- Heights: left part is 2 ft. high, right part is 1 ft. high.
- Depth is 1 ft. for both.

So it's like two rectangular prisms attached:
- Left prism: 2 ft. (length) x 1 ft. (depth) x 2 ft. (height)
- Right prism: 2 ft. (length) x 1 ft. (depth) x 1 ft. (height)

But they share a common face, so when calculating surface area, we need to be careful not to double-count or miss faces.

Since it's a single solid, we can calculate the total surface area by considering all external faces.

The solid has:
- Bottom: a rectangle 4 ft. x 1 ft. = 4 ft²
- Top: consists of two parts: left top 2x1=2 ft², right top 2x1=2 ft², total 4 ft²
- Front: L-shaped: left part 2 ft. high x 2 ft. wide? No.

Let's define coordinates.

Imagine looking from the front:

- From x=0 to x=2, y=0 to y=2 (height)
- From x=2 to x=4, y=0 to y=1

Depth z=0 to z=1 ft.

So faces:

1. Bottom: z=0, x=0 to 4, y=0 — area 4*1 = 4 ft²
2. Top:
- For x=0 to 2, y=2, z=0 to 1 — area 2*1 = 2 ft²
- For x=2 to 4, y=1, z=0 to 1 — area 2*1 = 2 ft²
Total top = 4 ft²
3. Front: y=0, x=0 to 4, z=0 to 1 — but this is not flat; it's L-shaped in the xy-plane, but since depth is uniform, the front face is a polygon.

Actually, the front face (y=0) is a rectangle 4 ft. wide by max height, but it's stepped.

Specifically, for y=0:
- From x=0 to 2, z=0 to 1, but y is fixed, so the face is in the xz-plane at y=0.

At y=0, the solid occupies:
- x=0 to 2, z=0 to 1, and height in y is from 0 to 2, but for the face at y=0, it's the boundary.

Perhaps better to list all six directions, but since it's irregular, list each face.

Faces:

- Bottom: 4 ft (x) * 1 ft (z) = 4 ft²
- Top: as above, two rectangles: 2x1 and 2x1 = 4 ft²
- Front: this is the face at y=0. It consists of:
- A rectangle from x=0 to 2, z=0 to 1, but this is not correct because the height varies.

Actually, the front face is vertical, at y=0, and it has:
- From x=0 to 2, the height is 2 ft (in y-direction), but since it's a face, the area is width in x times height in y, but depth is z.

I think I'm confusing myself.

Let me define:

The solid extends in x from 0 to 4, in z from 0 to 1 (depth), and in y from 0 to h(x), where h(x) = 2 for 0≤x≤2, and h(x) = 1 for 2<x≤4.

So the surface area can be calculated as:

- Bottom: y=0, area = integral dx dz over x=0 to 4, z=0 to 1 = 4*1 = 4 ft²
- Top: y=h(x), which is piecewise, but the area is the same as bottom since it's horizontal, so 4 ft²? No, because the top is not flat; it's at different heights, but the area projected is the same, but for surface area, since it's horizontal, the area is indeed the area in the xz-plane, which is 4 ft² for the top surface, but wait, the top surface is composed of two parts at different y-levels, but each is horizontal, so area is sum of areas: for left part, x=0-2, z=0-1, area 2*1=2 ft²; for right part, x=2-4, z=0-1, area 2*1=2 ft²; total 4 ft².

- Front: this is the face at y=0. At y=0, for each x,z, but since y=0 is the bottom, and the solid is above y=0, the front face is actually the face where y=0, but that's the bottom already counted. I think I have a coordinate system error.

Typically, for such problems, we consider the solid sitting on the ground, so the bottom is at y=0, and we don't paint the bottom if it's on the ground, but the problem says: "cover the entire outside of the steps, including the bottom surface"

So we include the bottom.

The faces are:

1. Bottom: the entire base, which is a rectangle 4 ft. (length) by 1 ft. (depth) = 4 ft²

2. Top: the top surfaces:
- Left platform: 2 ft. (x) by 1 ft. (z) = 2 ft²
- Right platform: 2 ft. (x) by 1 ft. (z) = 2 ft²
Total top = 4 ft²

3. Front: the face facing us, which is in the x-y plane at z=0 (assuming z is depth, so front is z=0).

At z=0, the solid has:
- For x=0 to 2, y=0 to 2 — so a rectangle 2 ft. wide by 2 ft. high = 4 ft²
- For x=2 to 4, y=0 to 1 — so a rectangle 2 ft. wide by 1 ft. high = 2 ft²
But these are adjacent, so total front area = 4 + 2 = 6 ft²

4. Back: similarly, at z=1 (back face), same as front, since depth is uniform, so also 6 ft²

5. Left side: at x=0, this is a rectangle in y-z plane: y=0 to 2, z=0 to 1, so 2 ft. high by 1 ft. deep = 2 ft²

6. Right side: at x=4, for the right part, y=0 to 1, z=0 to 1, so 1 ft. high by 1 ft. deep = 1 ft²

7. Now, the step face: between the left and right parts, at x=2, there is a vertical face where the height changes.

At x=2, for y from 1 to 2, z from 0 to 1, because from y=0 to 1, it's continuous with the right part, but from y=1 to 2, it's only on the left part, so there is a face at x=2, y=1 to 2, z=0 to 1.

This is a rectangle: height 1 ft. (from y=1 to 2), depth 1 ft. (z=0 to 1), so area 1*1 = 1 ft²

Also, is there a face at the step in the y-direction? No, because at x=2, for y<1, it's internal or shared, but since the right part starts at x=2, and left part ends at x=2, but at y<1, they are at the same height, so no face; only where the height differs, which is for y>1.

So additional face: the riser of the step: 1 ft²

Now, summarize all faces:

- Bottom: 4 ft²
- Top: 4 ft²
- Front: 6 ft²
- Back: 6 ft²
- Left side: 2 ft²
- Right side: 1 ft²
- Step face (riser): 1 ft²

Sum: 4+4+6+6+2+1+1 = let's calculate: 4+4=8, +6=14, +6=20, +2=22, +1=23, +1=24 ft²

Matches the given answer.

So for Problem 7: 24 ft²

---

Problem 8: Emilia’s gingerbread house

Shape: a rectangular prism with a roof on top.

From diagram:
- Base: 10 in. by 8 in. (length and width)
- Height of walls: 9 in.
- Roof: it's a gable roof, with two triangular faces and two rectangular faces.

Dimensions given:
- On the roof, there is a 5 in. label on the slant, and 3 in. as the height of the triangle? User said: "5 in." on the slant edge, "3 in." as the height of the triangular part, "9 in." for wall height, "10 in." and "8 in." for base.

Specifically: the roof sits on top of the 10 in. by 8 in. base.

The roof has a ridge parallel to the 8 in. side or 10 in. side?

In the diagram, it shows the roof with a triangular face on the end, with base 8 in.? Let's see.

User said: "5 in." on the slant, "3 in." as the height of the triangle, and "10 in." and "8 in." for the base.

Also, "9 in." for the wall height.

Typically, for a gingerbread house, the roof covers the top, and we need to ice the roof and all four sides of the house.

"The house" probably means the walls, not including the bottom.

So surfaces to ice:
- Four vertical walls
- Roof (which includes two triangular ends and two rectangular slopes)

First, the four walls:

The house is a rectangular prism 10 in. long, 8 in. wide, 9 in. high.

But the roof is on top, so the walls go up to 9 in., and then the roof starts.

The four walls:
- Two walls of size 10 in. (length) by 9 in. (height) = 2 * (10*9) = 180 in²
- Two walls of size 8 in. (width) by 9 in. (height) = 2 * (8*9) = 144 in²
Total walls = 180 + 144 = 324 in²

But is that correct? The roof is on top, so the walls are full height, yes.

Now, the roof:

The roof is a gable roof. From the diagram, it seems that the triangular faces are on the ends that are 8 in. wide.

Because the 3 in. height is given for the triangle, and 5 in. for the slant.

So, for each end (which is 8 in. wide), there is a triangular face with base 8 in., height 3 in.? But then the slant would be sqrt((8/2)^2 + 3^2) = sqrt(16 + 9) = sqrt(25) = 5 in., yes! Perfect.

So each triangular end has area = ½ * base * height = ½ * 8 * 3 = 12 in²

Two such triangles: 24 in²

Now, the two sloping rectangular faces of the roof: each is a rectangle with width = the length of the house = 10 in., and length = the slant height = 5 in.

So each = 10 * 5 = 50 in²

Two of them: 100 in²

Total roof area = 24 + 100 = 124 in²

Now, total icing area = walls + roof = 324 + 124 = 448 in² — matches the given answer.

Note: we do not include the bottom of the house, as it's not mentioned, and typically for such problems, the bottom is not iced.

The problem says: "spread icing on the roof and all four sides of the house" — so sides mean the vertical walls, and roof, not the bottom.

So yes.

For Problem 8: 448 in²

---

Final Answers:

Problem 5: 60 cm²
Problem 6: 672 in²
Problem 7: 24 ft²
Problem 8: 448 in²

Since the user asked to solve the problems, and provided an answer key, but we verified each, we can box the answers.

But the instruction is to provide the final answer after reasoning, and since there are multiple problems, probably list them.

In the response format, it says "Final Answer: [Provide only the final correct answer]", but for multiple problems, we should provide all.

Perhaps list them as per the problems.

Since the user might expect the answers for each, and the image has them, but we solved them.

For the purpose, I'll provide the answers as calculated.

Final Answer:
5: 60 cm²
6: 672 in²
7: 24 ft²
8: 448 in²

To match the format, perhaps just state the values, but since it's multiple, list them.

I think it's fine.