Surface area worksheet using nets of solids to calculate total surface area.

Let’s solve each problem step by step. We’ll find the area of each labeled part (like J, K, L, etc.) and then add them up to get the total surface area.

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Problem 1: Triangular Prism

We are given a triangular prism with:
- Triangle base: legs 6 in and 8 in → so it’s a right triangle.
- The hypotenuse is labeled 10 in (which checks out: 6² + 8² = 36 + 64 = 100 = 10²).
- The length of the prism (distance between triangles) is 20 in.

The net has:
- Two triangles: J and N
- Three rectangles: K, L, M

Let’s label them based on the net:

In the net:
- J and N are the two triangular ends.
- K, L, M are the three rectangular sides.

From the 3D shape:
- The triangle has base 6 in, height 8 in → Area = (1/2)*base*height = (1/2)*6*8 = 24 in²
→ So Area of J = 24, Area of N = 24

Now the rectangles:
Each rectangle has one side = 20 in (the length of the prism), and the other side = one side of the triangle.

So:
- Rectangle attached to side 6 in → Area = 6 * 20 = 120 in² → Let’s say this is K
- Rectangle attached to side 8 in → Area = 8 * 20 = 160 in² → Let’s say this is M
- Rectangle attached to hypotenuse 10 in → Area = 10 * 20 = 200 in² → This is L

Wait — let’s match the net layout.

Looking at the net:
It shows:
Top: triangle J
Middle row: K - L - M (three rectangles side by side)
Bottom: triangle N

So likely:
- K is attached to one leg of triangle → probably 6 in side → 6×20=120
- L is attached to hypotenuse → 10×20=200
- M is attached to other leg → 8×20=160

But actually, in the 3D drawing, the shaded face is the 8x20 rectangle? Wait, no — the 3D drawing shows:
- Front triangle: legs 6 and 8, hypotenuse 10
- The side going back is 20 in long.

Actually, looking again: the 3D figure shows:
- One rectangular face is 8 in by 20 in (shaded)
- Another is 6 in by 20 in (bottom edge)
- And the slanted one is 10 in by 20 in

In the net, the middle row is K-L-M. Typically, L is the center rectangle, which would be the one attached to the hypotenuse? Or maybe not.

Actually, let’s think differently. In the net, the triangles J and N are attached to the top and bottom of the middle strip.

Usually, for a triangular prism net, the three rectangles are arranged in a row, and the two triangles are attached to the ends of the middle rectangle or something.

But here, the net shows:

J
K L M
N

So J is above L, N is below L. That means L is the rectangle that connects to both triangles — so L must be the rectangle whose width equals the side of the triangle that the triangles are attached to.

In the 3D shape, the triangles are attached to the 6x8 right triangle faces. But which side of the triangle is connected to which rectangle?

Actually, in the net, since J is above L and N is below L, that suggests that L is the rectangle that shares its top and bottom edges with the triangles. So the height of rectangle L should equal the side of the triangle that is vertical in the net.

This is getting confusing. Let me use the dimensions given.

From the 3D figure:
- The triangular face has sides 6, 8, 10.
- The prism extends 20 units perpendicular to the triangle.

So the three rectangular faces have areas:
- 6 × 20 = 120
- 8 × 20 = 160
- 10 × 20 = 200

And the two triangular faces each have area (1/2)*6*8 = 24

Total surface area = 2*24 + 120 + 160 + 200 = 48 + 480 = 528 in²

Now assign to labels:

In the net:
- J and N are triangles → each 24
- K, L, M are rectangles.

Which is which? Looking at the net diagram:

The net has:
- Top: triangle J
- Middle: left rectangle K, middle rectangle L, right rectangle M
- Bottom: triangle N

Typically, in such nets, the middle rectangle L is the one that the triangles are attached to — meaning L corresponds to the side of the triangle that is "between" the two triangles in the net.

But in reality, for a triangular prism, any of the rectangles can be in the middle. However, since the triangles are drawn above and below L, it implies that L is the rectangle whose height matches the side of the triangle that is oriented vertically when the net is laid flat.

To avoid confusion, let's assume standard labeling from the 3D view.

In the 3D drawing, the front triangle has:
- Vertical leg: 8 in
- Horizontal leg: 6 in
- Hypotenuse: 10 in

The rectangular face that is vertical and on the side might be 8x20, the bottom is 6x20, and the slanted top is 10x20.

In the net, if we imagine unfolding:

Often, the net is drawn with the three rectangles in a row, and the triangles attached to the first and last rectangle or to the middle.

Here, since J is above L and N is below L, it suggests that L is the rectangle that is "central", and the triangles are attached to its top and bottom edges. That would mean that the dimension of L that is vertical in the net is the same as the side of the triangle that is being matched.

Perhaps it's easier to calculate all areas and assign based on size.

Areas of rectangles: 120, 160, 200

Areas of triangles: 24, 24

In the net, K, L, M are the three rectangles. Without specific indication, we can assign arbitrarily but consistently.

However, looking at the 3D figure, the shaded face is the 8x20 rectangle? The shading is on the face that is 8 in high and 20 in deep — so that's 160 in².

In the net, which rectangle is shaded? The problem doesn't show shading in the net, only in the 3D view.

Perhaps we should just compute the areas and list them as per common practice.

I recall that in many textbooks, for a triangular prism net like this, the middle rectangle L is often the one corresponding to the hypotenuse, but let's check the sizes.

Another way: the perimeter of the triangle is 6+8+10=24, times length 20 gives lateral area 480, plus two bases 48, total 528.

For assignment:

Let me assume that in the net:
- K is the rectangle attached to the 6-in side → 6*20=120
- L is attached to the 10-in side (hypotenuse) → 10*20=200
- M is attached to the 8-in side → 8*20=160

And J and N are the triangles, each 24.

This makes sense because in the net, L is in the middle, and if the hypotenuse is the longest side, it might be placed in the middle.

Moreover, in the 3D drawing, the face that is most visible might be the 8x20, but in the net, it's not specified.

I think it's safe to go with:

Area of J = 24
Area of K = 120 (assuming K is 6x20)
Area of L = 200 (10x20)
Area of M = 160 (8x20)
Area of N = 24

Surface Area = 24+120+200+160+24 = let's add: 24+24=48, 120+160=280, 280+200=480, 48+480=528

Yes.

But to confirm, let's see if there's another way.

Notice that in the 3D figure, the edge labeled 8 in is vertical, 6 in horizontal, 10 in diagonal.

When you unfold, the rectangle adjacent to the 8-in side will be 8x20, etc.

In the net, since J is above L, and assuming J is attached to the top of L, then the side of L that is vertical must match the side of the triangle that J is attached to.

If J is a triangle with base 6 and height 8, but when attached to a rectangle, it's attached along one side.

Typically, the triangle is attached along its base to the rectangle.

So if J is attached to the top of L, then the base of J is equal to the width of L.

What is the base of J? In the 3D figure, the triangle has sides 6,8,10. When we say "base", it could be any side, but usually for area calculation, we take the two legs.

But for attachment, in the net, the triangle is attached along one edge to the rectangle.

In this case, since the net shows J above L, and L is a rectangle, likely the bottom edge of J is attached to the top edge of L, so they share the same length.

Similarly, N is below L, so top edge of N attached to bottom edge of L.

That would imply that L has the same width as the side of the triangle that is shared.

But a triangle has three sides; which one is shared with L?

In a typical net for a triangular prism, the two triangles are attached to the two end rectangles, not the middle one. But here, both are attached to L, which is unusual.

Looking back at the net diagram:

It shows:

J
K L M
N

This suggests that J is attached to the top of L, N to the bottom of L, and K and M are attached to the left and right of L.

That means L is surrounded by J, N, K, M — which is impossible for a net because each face should be connected properly.

I think I misinterpreted the net.

Let me visualize: probably, the net is arranged as:

- Row 1: just J (triangle)
- Row 2: K, L, M (three rectangles in a row)
- Row 3: just N (triangle)

And J is attached to the top of L, N to the bottom of L, K to the left of L, M to the right of L.

But that would mean L is connected to four faces, which is fine for a net, but in reality, for a triangular prism, each rectangular face is connected to two triangles and two other rectangles? No.

Actually, in a triangular prism, there are 5 faces: 2 triangles and 3 rectangles.

Each rectangle is adjacent to two triangles and two other rectangles? No, each rectangle is adjacent to two triangles (at its ends) and to two other rectangles? Let's think.

Standard connectivity: each rectangular face shares an edge with two triangular faces (one at each end) and with two other rectangular faces? No.

Actually, in the prism, each rectangular face is bounded by two edges from the triangles and two edges from the length.

Perhaps it's better to accept that in this net, L is the central rectangle, and it is connected to J above, N below, K left, M right. But that would require L to have four neighbors, which is possible in a net, but geometrically, for a triangular prism, the three rectangles are in a band, and the triangles are on the ends.

I recall that a common net for a triangular prism is three rectangles in a row, and then one triangle attached to the top of the first rectangle, and the other triangle attached to the bottom of the last rectangle, or something like that.

But in this case, the net shows J above L and N below L, with K and M on sides.

Perhaps K and M are not both attached to L; maybe the net is:

Imagine: the three rectangles are in a row: K - L - M

Then J is attached to the top of K, and N is attached to the bottom of M, or something. But the diagram shows J above L and N below L.

Looking at the user's image description, it says:

For problem 1, the net is shown as:

J
K L M
N

With lines connecting, so likely J is connected to the top of L, N to the bottom of L, K to the left of L, M to the right of L.

But that would mean that L is a rectangle, and it has four adjacent faces, which is not typical for a triangular prism net because the triangular faces should be at the ends.

Unless... perhaps K and M are not full rectangles; but they are labeled as areas, so they are faces.

I think there might be a mistake in my assumption.

Another possibility: in some nets, for a triangular prism, you can have the two triangles on the same side, but here it's symmetric.

Let's calculate the areas without worrying about which is which, and then assign based on the 3D figure.

From the 3D figure:
- The triangular face: area = (1/2)*6*8 = 24 in². There are two, so 48 in².
- The three rectangular faces:
- One is 6 in by 20 in = 120 in²
- One is 8 in by 20 in = 160 in²
- One is 10 in by 20 in = 200 in²
- Total surface area = 24+24+120+160+200 = 528 in²

Now for the net labels:
- J and N are the two triangles, so each is 24.
- K, L, M are the three rectangles: 120, 160, 200.

To assign which is which, we need to see how they are arranged.

In the 3D figure, the face that is shaded is the 8x20 face, which is 160 in². In the net, is any face shaded? The problem doesn't say, so perhaps we can assign arbitrarily, but typically in such problems, the labeling corresponds to the position.

Perhaps from the net layout, L is the largest rectangle, since it's in the middle, and 200 is the largest, so L = 200.

Then K and M are 120 and 160.

In the 3D figure, the 6x20 face is the bottom, 8x20 is the side, 10x20 is the top slant.

In the net, if L is the top slant (10x20=200), then K and M are the other two.

But without more information, I'll assume:

Area of J = 24
Area of K = 120 (say, the 6x20)
Area of L = 200 (10x20)
Area of M = 160 (8x20)
Area of N = 24

Surface Area = 24+120+200+160+24 = 528

Let me double-check addition: 24+24=48, 120+160=280, 280+200=480, 48+480=528. Yes.

So for problem 1:
J=24, K=120, L=200, M=160, N=24, SA=528

But let's make sure about the assignment. Perhaps K is 8x20, etc. But since the problem doesn't specify which rectangle is which, and the answer will be the same sum, but for individual areas, we need to be consistent.

Looking back at the 3D drawing: it shows the triangle with legs 6 and 8, and the prism extending 20. The face that is vertical and on the left might be 8x20, the bottom 6x20, and the hypotenuse face 10x20.

In the net, if we consider that when unfolded, the rectangle corresponding to the 8-in side might be on the left or right.

But in the net diagram, K is on the left, L in middle, M on right.

Perhaps K is attached to the 6-in side, L to 10-in, M to 8-in, or vice versa.

I recall that in some standards, the net is labeled such that the rectangles correspond to the sides in order.

To resolve this, let's look at the second problem for clue, but it's different.

Another idea: in the net, the triangle J is above L, so the base of J is equal to the width of L. What is the base of J? If J is the triangle, and it's a right triangle with legs 6 and 8, but when attached to L, it is attached along one leg.

Suppose J is attached along the 8-in leg to L. Then the width of L would be 8 in, so L = 8*20 = 160.

Similarly, N is attached to the bottom of L, so also along the 8-in side, but that would mean both triangles are attached to the same rectangle along the same side, which is impossible because the rectangle has only one top and one bottom.

Unless L is oriented with height 20 and width 8, then attaching J to the top would mean J's base is 8 in, so if J has base 8 in, then its height would be 6 in, area (1/2)*8*6=24, same thing.

So if L has width 8 in (so area 8*20=160), then J is attached to it with base 8 in, so J's dimensions are base 8, height 6, area 24.

Similarly, N attached to bottom of L, same thing.

Then what about K and M? They are attached to the left and right of L.

L has height 20 in (since the prism length is 20), and width 8 in.

When you attach K to the left of L, K must share the height of L, which is 20 in, and its width would be the next side.

In the prism, after the 8-in side, the next side of the triangle is the 6-in or 10-in.

The triangle has sides 6,8,10. If we go around, from the 8-in side, the adjacent sides are 6-in and 10-in.

In the net, if L is the 8x20 rectangle, then K and M are attached to its left and right, so they must be the other two rectangles: 6x20 and 10x20.

But which is which? It depends on the orientation.

Typically, in the net, if you have the three rectangles in a row, the middle one is L, and K and M are on sides, but here K and M are on left and right of L, so the row is K-L-M, so L is middle.

So if L is 8x20=160, then K and M are 6x20=120 and 10x20=200.

Now, which is K and which is M? Probably K is left, M is right, but no specification.

However, in the 3D figure, the 6-in side is the bottom, 10-in is the hypotenuse.

When you unfold, if you start from the 8-in face, then unfolding to the left might give the 6-in face, to the right the 10-in face, or vice versa.

But to match the net, and since the problem likely expects specific values, perhaps we can assume that K is the 6x20, M is the 10x20, or something.

Let's calculate the surface area anyway, and for individual areas, we'll use the following logic:

From the 3D figure, the face that is "front" or shaded is the 8x20, which is 160. In the net, if L is the central rectangle, and it's often the one that is "main", but in this case, since J and N are attached to L, and if L is 8x20, then it makes sense.

Moreover, in many online sources, for a similar net, the middle rectangle is the one corresponding to the side that the triangles are attached to.

So let's set:

- L = 8 * 20 = 160 in² (since triangles are attached to it, and 8 in is a leg)
- Then K and M are the other two: 6*20=120 and 10*20=200

Now, which is K and which is M? In the net, K is on the left, M on the right. In the 3D figure, if we consider the triangle with 6 in horizontal, 8 in vertical, then when you unfold, the rectangle to the left of the 8x20 might be the 6x20, and to the right the 10x20, or vice versa.

Actually, if you have the triangle with vertices at (0,0), (6,0), (0,8), then the sides are from (0,0) to (6,0) = 6 in, (0,0) to (0,8) = 8 in, (6,0) to (0,8) = 10 in.

If you attach the 8x20 rectangle to the side from (0,0) to (0,8), then when you unfold, the rectangle attached to the left might be the one for the 6-in side, but it's messy.

Perhaps for simplicity, and since the problem is for students, they might expect:

Area of J = 24
Area of K = 120 (6x20)
Area of L = 200 (10x20) -- but earlier I said L should be 160 if triangles are attached to it.

I think I found a better way: in the net, the triangle J is above L, so the side of L that is horizontal (width) must equal the base of J.

What is the base of J? In the 3D figure, the triangle is shown with base 6 in and height 8 in, but when used in the net, the base could be any side.

However, in the area calculation, it's always 24, so for the attachment, the length of the side it's attached to determines the width of the rectangle.

So if J is attached to L along a side of length S, then L has width S, and area S*20.

Similarly for N.

But since both J and N are attached to L, that would require that L has the same width for both top and bottom, which it does, but the issue is that for a single rectangle L, it can only be attached to one triangle on top and one on bottom if they are on opposite sides, but in terms of the side length, it must be that the side of the triangle that J is attached to is the same as the side that N is attached to, which is true only if both are attached to the same side of the triangle, but that doesn't make sense because the triangle has three sides.

I think the only logical explanation is that in this net, L is not the rectangle that the triangles are attached to in the way I thought.

Let me search for a standard net.

Upon second thought, in the net shown:

J
K L M
N

This likely means that:
- J is attached to the top of L
- N is attached to the bottom of L
- K is attached to the left of L
- M is attached to the right of L

But for a triangular prism, this would imply that L is a rectangle, and it is surrounded by four faces, but a rectangular face in a prism is only adjacent to two triangular faces and two rectangular faces? No, in the 3D shape, each rectangular face is adjacent to two triangular faces (at its two ends) and to two other rectangular faces? Let's think.

In a triangular prism, take one rectangular face: it has four edges:
- Two edges are shared with the two triangular faces (one at each end)
- The other two edges are shared with the two adjacent rectangular faces.

So yes, each rectangular face is adjacent to two triangular faces and two rectangular faces.

In the net, when unfolded, a rectangular face can have up to four neighbors, but in this case, for the net to be flat, it's possible.

For example, if L is a rectangle, and you attach J to its top, N to its bottom, K to its left, M to its right, then when folded, J and N would be on opposite sides, but in a prism, the two triangular faces are parallel, so they should not be on the same "level".

This is confusing.

Perhaps the net is meant to be interpreted as:

The three rectangles are in a row: K - L - M

Then J is attached to the top of K, and N is attached to the bottom of M, or something. But the diagram shows J above L and N below L.

Looking at the user's input, it says "using the net", and for problem 1, the net is drawn with J on top, then K,L,M in a row, then N on bottom, with J connected to L, N connected to L, K connected to L, M connected to L.

To resolve this, I recall that in some nets for triangular prisms, you can have the two triangles on the same rectangle if it's a different configuration, but for a right triangular prism, it's standard.

Perhaps for this problem, we can calculate the areas as follows:

From the 3D figure, the areas are fixed.

Let me assign based on the following: in the net, the rectangle L is the one that is "central", and in many problems, it is the rectangle corresponding to the hypotenuse, so 10x20=200.

Then K and M are 6x20=120 and 8x20=160.

And J and N are 24 each.

And for the surface area, it's 528.

Moreover, in the answer, the individual areas may not be graded strictly, but the sum is important.

But to be precise, let's look at problem 2 for analogy.

Problem 2 is a square pyramid.

3D figure: base 10 ft x 10 ft, slant height 22 ft (from apex to midpoint of base side).

Net: a square W in center, with triangles U,V,X,Y on the four sides.

So for pyramid, the net has the base square, and four triangular faces.

In this case, W is the base, area 10*10=100 ft².

Each triangular face has base 10 ft, height 22 ft (slant height), so area (1/2)*10*22 = 110 ft² each.

There are four triangles: U,V,X,Y, each 110.

Surface area = base + 4*triangles = 100 + 4*110 = 100+440=540 ft².

And in the net, U,V,X,Y are the triangles, W is the square.

So for problem 2, it's clear.

Back to problem 1, perhaps in the net, L is the rectangle that is not attached to the triangles in the way I thought, but let's assume that for problem 1, the areas are:

J = 24 (triangle)
N = 24 (triangle)
K = 120 (6x20)
L = 200 (10x20)
M = 160 (8x20)

Or perhaps K=160, M=120, but I think it's arbitrary.

Another way: in the 3D figure, the edge labeled 8 in is vertical, 6 in horizontal, so the face that is vertical and on the side is 8x20=160, the bottom is 6x20=120, the top slant is 10x20=200.

In the net, if we consider that when unfolded, the bottom face might be K, the slant L, the side M, but it's guesswork.

Perhaps the problem intends for us to calculate the areas based on the net's geometry, but the net doesn't have dimensions labeled; only the 3D figure does.

So we must use the 3D figure's dimensions.

I think for the sake of time, I'll go with:

For problem 1:
- Area of J = 24 in²
- Area of K = 120 in² (assume it's the 6x20)
- Area of L = 200 in² (10x20)
- Area of M = 160 in² (8x20)
- Area of N = 24 in²
- Surface Area = 24+120+200+160+24 = 528 in²

And move on.

So for problem 1:
J=24, K=120, L=200, M=160, N=24, SA=528

Now problem 2: Square Pyramid

3D figure: base is square 10 ft by 10 ft.
Slant height is 22 ft (given as the height of the triangular face from apex to base).

Net:
- W is the square base
- U,V,X,Y are the four triangular faces

Area of W = 10 * 10 = 100 ft²

Each triangular face: base = 10 ft, height = 22 ft (slant height), so area = (1/2) * base * height = (1/2)*10*22 = 110 ft²

So:
Area of U = 110
Area of V = 110
Area of W = 100
Area of X = 110
Area of Y = 110

Surface Area = sum = 110+110+100+110+110 = let's calculate: 110*4 = 440, plus 100 = 540 ft²

Note: sometimes surface area excludes the base, but the problem says "surface area of solids", and for a pyramid, if it's closed, it includes the base. In the net, all faces are included, so yes, include base.

In the net, W is included, so surface area includes the base.

So SA = 540 ft²

Problem 3: Rectangular Prism (cuboid)

3D figure: dimensions 3 in, 10 in, 12 in.

So it's a box with length, width, height: let's say l=10 in, w=3 in, h=12 in. From the figure: one edge 3 in, one 10 in, one 12 in.

Net:
- A,B,C,D,E,F are the six faces.

Typically, for a cuboid net, it's cross-shaped or other.

Here, the net is shown as:

A
B
C D E
F

So likely:
- D is the front face
- B is top, F is bottom
- C is left, E is right
- A is back or something.

Standard assignment:

Assume:
- D is the face with dimensions 10 in by 12 in (say, front)
- Then B is top, which would be 10 in by 3 in (if 3 in is depth)
- F is bottom, same as top, 10x3
- C is left side, 3 in by 12 in
- E is right side, 3x12
- A is back, same as front, 10x12

From the 3D figure: it shows a box with edges labeled: one edge 3 in (depth), one 10 in (width), one 12 in (height).

So the three pairs of faces:
- Two faces of 10 in × 12 in = 120 in² each
- Two faces of 10 in × 3 in = 30 in² each
- Two faces of 12 in × 3 in = 36 in² each

Total surface area = 2*(120 + 30 + 36) = 2*186 = 372 in²

Now assign to labels in net:

Net:
A
B
C D E
F

Typically:
- D is the front face: say 10x12 = 120
- B is the top face: if top is 10x3, then 30
- F is the bottom face: 10x3 = 30
- C is the left face: 3x12 = 36
- E is the right face: 3x12 = 36
- A is the back face: same as front, 10x12 = 120

In the net, A is above B, so if B is top, A might be back, which is correct.

So:
Area of A = 120 (back)
Area of B = 30 (top)
Area of C = 36 (left)
Area of D = 120 (front)
Area of E = 36 (right)
Area of F = 30 (bottom)

Surface Area = 120+30+36+120+36+30 = let's add: 120+120=240, 30+30=60, 36+36=72, total 240+60=300, +72=372 in²

Perfect.

Now back to problem 1, to be consistent, perhaps in problem 1, we can assign similarly.

For problem 1, in the net:

J
K L M
N

With J and N triangles, K,L,M rectangles.

From the 3D figure, the two triangles are identical, area 24 each.

The three rectangles: 6x20=120, 8x20=160, 10x20=200.

In the net, since J is above L and N below L, and K left, M right, likely L is the rectangle that is "facing" us, which in the 3D figure might be the 8x20 or 10x20.

But to match the pyramid and cuboid, where the central face in the net is often the front or main face, in problem 2, W is the base, in problem 3, D is the front.

In problem 1, L is in the center, so perhaps L is the largest rectangle or the one corresponding to the hypotenuse.

I think it's safe to set L as 10x20=200, as it's the hypotenuse and often emphasized.

Then K and M are 120 and 160.

Let's say K is 6x20=120 (left), M is 8x20=160 (right), or vice versa.

In the 3D figure, the 6-in side is the bottom, 8-in is vertical, so if we consider the net, when unfolded, the left rectangle might be the bottom, right might be the side.

But for consistency, I'll set:

Area of J = 24
Area of K = 120 (6x20)
Area of L = 200 (10x20)
Area of M = 160 (8x20)
Area of N = 24
Surface Area = 528

And proceed.

So final answers:

Problem 1:
J=24, K=120, L=200, M=160, N=24, SA=528

Problem 2:
U=110, V=110, W=100, X=110, Y=110, SA=540

Problem 3:
A=120, B=30, C=36, D=120, E=36, F=30, SA=372

Now to write it neatly.

For problem 1, units are in², problem 2 ft², problem 3 in².

In the answer, we should include units, but since the blanks are for numbers, probably just numbers.

But in the final answer, we can state with units.

Since the problem asks for the values, and surface area, we'll provide the numbers.

So let's compile.

First, problem 1:

Area of J = 24
Area of K = 120
Area of L = 200
Area of M = 160
Area of N = 24
Surface Area = 528

All in square inches.

Problem 2:

Area of U = 110
Area of V = 110
Area of W = 100
Area of X = 110
Area of Y = 110
Surface Area = 540

Square feet.

Problem 3:

Area of A = 120
Area of B = 30
Area of C = 36
Area of D = 120
Area of E = 36
Area of F = 30
Surface Area = 372

Square inches.

Now for the final answer, since the user wants the solution, and to box the final answer, but there are multiple parts.

Probably, we need to provide all.

Since the instruction is to provide the final correct answer after reasoning, and it's a multi-part problem, we can list them.

So I'll write the answers for each part.

To confirm problem 1 once more.

I recall that in some sources, for a triangular prism with right triangle base legs a,b, hypotenuse c, length l, surface area is ab + l(a+b+c)

Here a=6, b=8, c=10, l=20, so ab=48, l(a+b+c)=20*24=480, total 528, yes.

And the two triangles are each (1/2)ab=24, so together 48, which is ab, yes.

Rectangles: a*l, b*l, c*l = 120,160,200.

In the net, if L is the rectangle for c=10, then 200, etc.

I think it's fine.

So final answers.