Surface Area using Nets Worksheets - Free Printable

Let’s solve each problem one by one. We’ll identify the shape from its net, then calculate the surface area using the given dimensions and π = 3.14. Round all answers to two decimal places.

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Problem 1:

The net has six rectangles:
- Two are 9 in × 6 in (top and bottom)
- Two are 9 in × 4 in (front and back)
- Two are 6 in × 4 in (left and right sides)

This is a rectangular prism.

Surface Area = sum of areas of all faces

= 2×(9×6) + 2×(9×4) + 2×(6×4)
= 2×54 + 2×36 + 2×24
= 108 + 72 + 48
= 228.00 in²

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Problem 2:

The net shows a square base (12 ft × 12 ft) with four identical triangles attached to each side. Each triangle has base 12 ft and height 8 ft (given as perpendicular height).

This is a square pyramid.

Surface Area = area of base + area of 4 triangular faces

Base area = 12 × 12 = 144 ft²

One triangle area = (1/2) × base × height = (1/2) × 12 × 8 = 48 ft²
Four triangles = 4 × 48 = 192 ft²

Total Surface Area = 144 + 192 = 336.00 ft²

*(Note: The slant heights labeled “10 ft” on the sides are not needed since we’re given the vertical height of the triangle — which is what we use for area.)*

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Problem 3:

The net has a pentagon in the center? Wait — actually, looking closely: it’s made of five triangles around a central point? No — wait, let’s re-examine.

Actually, this net has:

- One regular pentagon? No — wait, no pentagon. It’s composed of 5 congruent isosceles triangles meeting at a point? But there’s also a rectangle? Wait — no.

Wait — actually, this looks like a pentagonal pyramid? Let me check again.

Looking at the diagram: There is a central pentagon? Actually, no — the figure shows 5 triangles arranged around a central point, but they share edges. Actually, this is a regular pentagonal pyramid? Wait — no, because there’s no separate base shown.

Wait — correction: This net is for a regular pentagonal pyramid, where the base is a regular pentagon and 5 triangular faces meet at the apex.

But in the diagram, the base is not drawn separately — instead, the 5 triangles are shown connected along their bases? Actually, no — look: the inner part is a pentagon formed by the bases of the triangles? Actually, yes — the 5 triangles are attached to a central pentagon? Wait, no — in the drawing, the 5 triangles are arranged so that their bases form a pentagon? Actually, no — the diagram shows 5 triangles sharing a common vertex at the top, and their bases forming a pentagon at the bottom? But in the net, it's laid flat.

Actually, upon closer inspection: The net consists of 5 congruent isosceles triangles, each with base 8 yd and equal sides 5 yd, and a height of 3 yd (marked inside one triangle). Also, the base of each triangle is 8 yd, and they are arranged around a central point? No — actually, this is a pentagonal pyramid with a regular pentagon base? But the base isn’t shown as a separate polygon — instead, the 5 triangles are shown with their bases forming the perimeter of the base.

Wait — I think I misread. Actually, in this net, the central region is a pentagon? No — looking again: the figure shows 5 triangles radiating from a central point? That would be a cone-like shape, but it’s polygons.

Actually, this is a regular pentagonal pyramid where the base is a regular pentagon with side length 8 yd, and each lateral face is a triangle with base 8 yd and height 3 yd (the slant height? Or actual height?).

Wait — the diagram labels a height of 3 yd inside one triangle, perpendicular to the base of 8 yd. So that’s the height of the triangular face.

Also, the base of the pyramid is a regular pentagon with side 8 yd. But in the net, is the base included? In the diagram, the 5 triangles are shown, but there’s no separate pentagon — so perhaps this net does NOT include the base? But that can’t be — surface area usually includes all faces unless specified.

Wait — let’s count the faces: The net shows 5 triangles. If it’s a pyramid, it should have 5 triangular faces + 1 pentagonal base = 6 faces. But here only 5 shapes are drawn? Actually, no — looking again: the central region is bounded by the bases of the 5 triangles — that forms a pentagon! So the base is implied — it’s the inner pentagon formed by connecting the bases of the triangles.

So yes, this is a pentagonal pyramid with:

- Base: regular pentagon with side 8 yd
- 5 triangular faces, each with base 8 yd and height 3 yd

But wait — to find the area of the pentagonal base, we need more info. The diagram doesn’t give the apothem or area of the pentagon directly. However, notice that the 5 triangles are drawn with height 3 yd — but that’s the height of the lateral face, not related to the base area.

Actually, I think I made a mistake. Let me reinterpret.

Looking at the diagram again: It shows 5 identical isosceles triangles, each with two sides 5 yd, base 8 yd, and height 3 yd (from apex to base). These 5 triangles are arranged such that their bases form a closed loop — meaning the base of the pyramid is not a separate face; rather, this net is for a shape without a base? That doesn’t make sense.

Wait — another possibility: This is a regular pentagonal bipyramid? No.

Perhaps it’s a pyramid with a pentagonal base, and the net includes the base? But in the diagram, there is no separate pentagon — the 5 triangles are connected at their apexes? No.

Actually, I recall that sometimes nets for pyramids show the base in the center with triangles around it. In this case, the central region is a pentagon? Let me visualize: if you have a pentagon in the center, and attach a triangle to each side, that’s a pentagonal pyramid net.

In the diagram, the inner shape is a pentagon with side 8 yd, and attached to each side is a triangle with base 8 yd and height 3 yd. Yes! That makes sense. The "height 3 yd" is marked inside one triangle, perpendicular to the base.

So:

Shape: Pentagonal Pyramid

Surface Area = area of pentagonal base + area of 5 triangular faces

First, area of one triangular face = (1/2) × base × height = (1/2) × 8 × 3 = 12 yd²
Five triangles = 5 × 12 = 60 yd²

Now, area of regular pentagon with side length s = 8 yd.

Formula for area of regular pentagon: (1/4) × √(5(5+2√5)) × s²
Or approximately: 1.72048 × s²

But since this is for students, and the diagram might expect us to use the given data only — wait, the diagram doesn't provide the apothem or other info for the pentagon. However, notice that in the net, the pentagon is formed, but no dimensions for its area are given except side length.

But we can calculate the area of the regular pentagon if we know the side. Standard formula:

Area = (5/4) × s² × cot(π/5) = (5/4) × 64 × cot(36°)

cot(36°) ≈ 1.37638

So Area ≈ (5/4) × 64 × 1.37638 = (80) × 1.37638 ≈ 110.1104 yd²

But that seems messy, and the problem gives nice numbers — 5 yd, 8 yd, 3 yd. Perhaps the "height 3 yd" is not for the triangle, but for something else? Wait, no — it's clearly marked as the height of the triangle.

Another thought: Maybe this is not a pyramid with a pentagonal base, but a different shape. Let's count the faces: the net has 5 triangles. If it's a closed solid, it must have more faces. Unless it's an open shape, but surface area usually means total surface area of the closed solid.

Perhaps I misidentified. Let me look online or recall: a common net with 5 triangles is for a pentagonal pyramid, which has 6 faces. So the base must be included. In the diagram, the central region is the base — a pentagon with side 8 yd.

To find its area, we can divide the pentagon into 5 isosceles triangles from the center, but we don't have the apothem.

Notice that in the triangular faces, the height is given as 3 yd, but that's the slant height of the pyramid, not helpful for base area.

Perhaps the problem expects us to realize that the base is not included? But that would be unusual.

Wait — let's read the problem again: "Identify the shape that each net represents". For problem 3, the net shows 5 triangles arranged in a star-like fashion? No, in the diagram, it's 5 triangles sharing a common vertex at the top, and their bases forming a pentagon at the bottom — but in 2D net, it's laid out with the pentagon in the center and triangles on the sides.

Upon second thought, in many textbooks, when they show a net for a pentagonal pyramid, they draw the pentagon in the center and attach a triangle to each side. In this diagram, that's exactly what is shown: a central pentagon with 5 triangles attached to its sides. The "height 3 yd" is the height of each triangular face, measured from the base to the apex of the triangle.

So, to find the area of the pentagonal base, we need its area. Since it's regular with side 8 yd, we can use the formula:

Area = (1/4) * √(5(5+2√5)) * s^2

Calculate numerically:

s = 8

s^2 = 64

√5 ≈ 2.236

5+2*2.236 = 5+4.472 = 9.472

5*9.472 = 47.36

√47.36 ≈ 6.881

Then (1/4)*6.881*64 = (1.72025)*64 ≈ 110.096 yd²

Then area of 5 triangles = 5 * (1/2)*8*3 = 5*12 = 60 yd²

Total surface area = 110.096 + 60 = 170.096 ≈ 170.10 yd²

But this seems complicated for a student worksheet. Perhaps there's a mistake.

Another idea: maybe the "3 yd" is not the height of the triangle, but the apothem of the pentagon? But it's drawn inside the triangle, perpendicular to the base, so it's the height of the triangle.

Perhaps the shape is not a pyramid but a different solid. Let's think: if you have 5 triangles meeting at a point, it could be a cone, but it's polygons.

I recall that a regular pentagonal pyramid's net typically has the base and 5 triangles. Given that, and since the problem provides the height of the triangle, we must calculate the base area separately.

But let's check the other problems for consistency. Problem 4 is a cylinder, which is straightforward.

Perhaps for problem 3, the base is not included in the net? But that doesn't make sense for surface area.

Wait — looking back at the diagram description: "5 yd" on the sides of the triangles, "8 yd" on the bases, and "3 yd" as height inside one triangle. Also, the central region is a pentagon with side 8 yd.

Perhaps the problem expects us to use the formula for the area of the pentagon as (perimeter * apothem)/2, but we don't have the apothem.

Notice that in the triangular face, with base 8 yd and height 3 yd, the area is 12 yd², as before.

For the pentagon, if we assume it's regular, and we can find its area using the side length.

Standard value: area of regular pentagon with side s is approximately 1.72048 * s^2.

So 1.72048 * 64 = 110.11072

Plus 60 = 170.11072 ≈ 170.11 yd²

But let's see if there's a better way. Perhaps the "3 yd" is not used for the triangle area, but for something else. No, it's clearly the height of the triangle.

Another thought: maybe this is a triangular bipyramid or something, but it has 5 faces, which is not standard.

I think I have to go with the pentagonal pyramid interpretation.

So Shape: Pentagonal Pyramid

Surface Area = area of base + lateral area = area of regular pentagon + 5 * area of triangle

= [ (5/4) * s^2 * cot(π/5) ] + 5 * (1/2)*b*h

With s=8, b=8, h=3

cot(π/5) = cot(36°) = 1/tan(36°) ≈ 1/0.7265 ≈ 1.3764

So base area = (5/4) * 64 * 1.3764 = 80 * 1.3764 = 110.112

Lateral area = 5 * 12 = 60

Total = 170.112 ≈ 170.11 yd²

But to match the precision, and since the inputs are integers, perhaps they expect us to use a specific value.

Maybe the pentagon's area can be calculated by dividing it into 5 triangles from the center, but we need the radius or apothem.

Notice that in the lateral triangle, with base 8 and height 3, the slant edge is 5 yd (given), which checks out: half-base is 4, height 3, so hypotenuse = sqrt(4^2 + 3^2) = 5, yes.

For the base pentagon, the apothem a can be found from the formula: a = (s/2) / tan(π/5) = 4 / tan(36°) ≈ 4 / 0.7265 ≈ 5.505

Then area = (1/2) * perimeter * apothem = (1/2) * 40 * 5.505 = 20 * 5.505 = 110.1 yd²

Same as before.

So total surface area = 110.1 + 60 = 170.1 yd²

Rounded to two decimals: 170.10 yd² (since 110.1 is 110.10)

110.1 is 110.10, yes.

So I'll go with that.

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Problem 4:

The net shows two circles and one rectangle.

This is a cylinder.

Given: radius r = 7 m, height of rectangle = 9 m (which is the height of the cylinder).

Surface Area of cylinder = 2πr² + 2πrh

= 2πr(r + h)

Plug in r=7, h=9, π=3.14

First, 2πr² = 2 * 3.14 * 49 = 6.28 * 49

6.28 * 50 = 314, minus 6.28 = 307.72

Better: 2 * 3.14 = 6.28

6.28 * 49 = 6.28 * 50 - 6.28 = 314 - 6.28 = 307.72

Then 2πrh = 2 * 3.14 * 7 * 9 = 6.28 * 63

6.28 * 60 = 376.8, 6.28 * 3 = 18.84, total 395.64

Or: 2πrh = 2 * 3.14 * 63 = 6.28 * 63

63 * 6 = 378, 63 * 0.28 = 17.64, total 395.64

Total surface area = 307.72 + 395.64 = 703.36 m²

Using the formula 2πr(r+h) = 2*3.14*7*(7+9) = 6.28*7*16

First, 7*16=112

6.28*112 = 6.28*100=628, 6.28*12=75.36, total 703.36

Yes.

So 703.36 m²

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Now, summarizing:

1) Rectangular prism, SA = 228.00 in²

2) Square pyramid, SA = 336.00 ft²

3) Pentagonal pyramid, SA = 170.10 yd² (after calculation)

4) Cylinder, SA = 703.36 m²

For problem 3, I'm a bit unsure, but based on standard interpretation, it should be correct.

Perhaps the "3 yd" is the apothem of the pentagon, but it's drawn in the triangle, so unlikely.

Another possibility: in some nets, the base is not included, but the problem says "surface area", which usually means total surface area including all faces.

I think 170.10 is fine.

Let me double-check problem 2: square pyramid with base 12x12, and triangular faces with base 12 and height 8. Area per triangle 48, times 4 is 192, plus base 144, total 336. Correct.

Problem 1: rectangular prism 9x6x4. Faces: two 9x6=54, two 9x4=36, two 6x4=24. Sum 54+54+36+36+24+24= let's add: 54+54=108, 36+36=72, 24+24=48, total 108+72=180, +48=228. Correct.

Problem 4: cylinder r=7, h=9. SA=2πr² + 2πrh = 2*3.14*49 + 2*3.14*7*9 = 307.72 + 395.64 = 703.36. Correct.

For problem 3, perhaps the shape is a different one. Let me search my memory: is there a solid with 5 triangular faces? A triangular bipyramid has 6 faces. A square pyramid has 5 faces: 4 triangles and 1 square. Oh! Wait a minute!

In problem 3, if it's a square pyramid, but it has 5 triangles? No.

Look at the diagram description: it has 5 triangles, each with sides 5yd,5yd,8yd, and height 3yd. And they are arranged around a central point? But in a net, for a square pyramid, you have one square and four triangles.

Here, there are five triangles, so it must be a pentagonal pyramid.

Perhaps it's a cone, but it's polygons.

I think I have to stick with pentagonal pyramid.

Maybe the "central" part is not a face, but in reality, for a pentagonal pyramid, the net has the pentagon and five triangles, so six faces. In the diagram, if only five shapes are drawn, perhaps the base is missing, but that doesn't make sense.

Upon re-examining the user's image description: "3) [diagram with 5 triangles, each labeled 5yd on sides, 8yd on base, and 3yd height inside]"

And it's listed as a single net. In standard geometry, a net for a pentagonal pyramid includes the pentagonal base and five triangular faces, so six polygons. If only five are shown, perhaps the base is not included, but then it's not a closed solid.

Perhaps for this problem, they consider only the lateral surface area? But the instruction says "surface area", and for solids, it usually means total surface area.

Let's look at problem 2: it has a square and four triangles, so five faces, and we included the base.

Similarly, for problem 3, if it's a pentagonal pyramid, it should have six faces.

But in the diagram, how many shapes are there? The user said "5) " but it's problem 3, and described as having 5 triangles. Perhaps the central region is the base, and it's counted as one face, even though not labeled as a separate polygon.

In many diagrams, the base is shown as the central polygon, and triangles attached, so it's understood.

So I think my calculation is correct.

To confirm, let's calculate the area of the pentagon using another method.

A regular pentagon can be divided into 5 isosceles triangles from the center, each with two sides equal to the radius R, and base 8 yd.

The angle at center is 72 degrees.

So area of one such triangle = (1/2) * R * R * sin(72°)

But we don't have R.

From the side length, the radius R = s / (2*sin(36°)) = 8 / (2*0.5878) ≈ 8 / 1.1756 ≈ 6.805

Then area of one triangle = (1/2)*R^2*sin(72°) = (1/2)*(46.308)*0.9511 ≈ (23.154)*0.9511 ≈ 22.02

Times 5 = 110.1, same as before.

So yes.

Thus, total SA = 110.1 + 60 = 170.1 yd²

Rounded to two decimals: 170.10 yd²

Okay.

Final Answers:

1) Shape: Rectangular Prism, Surface Area = 228.00 in²

2) Shape: Square Pyramid, Surface Area = 336.00 ft²

3) Shape: Pentagonal Pyramid, Surface Area = 170.10 yd²

4) Shape: Cylinder, Surface Area = 703.36 m²

Now, for the response, since ExplainAnswer is false, etc., I just need to provide the final answer in the specified format.

But the Final Answer section should contain the answers for all parts.

So I'll list them clearly.