Surface Area -3 With Nets online exercise for - Free Printable

Let’s solve each problem step by step. We’ll find the area of each labeled face in the net, 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 dimensions:
- Triangle base = 6 ft
- Triangle height = 8 ft (right triangle)
- Hypotenuse of triangle = 10 ft (given)
- Length of prism (depth) = 9 ft

The net has 5 faces: J, K, L, M, N

From the 3D shape and net:

- Face N is the right triangle on the front → same as the back triangle (J)
- Face J is the other triangle (back) → same size as N
- Face K is the rectangle on the left side → height = 8 ft, width = 9 ft
- Face L is the bottom rectangle → base = 6 ft, width = 9 ft
- Face M is the slanted top rectangle → hypotenuse = 10 ft, width = 9 ft

Now calculate areas:

→ Area of triangle = (base × height) ÷ 2
So for J and N:
Area = (6 × 8) ÷ 2 = 48 ÷ 2 = 24 ft²

→ Area of rectangle = length × width

K: 8 ft × 9 ft = 72 ft²
L: 6 ft × 9 ft = 54 ft²
M: 10 ft × 9 ft = 90 ft²

Now add all areas for surface area:

J + K + L + M + N = 24 + 72 + 54 + 90 + 24
= (24 + 24) + 72 + 54 + 90
= 48 + 72 = 120; 120 + 54 = 174; 174 + 90 = 264 ft²

✔ Check: Two triangles (24×2=48), three rectangles (72+54+90=216); 48+216=264 ✔️

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Problem 2: Square Pyramid

Base is a square: 16 yd × 16 yd
Each triangular face has base = 16 yd, and slant height = 15 yd (given from apex to midpoint of base edge)

Net has 5 faces: U, V, W, X, Y

W is the square base → 16 yd × 16 yd
U, V, X, Y are the four identical triangular faces

Area of square W = 16 × 16 = 256 yd²

Area of one triangle = (base × height) ÷ 2 = (16 × 15) ÷ 2 = 240 ÷ 2 = 120 yd²

There are 4 triangles → 4 × 120 = 480 yd²

Total surface area = 256 + 480 = 736 yd²

Assigning labels:

In the net, W is center square. U, V, X, Y are the four triangles around it — all same size.

So:

Area of U = 120
Area of V = 120
Area of W = 256
Area of X = 120
Area of Y = 120

Sum: 120×4 = 480 + 256 = 736 ✔️

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Problem 3: Rectangular Prism (Box)

Dimensions:
Length = 10 in
Width = 5 in
Height = 12 in

Net shows 6 faces: A, B, C, D, E, F

From the 3D box and net layout:

Typical labeling for such a net:

- A and F are top and bottom → both are 10 in × 5 in
- B and D are front and back → both are 10 in × 12 in
- C and E are left and right sides → both are 5 in × 12 in

Check the net diagram:

Looking at the cross-shaped net:

- Top row: A (top face)
- Middle column: B (front?), D (bottom?), F (back?) — wait, let's map carefully.

Actually, standard net for rectangular prism:

Usually:

- Center square/rectangle is front or bottom.
But looking at the 3D drawing:

Front face is labeled F → 10 in wide, 12 in tall → so F = 10×12
Top face is A → 10 in long, 5 in deep → A = 10×5
Side face is C → 5 in deep, 12 in tall → C = 5×12

In the net:

It’s arranged as:

A
B
C D E
F

This suggests:

- D is the bottom face? Or front?

Better approach: In a rectangular prism, opposite faces are equal.

So we have three pairs:

Pair 1: 10 in × 5 in → two faces (top and bottom) → let’s say A and F? But in 3D, A is top, F is front — not opposite.

Wait — look again at 3D drawing:

Labelled faces:

- A: top → 10 in (length) × 5 in (width) → area = 50 in²
- F: front → 10 in (length) × 12 in (height) → area = 120 in²
- C: left side → 5 in (width) × 12 in (height) → area = 60 in²

Opposite faces:

- Back face (opposite F) = also 10×12 = 120 → that should be B or D?
- Right side (opposite C) = 5×12 = 60 → that should be E
- Bottom (opposite A) = 10×5 = 50 → that should be D or F? Wait, F is already used.

Actually, in the net:

Positions:

Row 1: A
Row 2: B
Row 3: C - D - E
Row 4: F

Standard interpretation:

When folded:

- D is the bottom face
- B is the back face
- F is the front face? No — usually front is middle.

Alternative method: Just compute all six faces using dimensions.

Rectangular prism has:

- Two faces of 10×5 → area = 50 each → total 100
- Two faces of 10×12 → area = 120 each → total 240
- Two faces of 5×12 → area = 60 each → total 120

Total surface area = 100 + 240 + 120 = 460 in²

Now assign to letters based on typical net folding:

Assume:

- A = top = 10×5 = 50
- F = bottom = 10×5 = 50 (since it’s below D, likely bottom)
- B = back = 10×12 = 120
- D = front = 10×12 = 120? But in 3D, F is front — conflict.

Wait — in the 3D drawing, F is the large front face → 10x12 → so in net, which face corresponds to front?

Looking at net: when folded, if D is the center, then:

- Above D is B → becomes back
- Below D is F → becomes front? But in 3D, F is front — yes!
- Left of D is C → left side
- Right of D is E → right side
- Above B is A → top

So:

- A = top = 10×5 = 50
- B = back = 10×12 = 120
- C = left = 5×12 = 60
- D = bottom = 10×5 = 50? But bottom should be same as top — yes, but in 3D, bottom isn’t labeled, but in net D is between C and E, below B, above F.

Actually, when you fold the net:

Imagine D is the bottom face.

Then:

- Fold up C, E, B, F to form sides.
- Then fold A down to cover top.

But in 3D drawing, F is the front face — so perhaps F is not bottom.

To avoid confusion, let’s use the fact that in the net, the faces are arranged so that:

- Faces adjacent in net share an edge.

But simpler: since we know there are three pairs, and we can match by size.

From 3D:

- Face A: top → 10x5 → 50
- Face C: side → 5x12 → 60
- Face F: front → 10x12 → 120

In net:

A is alone on top → likely top face → 50

F is at bottom → likely bottom face → but bottom should be same as top → 50? But in 3D, F is front — contradiction.

I think there’s a mislabel assumption.

Look back at user’s image description:

In 3D box:

- A is top
- C is left side
- F is front

In net:

The net is drawn as:

[A]
[B]
[C][D][E]
[F]

When folded into a box:

- D is the bottom
- B is the back
- F is the front
- C is left
- E is right
- A is top

Yes! That makes sense.

So:

- A = top = 10 in (length) × 5 in (width) = 50 in²
- B = back = 10 in (length) × 12 in (height) = 120 in²
- C = left = 5 in (width) × 12 in (height) = 60 in²
- D = bottom = 10 in × 5 in = 50 in²
- E = right = 5 in × 12 in = 60 in²
- F = front = 10 in × 12 in = 120 in²

Perfect — matches 3D labeling: F is front, A is top, C is left.

Now verify totals:

A: 50
B: 120
C: 60
D: 50
E: 60
F: 120

Sum: 50+120=170; +60=230; +50=280; +60=340; +120=460 in² ✔️

And pairs:
Top/bottom: A&D = 50+50=100
Front/back: F&B = 120+120=240
Left/right: C&E = 60+60=120
Total: 100+240+120=460 ✔️

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Final Answers:

Problem 1:

Area of J = 24 ft²
Area of K = 72 ft²
Area of L = 54 ft²
Area of M = 90 ft²
Area of N = 24 ft²
Surface Area = 264 ft²

Problem 2:

Area of U = 120 yd²
Area of V = 120 yd²
Area of W = 256 yd²
Area of X = 120 yd²
Area of Y = 120 yd²
Surface Area = 736 yd²

Problem 3:

Area of A = 50 in²
Area of B = 120 in²
Area of C = 60 in²
Area of D = 50 in²
Area of E = 60 in²
Area of F = 120 in²
Surface Area = 460 in²

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Final Answer:
1)
Area of J = 24 ft²
Area of K = 72 ft²
Area of L = 54 ft²
Area of M = 90 ft²
Area of N = 24 ft²
Surface Area = 264 ft²

2)
Area of U = 120 yd²
Area of V = 120 yd²
Area of W = 256 yd²
Area of X = 120 yd²
Area of Y = 120 yd²
Surface Area = 736 yd²

3)
Area of A = 50 in²
Area of B = 120 in²
Area of C = 60 in²
Area of D = 50 in²
Area of E = 60 in²
Area of F = 120 in²
Surface Area = 460 in²