Surface Area Worksheets - Free Printable

Let’s solve each problem step by step. We’ll find the surface area of each solid by adding up the areas of all its faces, using the nets provided.

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Problem 1: Rectangular Prism (with triangular ends)

Looking at the net:
- There are two triangles (front and back)
- Three rectangles on the sides
- One rectangle on top/bottom? Wait — actually, looking closely, this is a triangular prism.

The net shows:
- Two identical right triangles (bases)
- Three rectangles connecting them

From the diagram:
- Triangle base = 6 units, height = 8 units → Area of one triangle = (1/2) × 6 × 8 = 24
→ So two triangles = 24 × 2 = 48

Now the three rectangles:
- One rectangle: 6 × 10 = 60 (this is the bottom face, matching the triangle’s base)
- One rectangle: 8 × 10 = 80 (matching the triangle’s height side)
- One rectangle: ? The hypotenuse of the triangle is needed.

Wait — we need to check if it’s a right triangle with legs 6 and 8 → then hypotenuse = √(6² + 8²) = √(36+64) = √100 = 10. Oh! So the third side is also 10? That can’t be — wait, no: in the net, the middle rectangle is labeled “10” as length, and the triangle has sides 6, 8, and... let me recheck.

Actually, looking again: the triangle has base 6, height 8, and the slant side is not given — but in the net, the rectangle attached to the hypotenuse is labeled “10”. So yes, the triangle must have sides 6, 8, 10 — which is a right triangle (since 6-8-10 is a Pythagorean triple).

So the three rectangular faces:
- Rectangle 1: 6 × 10 = 60
- Rectangle 2: 8 × 10 = 80
- Rectangle 3: 10 × 10 = 100? Wait — no! The prism’s length (the distance between the two triangular bases) is 10. So each rectangle has one side = 10 (the length of the prism), and the other side = the side of the triangle.

So:
- Rectangle along base 6: 6 × 10 = 60
- Rectangle along height 8: 8 × 10 = 80
- Rectangle along hypotenuse 10: 10 × 10 = 100

Total for rectangles: 60 + 80 + 100 = 240

Plus two triangles: 24 × 2 = 48

Total surface area = 240 + 48 = 288

But wait — let me double-check the diagram labels. In the image, the triangle is drawn with base 6, height 8, and the rectangle attached to the hypotenuse is labeled “10”, meaning the prism’s length is 10. Yes, that matches.

So:

Area of A (triangle): 24
Area of B (rectangle 6x10): 60
Area of C (rectangle 8x10): 80
Area of D (rectangle 10x10): 100
Area of E (other triangle): 24

Surface Area = 24 + 60 + 80 + 100 + 24 = 288

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Problem 2: Square Pyramid

Net shows:
- One square base
- Four identical triangular faces

From diagram:
- Square base: side = 6 → area = 6 × 6 = 36
- Each triangle: base = 6, height = 5 (labeled on the triangle)

Area of one triangle = (1/2) × 6 × 5 = 15
Four triangles = 15 × 4 = 60

Total surface area = 36 + 60 = 96

Check labels:
A = square base = 36
B, C, D, E = four triangles = each 15

So:

Area of A: 36
Area of B: 15
Area of C: 15
Area of D: 15
Area of E: 15
Surface Area = 36 + 15×4 = 96

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Problem 3: Cube

Net shows a cube unfolded into a cross shape.

All faces are squares. From diagram, each square has side = 4.

Area of one face = 4 × 4 = 16

Cube has 6 faces → total surface area = 16 × 6 = 96

In the net, there are 6 squares labeled A through F.

Each should be 16.

So:

Area of A: 16
Area of B: 16
Area of C: 16
Area of D: 16
Area of E: 16
Area of F: 16
Surface Area = 16 × 6 = 96

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Final Answer:

Problem 1:
Area of A = 24
Area of B = 60
Area of C = 80
Area of D = 100
Area of E = 24
Surface Area = 288

Problem 2:
Area of A = 36
Area of B = 15
Area of C = 15
Area of D = 15
Area of E = 15
Surface Area = 96

Problem 3:
Area of A = 16
Area of B = 16
Area of C = 16
Area of D = 16
Area of E = 16
Area of F = 16
Surface Area = 96