Geometric shapes and their 2D net diagrams for educational purposes.
Diagrams of various 3D geometric shapes including a rectangular prism, cylinder, cube, and pyramid, with corresponding 2D net representations.
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Show Answer Key & Explanations
Step-by-step solution for: Perimeter, surface area and volume worksheet
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Show Answer Key & Explanations
Step-by-step solution for: Perimeter, surface area and volume worksheet
Let’s solve each problem step by step. We’ll calculate surface area and volume for each shape using the correct formulas.
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Problem 1: Rectangle (2D)
Dimensions: length = 5 cm, width = 2 cm
→ This is a flat rectangle, so we only find area, not surface area or volume.
Area of rectangle = length × width
= 5 cm × 2 cm = 10 cm²
*(Note: Since it’s 2D, no volume.)*
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Problem 2: Rectangle (2D)
Dimensions: length = 7 m, width = 3 m
Area = 7 m × 3 m = 21 m²
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Problem 3: L-shaped figure (2D composite)
Break into two rectangles:
- Big rectangle: 8 in × 4 in = 32 in²
- Small rectangle sticking out: 2 in × 2 in = 4 in²
But wait — looking at the diagram, it’s an L-shape made from:
- A vertical part: 6 in tall × 2 in wide → 12 in²
- A horizontal base: 8 in long × 2 in high → but overlaps with vertical part? Let’s recheck.
Actually, better way: Think of it as one big rectangle minus a missing corner.
Big rectangle if full: 8 in wide × 6 in tall = 48 in²
Missing piece: top right corner that’s cut out — size? From diagram: from x=6 to 8 (width 2), y=4 to 6 (height 2) → so missing 2×2 = 4 in²
So area = 48 - 4 = 44 in²
Wait — let me double-check with addition method:
Left column: 2 in wide × 6 in tall = 12 in²
Bottom row: from x=2 to 8 (6 in long) × 2 in high = 12 in²
Top middle: from x=2 to 6 (4 in long) × 2 in high = 8 in²
Total = 12 + 12 + 8 = 32 in²? That doesn’t match.
Let me label coordinates based on diagram:
Assume bottom-left is (0,0). Shape goes:
- Right 8 in → then up 2 in → left 2 in → up 2 in → left 4 in → down 6 in → back to start.
Better: Divide into three parts:
1. Bottom rectangle: 8 in × 2 in = 16 in²
2. Middle left rectangle: 2 in × 4 in = 8 in² (from y=2 to y=6, x=0 to 2)
3. Top middle rectangle: 4 in × 2 in = 8 in² (from y=4 to y=6, x=2 to 6)
Total = 16 + 8 + 8 = 32 in²
Yes! So area = 32 in²
*(No volume — it’s 2D.)*
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Problem 4: Rectangular Prism (3D)
Dimensions: length = 6 ft, width = 3 ft, height = 4 ft
Surface Area of rectangular prism = 2(lw + lh + wh)
= 2[(6×3) + (6×4) + (3×4)]
= 2[18 + 24 + 12] = 2[54] = 108 ft²
Volume = l × w × h = 6 × 3 × 4 = 72 ft³
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Problem 5: Cylinder (3D)
Radius = 3 cm, Height = 9 cm
Surface Area of cylinder = 2πr² + 2πrh
= 2π(3)² + 2π(3)(9)
= 2π(9) + 2π(27)
= 18π + 54π = 72π ≈ 72 × 3.14 = 226.08 cm²
Volume = πr²h = π(9)(9) = 81π ≈ 81 × 3.14 = 254.34 cm³
*(We’ll use π ≈ 3.14 unless told otherwise.)*
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Problem 6: Cylinder (3D)
Diameter = 10 cm → radius = 5 cm, Height = 12 cm
Surface Area = 2πr² + 2πrh
= 2π(25) + 2π(5)(12)
= 50π + 120π = 170π ≈ 170 × 3.14 = 533.8 cm²
Volume = πr²h = π(25)(12) = 300π ≈ 300 × 3.14 = 942 cm³
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Problem 7: Rectangular Prism (3D)
Length = 10 m, Width = 5 m, Height = 8 m
Surface Area = 2(lw + lh + wh)
= 2[(10×5) + (10×8) + (5×8)]
= 2[50 + 80 + 40] = 2[170] = 340 m²
Volume = 10 × 5 × 8 = 400 m³
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Problem 8: Cylinder (3D)
Diameter = 8 cm → radius = 4 cm, Height = 12 cm
Surface Area = 2πr² + 2πrh
= 2π(16) + 2π(4)(12)
= 32π + 96π = 128π ≈ 128 × 3.14 = 401.92 cm²
Volume = πr²h = π(16)(12) = 192π ≈ 192 × 3.14 = 602.88 cm³
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Problem 9: Square Pyramid (3D)
Base side = 6 in, Slant height = 5 in, Height = 4 in (given in diagram)
First, Surface Area = Base Area + Lateral Area
Base is square: 6 × 6 = 36 in²
Lateral area = 4 triangles. Each triangle has base 6 in, height = slant height = 5 in
Area of one triangle = (1/2) × 6 × 5 = 15 in²
Four triangles = 4 × 15 = 60 in²
Total Surface Area = 36 + 60 = 96 in²
Volume of pyramid = (1/3) × Base Area × Height
= (1/3) × 36 × 4 = (1/3) × 144 = 48 in³
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Problem 10: Triangular Prism (3D)
It says “right triangular prism” with legs 3 cm and 4 cm, length (prism height) = 10 cm
First, base is a right triangle with legs 3 cm and 4 cm → hypotenuse = √(3²+4²) = 5 cm (Pythagorean triple)
Surface Area = 2 × (base triangle area) + lateral faces
Base triangle area = (1/2) × 3 × 4 = 6 cm² → two bases = 12 cm²
Lateral faces: three rectangles:
- One 3 cm × 10 cm = 30 cm²
- One 4 cm × 10 cm = 40 cm²
- One 5 cm × 10 cm = 50 cm²
Total lateral = 30 + 40 + 50 = 120 cm²
Total Surface Area = 12 + 120 = 132 cm²
Volume = Base Area × Length = 6 × 10 = 60 cm³
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Now let’s compile all answers clearly.
Final Answer:
1. Area = 10 cm²
2. Area = 21 m²
3. Area = 32 in²
4. Surface Area = 108 ft², Volume = 72 ft³
5. Surface Area ≈ 226.08 cm², Volume ≈ 254.34 cm³
6. Surface Area ≈ 533.8 cm², Volume ≈ 942 cm³
7. Surface Area = 340 m², Volume = 400 m³
8. Surface Area ≈ 401.92 cm², Volume ≈ 602.88 cm³
9. Surface Area = 96 in², Volume = 48 in³
10. Surface Area = 132 cm², Volume = 60 cm³
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Problem 1: Rectangle (2D)
Dimensions: length = 5 cm, width = 2 cm
→ This is a flat rectangle, so we only find area, not surface area or volume.
Area of rectangle = length × width
= 5 cm × 2 cm = 10 cm²
*(Note: Since it’s 2D, no volume.)*
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Problem 2: Rectangle (2D)
Dimensions: length = 7 m, width = 3 m
Area = 7 m × 3 m = 21 m²
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Problem 3: L-shaped figure (2D composite)
Break into two rectangles:
- Big rectangle: 8 in × 4 in = 32 in²
- Small rectangle sticking out: 2 in × 2 in = 4 in²
But wait — looking at the diagram, it’s an L-shape made from:
- A vertical part: 6 in tall × 2 in wide → 12 in²
- A horizontal base: 8 in long × 2 in high → but overlaps with vertical part? Let’s recheck.
Actually, better way: Think of it as one big rectangle minus a missing corner.
Big rectangle if full: 8 in wide × 6 in tall = 48 in²
Missing piece: top right corner that’s cut out — size? From diagram: from x=6 to 8 (width 2), y=4 to 6 (height 2) → so missing 2×2 = 4 in²
So area = 48 - 4 = 44 in²
Wait — let me double-check with addition method:
Left column: 2 in wide × 6 in tall = 12 in²
Bottom row: from x=2 to 8 (6 in long) × 2 in high = 12 in²
Top middle: from x=2 to 6 (4 in long) × 2 in high = 8 in²
Total = 12 + 12 + 8 = 32 in²? That doesn’t match.
Let me label coordinates based on diagram:
Assume bottom-left is (0,0). Shape goes:
- Right 8 in → then up 2 in → left 2 in → up 2 in → left 4 in → down 6 in → back to start.
Better: Divide into three parts:
1. Bottom rectangle: 8 in × 2 in = 16 in²
2. Middle left rectangle: 2 in × 4 in = 8 in² (from y=2 to y=6, x=0 to 2)
3. Top middle rectangle: 4 in × 2 in = 8 in² (from y=4 to y=6, x=2 to 6)
Total = 16 + 8 + 8 = 32 in²
Yes! So area = 32 in²
*(No volume — it’s 2D.)*
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Problem 4: Rectangular Prism (3D)
Dimensions: length = 6 ft, width = 3 ft, height = 4 ft
Surface Area of rectangular prism = 2(lw + lh + wh)
= 2[(6×3) + (6×4) + (3×4)]
= 2[18 + 24 + 12] = 2[54] = 108 ft²
Volume = l × w × h = 6 × 3 × 4 = 72 ft³
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Problem 5: Cylinder (3D)
Radius = 3 cm, Height = 9 cm
Surface Area of cylinder = 2πr² + 2πrh
= 2π(3)² + 2π(3)(9)
= 2π(9) + 2π(27)
= 18π + 54π = 72π ≈ 72 × 3.14 = 226.08 cm²
Volume = πr²h = π(9)(9) = 81π ≈ 81 × 3.14 = 254.34 cm³
*(We’ll use π ≈ 3.14 unless told otherwise.)*
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Problem 6: Cylinder (3D)
Diameter = 10 cm → radius = 5 cm, Height = 12 cm
Surface Area = 2πr² + 2πrh
= 2π(25) + 2π(5)(12)
= 50π + 120π = 170π ≈ 170 × 3.14 = 533.8 cm²
Volume = πr²h = π(25)(12) = 300π ≈ 300 × 3.14 = 942 cm³
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Problem 7: Rectangular Prism (3D)
Length = 10 m, Width = 5 m, Height = 8 m
Surface Area = 2(lw + lh + wh)
= 2[(10×5) + (10×8) + (5×8)]
= 2[50 + 80 + 40] = 2[170] = 340 m²
Volume = 10 × 5 × 8 = 400 m³
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Problem 8: Cylinder (3D)
Diameter = 8 cm → radius = 4 cm, Height = 12 cm
Surface Area = 2πr² + 2πrh
= 2π(16) + 2π(4)(12)
= 32π + 96π = 128π ≈ 128 × 3.14 = 401.92 cm²
Volume = πr²h = π(16)(12) = 192π ≈ 192 × 3.14 = 602.88 cm³
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Problem 9: Square Pyramid (3D)
Base side = 6 in, Slant height = 5 in, Height = 4 in (given in diagram)
First, Surface Area = Base Area + Lateral Area
Base is square: 6 × 6 = 36 in²
Lateral area = 4 triangles. Each triangle has base 6 in, height = slant height = 5 in
Area of one triangle = (1/2) × 6 × 5 = 15 in²
Four triangles = 4 × 15 = 60 in²
Total Surface Area = 36 + 60 = 96 in²
Volume of pyramid = (1/3) × Base Area × Height
= (1/3) × 36 × 4 = (1/3) × 144 = 48 in³
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Problem 10: Triangular Prism (3D)
It says “right triangular prism” with legs 3 cm and 4 cm, length (prism height) = 10 cm
First, base is a right triangle with legs 3 cm and 4 cm → hypotenuse = √(3²+4²) = 5 cm (Pythagorean triple)
Surface Area = 2 × (base triangle area) + lateral faces
Base triangle area = (1/2) × 3 × 4 = 6 cm² → two bases = 12 cm²
Lateral faces: three rectangles:
- One 3 cm × 10 cm = 30 cm²
- One 4 cm × 10 cm = 40 cm²
- One 5 cm × 10 cm = 50 cm²
Total lateral = 30 + 40 + 50 = 120 cm²
Total Surface Area = 12 + 120 = 132 cm²
Volume = Base Area × Length = 6 × 10 = 60 cm³
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Now let’s compile all answers clearly.
Final Answer:
1. Area = 10 cm²
2. Area = 21 m²
3. Area = 32 in²
4. Surface Area = 108 ft², Volume = 72 ft³
5. Surface Area ≈ 226.08 cm², Volume ≈ 254.34 cm³
6. Surface Area ≈ 533.8 cm², Volume ≈ 942 cm³
7. Surface Area = 340 m², Volume = 400 m³
8. Surface Area ≈ 401.92 cm², Volume ≈ 602.88 cm³
9. Surface Area = 96 in², Volume = 48 in³
10. Surface Area = 132 cm², Volume = 60 cm³
Parent Tip: Review the logic above to help your child master the concept of area and volume worksheet.
