Math worksheet for calculating the volume of pyramids, cones, and spheres with labeled dimensions and answer spaces.

Let’s go through each problem one by one, using the correct volume formulas:

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Formulas to remember:

- Pyramid (rectangular base):
Volume = (1/3) × length × width × height

- Cone:
Volume = (1/3) × π × r² × h

- Sphere:
Volume = (4/3) × π × r³

- Hemisphere (half a sphere):
Volume = (2/3) × π × r³

Use π ≈ 3.1416 unless told otherwise. Round final answers to 2 decimal places.

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Problem 1: Square pyramid

Base = 4 cm × 4 cm, Height = 9 cm
V = (1/3) × 4 × 4 × 9 = (1/3) × 144 = 48.00 cm³

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Problem 2: Rectangular pyramid

Length = 12 cm, Width = 6 cm, Height = 15 cm
V = (1/3) × 12 × 6 × 15 = (1/3) × 1080 = 360.00 cm³

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Problem 3: Rectangular pyramid

Length = 5 cm, Width = 7 cm, Height = 20 cm
V = (1/3) × 5 × 7 × 20 = (1/3) × 700 ≈ 233.33 cm³

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Problem 4: Composite shape — square pyramid on top of rectangular prism? Wait — looking at diagram: it’s a pyramid with rectangular base 3cm x 2cm and height 7cm? But also has a “step” — actually, from labels: base is 3cm by 2cm, height is 7cm — so just a pyramid.

Wait — re-examining: The figure shows a pyramid sitting on a flat base that’s 3cm by 2cm, and vertical height is 7cm. So yes, pyramid only.

V = (1/3) × 3 × 2 × 7 = (1/3) × 42 = 14.00 cm³

*(Note: Some might think it's two shapes, but label says "pyramid" and dimensions match single pyramid)*

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Problem 5: Cone

Radius = 3 cm, Height = 9 cm
V = (1/3) × π × 3² × 9 = (1/3) × π × 9 × 9 = (1/3) × 81π = 27π ≈ 27 × 3.1416 ≈ 84.82 cm³

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Problem 6: Cone

Radius = 1.5 cm, Height = 7 cm
V = (1/3) × π × (1.5)² × 7 = (1/3) × π × 2.25 × 7 = (1/3) × 15.75π ≈ 5.25π ≈ 5.25 × 3.1416 ≈ 16.49 cm³

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Problem 7: Cone

Diameter = 20 cm → Radius = 10 cm, Height = 24 cm
V = (1/3) × π × 10² × 24 = (1/3) × π × 100 × 24 = (1/3) × 2400π = 800π ≈ 800 × 3.1416 ≈ 2513.27 cm³

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Problem 8: Cone

Given: slant height = 9.5 mm, radius = 7.2 mm / 2? Wait — no: label says “7.2 mm” across the top — that’s diameter? Or radius?

Looking: “7.2 mm” is written across the circular base — likely diameter. Also, height is given as 7.7 mm? And slant height 9.5 mm.

But for cone volume, we need radius and perpendicular height.

If 7.2 mm is diameter → radius = 3.6 mm
Height = 7.7 mm (given vertically)

Check if Pythagoras holds: r=3.6, h=7.7 → slant should be √(3.6² + 7.7²) = √(12.96 + 59.29) = √72.25 = 8.5 — but labeled slant is 9.5 mm. Inconsistency?

Wait — perhaps 7.2 mm is radius? Let’s check:

If radius = 7.2 mm, height = 7.7 mm → slant = √(7.2² + 7.7²) = √(51.84 + 59.29) = √111.13 ≈ 10.54 — not 9.5.

Alternatively, maybe 7.2 mm is diameter, and 7.7 mm is NOT the height? But arrow points vertically — probably height.

Actually, in many diagrams, they give you what you need. Since both radius and height are needed, and 7.2 mm is shown across the circle, let’s assume it’s diameter, so radius = 3.6 mm. Height = 7.7 mm.

Even though slant doesn’t match exactly, we’ll use given r and h for volume.

V = (1/3) × π × (3.6)² × 7.7 = (1/3) × π × 12.96 × 7.7

First: 12.96 × 7.7 = 99.792
Then: 99.792 / 3 = 33.264
Times π ≈ 33.264 × 3.1416 ≈ 104.50 mm³

*(We ignore the slant height since it’s not needed for volume if we have r and h)*

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Problem 9: Sphere

Diameter = 10 cm → radius = 5 cm
V = (4/3) × π × 5³ = (4/3) × π × 125 = (500/3)π ≈ 166.6667 × 3.1416 ≈ 523.60 cm³

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Problem 10: Sphere

Radius = 2 cm
V = (4/3) × π × 8 = (32/3)π ≈ 10.6667 × 3.1416 ≈ 33.51 cm³

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Problem 11: Sphere

Diameter = 22 cm → radius = 11 cm
V = (4/3) × π × 1331 = (5324/3)π ≈ 1774.6667 × 3.1416 ≈ 5575.28 cm³

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Problem 12: Sphere

Diameter = 17 mm → radius = 8.5 mm
V = (4/3) × π × (8.5)³ = (4/3) × π × 614.125 ≈ (4/3) × 614.125 = 818.8333 × π ≈ 818.8333 × 3.1416 ≈ 2572.44 mm³

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Problem 13: Hemisphere

Diameter = 7 cm → radius = 3.5 cm
Volume of hemisphere = (2/3) × π × r³ = (2/3) × π × (3.5)³
3.5³ = 42.875
(2/3) × 42.875 = 28.5833
× π ≈ 28.5833 × 3.1416 ≈ 89.80 cm³

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Problem 14: Hemispherical bowl? Same as hemisphere.

Diameter = 8.4 cm → radius = 4.2 cm
V = (2/3) × π × (4.2)³
4.2³ = 74.088
(2/3) × 74.088 = 49.392
× π ≈ 49.392 × 3.1416 ≈ 155.18 cm³

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Problem 15: Square pyramid? Base is 6cm x 6cm? Diagram shows base edges 6cm and 6cm, height 15cm.

Yes — square base 6x6, height 15
V = (1/3) × 6 × 6 × 15 = (1/3) × 540 = 180.00 cm³

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Problem 16: Cone on top of hemisphere? Composite shape.

Cone part: radius = 4 mm, height = 12 mm
Hemisphere part: same radius 4 mm

Volume cone = (1/3) × π × 4² × 12 = (1/3) × π × 16 × 12 = (1/3) × 192π = 64π
Volume hemisphere = (2/3) × π × 4³ = (2/3) × π × 64 = (128/3)π ≈ 42.6667π

Total = 64π + 42.6667π = 106.6667π ≈ 106.6667 × 3.1416 ≈ 335.10 mm³

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Now compiling all answers:

1) 48.00
2) 360.00
3) 233.33
4) 14.00
5) 84.82
6) 16.49
7) 2513.27
8) 104.50
9) 523.60
10) 33.51
11) 5575.28
12) 2572.44
13) 89.80
14) 155.18
15) 180.00
16) 335.10

All rounded to 2 decimal places as instructed.

Final Answer:
1) 48.00 cm³
2) 360.00 cm³
3) 233.33 cm³
4) 14.00 cm³
5) 84.82 cm³
6) 16.49 cm³
7) 2513.27 cm³
8) 104.50 mm³
9) 523.60 cm³
10) 33.51 cm³
11) 5575.28 cm³
12) 2572.44 mm³
13) 89.80 cm³
14) 155.18 cm³
15) 180.00 cm³
16) 335.10 mm³