Math worksheet for calculating the volume of pyramids, cones, and spheres, including practice problems with diagrams and space for answers.

Let’s solve each problem one by one. We’ll use the correct volume formulas and round to 2 decimal places when needed.

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Formulas we need:

- Pyramid (square or rectangular base):
Volume = (1/3) × Base Area × Height
For square base: Base Area = side × side
For rectangular base: Base Area = length × width

- Cone:
Volume = (1/3) × π × r² × h

- Sphere:
Volume = (4/3) × π × r³

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

We’ll use π ≈ 3.1416 for calculations.

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Problem 1: Square Pyramid

Base = 4 cm × 4 cm, Height = 9 cm
Volume = (1/3) × (4×4) × 9 = (1/3) × 16 × 9 = 48 cm³
→ 48.00

Problem 2: Rectangular Pyramid

Base = 12 cm × 6 cm, Height = 15 cm
Volume = (1/3) × (12×6) × 15 = (1/3) × 72 × 15 = 360 cm³
→ 360.00

Problem 3: Rectangular Pyramid

Base = 5 cm × 7 cm, Height = 20 cm
Volume = (1/3) × (5×7) × 20 = (1/3) × 35 × 20 ≈ 233.333... → 233.33

Problem 4: Composite Shape — Pyramid on top of Cube? Wait — actually, it's a pyramid with a rectangular base inside a box? Let’s look again.

Actually, looking at diagram 4: It shows a pyramid with a rectangular base (3 cm × 3 cm) and height 7 cm, but there’s also a cube below? No — wait, the dashed lines suggest the full height is 7 cm, and the base is 3 cm × 3 cm. The “2 cm” might be misleading — actually, re-examining: the figure is a pyramid with base 3 cm × 3 cm and total height 7 cm. The “2 cm” is probably part of the internal structure, but since the apex is 7 cm above the base, we use height = 7 cm.

So: Volume = (1/3) × (3×3) × 7 = (1/3) × 9 × 7 = 21 cm³
→ 21.00

*(Note: If the 2 cm was meant to be something else, but based on standard interpretation, height from base to apex is 7 cm.)*

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Problem 5: Cone

Radius = 3 cm, Height = 9 cm
Volume = (1/3) × π × 3² × 9 = (1/3) × π × 9 × 9 = 27π ≈ 27 × 3.1416 ≈ 84.8232 → 84.82

Problem 6: Cone

Radius = 1.5 cm, Height = 7 cm
Volume = (1/3) × π × (1.5)² × 7 = (1/3) × π × 2.25 × 7 = (1/3) × 15.75 × π = 5.25π ≈ 5.25 × 3.1416 ≈ 16.4934 → 16.49

Problem 7: Cone

Diameter = 20 cm → Radius = 10 cm, Height = 24 cm
Volume = (1/3) × π × 10² × 24 = (1/3) × π × 100 × 24 = 800π ≈ 800 × 3.1416 ≈ 2513.28 → 2513.28

Problem 8: Cone (inverted)

Given: Slant height? Wait — no, they give vertical height? Actually, diagram shows:

It says 7.7 mm (vertical height?), 8.5 mm (slant?), and diameter 7.2 mm → radius = 3.6 mm.

But in cone volume, we need perpendicular height. The 7.7 mm is labeled as vertical (from apex to center of base), so that’s the height.

So: r = 3.6 mm, h = 7.7 mm
Volume = (1/3) × π × (3.6)² × 7.7
First, 3.6² = 12.96
Then, 12.96 × 7.7 = 99.792
Then, (1/3) × 99.792 = 33.264
Then, 33.264 × π ≈ 33.264 × 3.1416 ≈ 104.50 → 104.50

*(Check: 33.264 * 3.1416 = let’s compute: 33 * 3.1416 = 103.6728, 0.264*3.1416≈0.829, total ≈104.50 — yes)*

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Problem 9: Sphere

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

Problem 10: Sphere

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

Problem 11: Sphere

Diameter = 22 cm → Radius = 11 cm
Volume = (4/3) × π × 11³ = (4/3) × π × 1331 = (5324/3)π ≈ 1774.6667 × 3.1416 ≈ 5575.28 → 5575.28

*(Calculation: 1774.6667 * 3.1416 ≈ let’s do 1774.6667 * 3 = 5324, 1774.6667 * 0.1416 ≈ 251.28, total ≈ 5575.28 — correct)*

Problem 12: Sphere

Diameter = 17 mm → Radius = 8.5 mm
Volume = (4/3) × π × (8.5)³
8.5³ = 8.5 × 8.5 = 72.25; 72.25 × 8.5 = 614.125
Then, (4/3) × 614.125 = (2456.5)/3 ≈ 818.8333
Then, 818.8333 × π ≈ 818.8333 × 3.1416 ≈ 2572.47 → 2572.47

*(Check: 818.8333 * 3.1416 ≈ 800*3.1416=2513.28, 18.8333*3.1416≈59.19, total≈2572.47 — good)*

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Problem 13: Hemisphere

Diameter = 7 cm → Radius = 3.5 cm
Volume = (2/3) × π × (3.5)³
3.5³ = 42.875
(2/3) × 42.875 = 85.75 / 3 ≈ 28.5833
28.5833 × π ≈ 28.5833 × 3.1416 ≈ 89.80 → 89.80

Problem 14: Hemisphere

Diameter = 8.4 cm → Radius = 4.2 cm
Volume = (2/3) × π × (4.2)³
4.2³ = 4.2 × 4.2 = 17.64; 17.64 × 4.2 = 74.088
(2/3) × 74.088 = 148.176 / 3 = 49.392
49.392 × π ≈ 49.392 × 3.1416 ≈ 155.17 → 155.17

Problem 15: Square Pyramid

Base = 6 cm × 6 cm, Height = 15 cm
Volume = (1/3) × (6×6) × 15 = (1/3) × 36 × 15 = 180 cm³ → 180.00

Problem 16: Composite — Cone on top of Hemisphere?

Wait — diagram 16: It shows a cone sitting on a hemisphere. But the question is to find the volume of the whole shape? Or just the cone? Looking at the label: it says “Volume =” under the entire figure. Also, dimensions: cone height = 12 mm, hemisphere radius = 4 mm (since diameter of base is 8 mm? Wait — it says “4 mm” pointing to radius of hemisphere, and cone has same base radius.

Actually, the figure shows a cone with height 12 mm and base radius 4 mm, attached to a hemisphere of radius 4 mm. So total volume = volume of cone + volume of hemisphere.

Cone: V_cone = (1/3)πr²h = (1/3)π(16)(12) = 64π
Hemisphere: V_hemi = (2/3)πr³ = (2/3)π(64) = 128/3 π ≈ 42.6667π
Total = 64π + 42.6667π = 106.6667π ≈ 106.6667 × 3.1416 ≈ 335.10 → 335.10

*(Check: 106.6667 * 3.1416 = 100*3.1416=314.16, 6.6667*3.1416≈20.94, total≈335.10 — correct)*

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Now, compiling all answers:

1) 48.00
2) 360.00
3) 233.33
4) 21.00
5) 84.82
6) 16.49
7) 2513.28
8) 104.50
9) 523.60
10) 33.51
11) 5575.28
12) 2572.47
13) 89.80
14) 155.17
15) 180.00
16) 335.10

Final Answer:
1) 48.00
2) 360.00
3) 233.33
4) 21.00
5) 84.82
6) 16.49
7) 2513.28
8) 104.50
9) 523.60
10) 33.51
11) 5575.28
12) 2572.47
13) 89.80
14) 155.17
15) 180.00
16) 335.10