Geometry - Volume and Surface Area - Volume of Pyramids Cones and ... - Free Printable
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Step-by-step solution for: Geometry - Volume and Surface Area - Volume of Pyramids Cones and ...
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Show Answer Key & Explanations
Step-by-step solution for: Geometry - Volume and Surface Area - Volume of Pyramids Cones and ...
To solve the problems involving the volumes of pyramids, cones, and spheres, we will use the following formulas:
1. Volume of a Pyramid:
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
2. Volume of a Cone:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the base and \( h \) is the height.
3. Volume of a Sphere:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
4. Volume of a Hemisphere:
\[
V = \frac{2}{3} \pi r^3
\]
where \( r \) is the radius of the hemisphere.
Let's solve each problem step by step.
---
- Base Side: 4 cm
- Height: 9 cm
The base area of the square is:
\[
\text{Base Area} = 4 \times 4 = 16 \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times 16 \times 9 = \frac{1}{3} \times 144 = 48 \, \text{cm}^3
\]
Answer: \( V = 48.00 \, \text{cm}^3 \)
---
- Base Side: 6 cm
- Height: 15 cm
The base is an equilateral triangle with side length 6 cm. The area of an equilateral triangle is:
\[
\text{Base Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times 9\sqrt{3} \times 15 = \frac{1}{3} \times 135\sqrt{3} = 45\sqrt{3} \approx 77.94 \, \text{cm}^3
\]
Answer: \( V = 77.94 \, \text{cm}^3 \)
---
- Base Side: 7 cm
- Height: 10 cm
The base is an equilateral triangle with side length 7 cm. The area of an equilateral triangle is:
\[
\text{Base Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 7^2 = \frac{\sqrt{3}}{4} \times 49 = \frac{49\sqrt{3}}{4} \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times \frac{49\sqrt{3}}{4} \times 10 = \frac{1}{3} \times \frac{490\sqrt{3}}{4} = \frac{490\sqrt{3}}{12} \approx 71.48 \, \text{cm}^3
\]
Answer: \( V = 71.48 \, \text{cm}^3 \)
---
- Base Dimensions: 3 cm × 2 cm
- Height: 7 cm
The base area of the rectangle is:
\[
\text{Base Area} = 3 \times 2 = 6 \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times 6 \times 7 = \frac{1}{3} \times 42 = 14 \, \text{cm}^3
\]
Answer: \( V = 14.00 \, \text{cm}^3 \)
---
- Radius: 3 cm
- Height: 9 cm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (9) = \frac{1}{3} \pi \times 9 \times 9 = 27\pi \approx 84.82 \, \text{cm}^3
\]
Answer: \( V = 84.82 \, \text{cm}^3 \)
---
- Radius: 1.5 cm
- Height: 7 cm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1.5)^2 (7) = \frac{1}{3} \pi \times 2.25 \times 7 = \frac{1}{3} \pi \times 15.75 = 5.25\pi \approx 16.50 \, \text{cm}^3
\]
Answer: \( V = 16.50 \, \text{cm}^3 \)
---
- Radius: 10 cm (since diameter is 20 cm)
- Height: 24 cm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (10)^2 (24) = \frac{1}{3} \pi \times 100 \times 24 = 800\pi \approx 2513.27 \, \text{cm}^3
\]
Answer: \( V = 2513.27 \, \text{cm}^3 \)
---
- Radius: 3.6 mm (since diameter is 7.2 mm)
- Height: 7.7 mm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3.6)^2 (7.7) = \frac{1}{3} \pi \times 12.96 \times 7.7 = \frac{1}{3} \pi \times 100.032 \approx 104.72 \, \text{mm}^3
\]
Answer: \( V = 104.72 \, \text{mm}^3 \)
---
- Radius: 5 cm (since diameter is 10 cm)
Using the sphere volume formula:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (5)^3 = \frac{4}{3} \pi \times 125 = \frac{500}{3} \pi \approx 523.60 \, \text{cm}^3
\]
Answer: \( V = 523.60 \, \text{cm}^3 \)
---
- Radius: 1 cm (since diameter is 2 cm)
Using the sphere volume formula:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \approx 4.19 \, \text{cm}^3
\]
Answer: \( V = 4.19 \, \text{cm}^3 \)
---
- Radius: 11 cm (since diameter is 22 cm)
Using the sphere volume formula:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (11)^3 = \frac{4}{3} \pi \times 1331 = \frac{5324}{3} \pi \approx 5575.28 \, \text{cm}^3
\]
Answer: \( V = 5575.28 \, \text{cm}^3 \)
---
- Radius: 8.5 mm (since diameter is 17 mm)
Using the sphere volume formula:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (8.5)^3 = \frac{4}{3} \pi \times 614.125 = \frac{2456.5}{3} \pi \approx 2651.73 \, \text{mm}^3
\]
Answer: \( V = 2651.73 \, \text{mm}^3 \)
---
- Radius: 3.5 cm (since diameter is 7 cm)
Using the hemisphere volume formula:
\[
V = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3.5)^3 = \frac{2}{3} \pi \times 42.875 = \frac{85.75}{3} \pi \approx 89.81 \, \text{cm}^3
\]
Answer: \( V = 89.81 \, \text{cm}^3 \)
---
- Radius: 4.2 cm (since diameter is 8.4 cm)
Using the hemisphere volume formula:
\[
V = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (4.2)^3 = \frac{2}{3} \pi \times 74.088 = \frac{148.176}{3} \pi \approx 155.46 \, \text{cm}^3
\]
Answer: \( V = 155.46 \, \text{cm}^3 \)
---
- Base Side: 6 cm
- Height: 15 cm
The base is an equilateral triangle with side length 6 cm. The area of an equilateral triangle is:
\[
\text{Base Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times 9\sqrt{3} \times 15 = \frac{1}{3} \times 135\sqrt{3} = 45\sqrt{3} \approx 77.94 \, \text{cm}^3
\]
Answer: \( V = 77.94 \, \text{cm}^3 \)
---
- Radius: 2 mm
- Height: 4 mm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2)^2 (4) = \frac{1}{3} \pi \times 4 \times 4 = \frac{16}{3} \pi \approx 16.76 \, \text{mm}^3
\]
Answer: \( V = 16.76 \, \text{mm}^3 \)
---
\[
\boxed{
\begin{array}{ll}
1) & 48.00 \, \text{cm}^3 \\
2) & 77.94 \, \text{cm}^3 \\
3) & 71.48 \, \text{cm}^3 \\
4) & 14.00 \, \text{cm}^3 \\
5) & 84.82 \, \text{cm}^3 \\
6) & 16.50 \, \text{cm}^3 \\
7) & 2513.27 \, \text{cm}^3 \\
8) & 104.72 \, \text{mm}^3 \\
9) & 523.60 \, \text{cm}^3 \\
10) & 4.19 \, \text{cm}^3 \\
11) & 5575.28 \, \text{cm}^3 \\
12) & 2651.73 \, \text{mm}^3 \\
13) & 89.81 \, \text{cm}^3 \\
14) & 155.46 \, \text{cm}^3 \\
15) & 77.94 \, \text{cm}^3 \\
16) & 16.76 \, \text{mm}^3 \\
\end{array}
}
\]
1. Volume of a Pyramid:
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
2. Volume of a Cone:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the base and \( h \) is the height.
3. Volume of a Sphere:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
4. Volume of a Hemisphere:
\[
V = \frac{2}{3} \pi r^3
\]
where \( r \) is the radius of the hemisphere.
Let's solve each problem step by step.
---
1. Pyramid (Square Base)
- Base Side: 4 cm
- Height: 9 cm
The base area of the square is:
\[
\text{Base Area} = 4 \times 4 = 16 \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times 16 \times 9 = \frac{1}{3} \times 144 = 48 \, \text{cm}^3
\]
Answer: \( V = 48.00 \, \text{cm}^3 \)
---
2. Pyramid (Triangular Base)
- Base Side: 6 cm
- Height: 15 cm
The base is an equilateral triangle with side length 6 cm. The area of an equilateral triangle is:
\[
\text{Base Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times 9\sqrt{3} \times 15 = \frac{1}{3} \times 135\sqrt{3} = 45\sqrt{3} \approx 77.94 \, \text{cm}^3
\]
Answer: \( V = 77.94 \, \text{cm}^3 \)
---
3. Pyramid (Triangular Base)
- Base Side: 7 cm
- Height: 10 cm
The base is an equilateral triangle with side length 7 cm. The area of an equilateral triangle is:
\[
\text{Base Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 7^2 = \frac{\sqrt{3}}{4} \times 49 = \frac{49\sqrt{3}}{4} \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times \frac{49\sqrt{3}}{4} \times 10 = \frac{1}{3} \times \frac{490\sqrt{3}}{4} = \frac{490\sqrt{3}}{12} \approx 71.48 \, \text{cm}^3
\]
Answer: \( V = 71.48 \, \text{cm}^3 \)
---
4. Pyramid (Rectangular Base)
- Base Dimensions: 3 cm × 2 cm
- Height: 7 cm
The base area of the rectangle is:
\[
\text{Base Area} = 3 \times 2 = 6 \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times 6 \times 7 = \frac{1}{3} \times 42 = 14 \, \text{cm}^3
\]
Answer: \( V = 14.00 \, \text{cm}^3 \)
---
5. Cone
- Radius: 3 cm
- Height: 9 cm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (9) = \frac{1}{3} \pi \times 9 \times 9 = 27\pi \approx 84.82 \, \text{cm}^3
\]
Answer: \( V = 84.82 \, \text{cm}^3 \)
---
6. Cone
- Radius: 1.5 cm
- Height: 7 cm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1.5)^2 (7) = \frac{1}{3} \pi \times 2.25 \times 7 = \frac{1}{3} \pi \times 15.75 = 5.25\pi \approx 16.50 \, \text{cm}^3
\]
Answer: \( V = 16.50 \, \text{cm}^3 \)
---
7. Cone
- Radius: 10 cm (since diameter is 20 cm)
- Height: 24 cm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (10)^2 (24) = \frac{1}{3} \pi \times 100 \times 24 = 800\pi \approx 2513.27 \, \text{cm}^3
\]
Answer: \( V = 2513.27 \, \text{cm}^3 \)
---
8. Cone
- Radius: 3.6 mm (since diameter is 7.2 mm)
- Height: 7.7 mm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3.6)^2 (7.7) = \frac{1}{3} \pi \times 12.96 \times 7.7 = \frac{1}{3} \pi \times 100.032 \approx 104.72 \, \text{mm}^3
\]
Answer: \( V = 104.72 \, \text{mm}^3 \)
---
9. Sphere
- Radius: 5 cm (since diameter is 10 cm)
Using the sphere volume formula:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (5)^3 = \frac{4}{3} \pi \times 125 = \frac{500}{3} \pi \approx 523.60 \, \text{cm}^3
\]
Answer: \( V = 523.60 \, \text{cm}^3 \)
---
10. Sphere
- Radius: 1 cm (since diameter is 2 cm)
Using the sphere volume formula:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \approx 4.19 \, \text{cm}^3
\]
Answer: \( V = 4.19 \, \text{cm}^3 \)
---
11. Sphere
- Radius: 11 cm (since diameter is 22 cm)
Using the sphere volume formula:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (11)^3 = \frac{4}{3} \pi \times 1331 = \frac{5324}{3} \pi \approx 5575.28 \, \text{cm}^3
\]
Answer: \( V = 5575.28 \, \text{cm}^3 \)
---
12. Sphere
- Radius: 8.5 mm (since diameter is 17 mm)
Using the sphere volume formula:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (8.5)^3 = \frac{4}{3} \pi \times 614.125 = \frac{2456.5}{3} \pi \approx 2651.73 \, \text{mm}^3
\]
Answer: \( V = 2651.73 \, \text{mm}^3 \)
---
13. Hemisphere
- Radius: 3.5 cm (since diameter is 7 cm)
Using the hemisphere volume formula:
\[
V = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3.5)^3 = \frac{2}{3} \pi \times 42.875 = \frac{85.75}{3} \pi \approx 89.81 \, \text{cm}^3
\]
Answer: \( V = 89.81 \, \text{cm}^3 \)
---
14. Hemisphere
- Radius: 4.2 cm (since diameter is 8.4 cm)
Using the hemisphere volume formula:
\[
V = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (4.2)^3 = \frac{2}{3} \pi \times 74.088 = \frac{148.176}{3} \pi \approx 155.46 \, \text{cm}^3
\]
Answer: \( V = 155.46 \, \text{cm}^3 \)
---
15. Pyramid (Triangular Base)
- Base Side: 6 cm
- Height: 15 cm
The base is an equilateral triangle with side length 6 cm. The area of an equilateral triangle is:
\[
\text{Base Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \, \text{cm}^2
\]
Using the pyramid volume formula:
\[
V = \frac{1}{3} \times 9\sqrt{3} \times 15 = \frac{1}{3} \times 135\sqrt{3} = 45\sqrt{3} \approx 77.94 \, \text{cm}^3
\]
Answer: \( V = 77.94 \, \text{cm}^3 \)
---
16. Cone
- Radius: 2 mm
- Height: 4 mm
Using the cone volume formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2)^2 (4) = \frac{1}{3} \pi \times 4 \times 4 = \frac{16}{3} \pi \approx 16.76 \, \text{mm}^3
\]
Answer: \( V = 16.76 \, \text{mm}^3 \)
---
Final Answers
\[
\boxed{
\begin{array}{ll}
1) & 48.00 \, \text{cm}^3 \\
2) & 77.94 \, \text{cm}^3 \\
3) & 71.48 \, \text{cm}^3 \\
4) & 14.00 \, \text{cm}^3 \\
5) & 84.82 \, \text{cm}^3 \\
6) & 16.50 \, \text{cm}^3 \\
7) & 2513.27 \, \text{cm}^3 \\
8) & 104.72 \, \text{mm}^3 \\
9) & 523.60 \, \text{cm}^3 \\
10) & 4.19 \, \text{cm}^3 \\
11) & 5575.28 \, \text{cm}^3 \\
12) & 2651.73 \, \text{mm}^3 \\
13) & 89.81 \, \text{cm}^3 \\
14) & 155.46 \, \text{cm}^3 \\
15) & 77.94 \, \text{cm}^3 \\
16) & 16.76 \, \text{mm}^3 \\
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of volume of pyramids and cones worksheet answers.
