Math worksheet for calculating volume and surface area of cones and spheres.

Problem: Calculate the volume, surface area, or radius of the cones and spheres as required.

#### Section A: Spheres

We will use the following formulas for spheres:
- Volume of a sphere: \( V = \frac{4}{3} \pi r^3 \)
- Surface area of a sphere: \( A = 4 \pi r^2 \)

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1) Sphere with radius \( r = 2 \, \text{cm} \)

#### Volume:
\[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \]
\[ V \approx \frac{32}{3} \times 3.1416 \approx 33.51 \, \text{cm}^3 \]

#### Surface Area:
\[ A = 4 \pi r^2 = 4 \pi (2)^2 = 4 \pi (4) = 16 \pi \]
\[ A \approx 16 \times 3.1416 \approx 50.27 \, \text{cm}^2 \]

Answers:
- Volume = \( 33.51 \, \text{cm}^3 \)
- Surface area = \( 50.27 \, \text{cm}^2 \)

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2) Sphere with diameter \( d = 14 \, \text{mm} \)

First, find the radius:
\[ r = \frac{d}{2} = \frac{14}{2} = 7 \, \text{mm} \]

#### Volume:
\[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (7)^3 = \frac{4}{3} \pi (343) = \frac{1372}{3} \pi \]
\[ V \approx \frac{1372}{3} \times 3.1416 \approx 1436.76 \, \text{mm}^3 \]

#### Surface Area:
\[ A = 4 \pi r^2 = 4 \pi (7)^2 = 4 \pi (49) = 196 \pi \]
\[ A \approx 196 \times 3.1416 \approx 615.75 \, \text{mm}^2 \]

Answers:
- Volume = \( 1436.76 \, \text{mm}^3 \)
- Surface area = \( 615.75 \, \text{mm}^2 \)

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3) Sphere with volume \( V = 180 \, \text{cm}^3 \)

Use the volume formula to solve for the radius:
\[ V = \frac{4}{3} \pi r^3 \]
\[ 180 = \frac{4}{3} \pi r^3 \]
\[ r^3 = \frac{180 \times 3}{4 \pi} = \frac{540}{4 \pi} = \frac{135}{\pi} \]
\[ r^3 \approx \frac{135}{3.1416} \approx 42.97 \]
\[ r \approx \sqrt[3]{42.97} \approx 3.50 \, \text{cm} \]

Answer:
- Radius = \( 3.50 \, \text{cm} \)

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4) Sphere with surface area \( A = 25 \, \text{mm}^2 \)

Use the surface area formula to solve for the radius:
\[ A = 4 \pi r^2 \]
\[ 25 = 4 \pi r^2 \]
\[ r^2 = \frac{25}{4 \pi} \]
\[ r^2 \approx \frac{25}{4 \times 3.1416} \approx \frac{25}{12.5664} \approx 1.99 \]
\[ r \approx \sqrt{1.99} \approx 1.41 \, \text{mm} \]

Answer:
- Radius = \( 1.41 \, \text{mm} \)

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#### Section B: Cones

We will use the following formulas for cones:
- Volume of a cone: \( V = \frac{1}{3} \pi r^2 h \)
- Curved surface area of a cone: \( A_{\text{curved}} = \pi r l \) (where \( l \) is the slant height)
- Total surface area of a cone: \( A_{\text{total}} = \pi r l + \pi r^2 \)

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5) Cone with \( r = 5 \, \text{cm} \), \( h = 12 \, \text{cm} \), and \( l = 13 \, \text{cm} \)

#### Curved Surface Area:
\[ A_{\text{curved}} = \pi r l = \pi (5)(13) = 65 \pi \]
\[ A_{\text{curved}} \approx 65 \times 3.1416 \approx 204.20 \, \text{cm}^2 \]

#### Total Surface Area:
\[ A_{\text{total}} = \pi r l + \pi r^2 = 65 \pi + \pi (5)^2 = 65 \pi + 25 \pi = 90 \pi \]
\[ A_{\text{total}} \approx 90 \times 3.1416 \approx 282.74 \, \text{cm}^2 \]

Answers:
- Curved surface area = \( 204.20 \, \text{cm}^2 \)
- Total surface area = \( 282.74 \, \text{cm}^2 \)

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6) Cone with \( r = 7 \, \text{cm} \) and \( l = 25 \, \text{cm} \)

First, find the height \( h \) using the Pythagorean theorem:
\[ l^2 = r^2 + h^2 \]
\[ 25^2 = 7^2 + h^2 \]
\[ 625 = 49 + h^2 \]
\[ h^2 = 625 - 49 = 576 \]
\[ h = \sqrt{576} = 24 \, \text{cm} \]

#### Volume:
\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (7)^2 (24) = \frac{1}{3} \pi (49)(24) = \frac{1}{3} \pi (1176) = 392 \pi \]
\[ V \approx 392 \times 3.1416 \approx 1231.50 \, \text{cm}^3 \]

#### Total Surface Area:
\[ A_{\text{total}} = \pi r l + \pi r^2 = \pi (7)(25) + \pi (7)^2 = 175 \pi + 49 \pi = 224 \pi \]
\[ A_{\text{total}} \approx 224 \times 3.1416 \approx 703.72 \, \text{cm}^2 \]

Answers:
- Volume = \( 1231.50 \, \text{cm}^3 \)
- Total surface area = \( 703.72 \, \text{cm}^2 \)

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7) Cone with \( r = 7.2 \, \text{cm} \) and \( h = 9.6 \, \text{cm} \)

First, find the slant height \( l \) using the Pythagorean theorem:
\[ l^2 = r^2 + h^2 \]
\[ l^2 = (7.2)^2 + (9.6)^2 = 51.84 + 92.16 = 144 \]
\[ l = \sqrt{144} = 12 \, \text{cm} \]

#### Volume:
\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (7.2)^2 (9.6) = \frac{1}{3} \pi (51.84)(9.6) = \frac{1}{3} \pi (497.664) = 165.888 \pi \]
\[ V \approx 165.888 \times 3.1416 \approx 521.23 \, \text{cm}^3 \]

#### Total Surface Area:
\[ A_{\text{total}} = \pi r l + \pi r^2 = \pi (7.2)(12) + \pi (7.2)^2 = 86.4 \pi + 51.84 \pi = 138.24 \pi \]
\[ A_{\text{total}} \approx 138.24 \times 3.1416 \approx 434.01 \, \text{cm}^2 \]

Answers:
- Volume = \( 521.23 \, \text{cm}^3 \)
- Total surface area = \( 434.01 \, \text{cm}^2 \)

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8) Cone on top of a hemisphere with \( r = 7.5 \, \text{cm} \), \( h_{\text{cone}} = 23 \, \text{cm} \), and \( h_{\text{hemisphere}} = 15 \, \text{cm} \)

#### Volume of the cone:
\[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}} = \frac{1}{3} \pi (7.5)^2 (23) = \frac{1}{3} \pi (56.25)(23) = \frac{1}{3} \pi (1293.75) = 431.25 \pi \]
\[ V_{\text{cone}} \approx 431.25 \times 3.1416 \approx 1354.47 \, \text{cm}^3 \]

#### Volume of the hemisphere:
\[ V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (7.5)^3 = \frac{2}{3} \pi (421.875) = 281.25 \pi \]
\[ V_{\text{hemisphere}} \approx 281.25 \times 3.1416 \approx 883.57 \, \text{cm}^3 \]

#### Total Volume:
\[ V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} = 431.25 \pi + 281.25 \pi = 712.5 \pi \]
\[ V_{\text{total}} \approx 712.5 \times 3.1416 \approx 2238.04 \, \text{cm}^3 \]

#### Surface Area of the cone (excluding the base):
\[ A_{\text{cone}} = \pi r l \]
First, find the slant height \( l \) of the cone:
\[ l = \sqrt{r^2 + h_{\text{cone}}^2} = \sqrt{(7.5)^2 + (23)^2} = \sqrt{56.25 + 529} = \sqrt{585.25} \approx 24.2 \, \text{cm} \]
\[ A_{\text{cone}} = \pi (7.5)(24.2) \approx 181.5 \pi \]
\[ A_{\text{cone}} \approx 181.5 \times 3.1416 \approx 569.97 \, \text{cm}^2 \]

#### Surface Area of the hemisphere (excluding the base):
\[ A_{\text{hemisphere}} = 2 \pi r^2 = 2 \pi (7.5)^2 = 2 \pi (56.25) = 112.5 \pi \]
\[ A_{\text{hemisphere}} \approx 112.5 \times 3.1416 \approx 353.43 \, \text{cm}^2 \]

#### Total Surface Area:
\[ A_{\text{total}} = A_{\text{cone}} + A_{\text{hemisphere}} = 181.5 \pi + 112.5 \pi = 294 \pi \]
\[ A_{\text{total}} \approx 294 \times 3.1416 \approx 923.40 \, \text{cm}^2 \]

Answers:
- Volume = \( 2238.04 \, \text{cm}^3 \)
- Total surface area = \( 923.40 \, \text{cm}^2 \)

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Final Answers:

1. Volume = \( 33.51 \, \text{cm}^3 \), Surface area = \( 50.27 \, \text{cm}^2 \)
2. Volume = \( 1436.76 \, \text{mm}^3 \), Surface area = \( 615.75 \, \text{mm}^2 \)
3. Radius = \( 3.50 \, \text{cm} \)
4. Radius = \( 1.41 \, \text{mm} \)
5. Curved surface area = \( 204.20 \, \text{cm}^2 \), Total surface area = \( 282.74 \, \text{cm}^2 \)
6. Volume = \( 1231.50 \, \text{cm}^3 \), Total surface area = \( 703.72 \, \text{cm}^2 \)
7. Volume = \( 521.23 \, \text{cm}^3 \), Total surface area = \( 434.01 \, \text{cm}^2 \)
8. Volume = \( 2238.04 \, \text{cm}^3 \), Total surface area = \( 923.40 \, \text{cm}^2 \)

\[
\boxed{
\begin{array}{l}
1. \text{Volume} = 33.51 \, \text{cm}^3, \text{Surface area} = 50.27 \, \text{cm}^2 \\
2. \text{Volume} = 1436.76 \, \text{mm}^3, \text{Surface area} = 615.75 \, \text{mm}^2 \\
3. \text{Radius} = 3.50 \, \text{cm} \\
4. \text{Radius} = 1.41 \, \text{mm} \\
5. \text{Curved surface area} = 204.20 \, \text{cm}^2, \text{Total surface area} = 282.74 \, \text{cm}^2 \\
6. \text{Volume} = 1231.50 \, \text{cm}^3, \text{Total surface area} = 703.72 \, \text{cm}^2 \\
7. \text{Volume} = 521.23 \, \text{cm}^3, \text{Total surface area} = 434.01 \, \text{cm}^2 \\
8. \text{Volume} = 2238.04 \, \text{cm}^3, \text{Total surface area} = 923.40 \, \text{cm}^2
\end{array}
}
\]