Worksheet for calculating surface area and volume of geometric shapes.

To solve the problem of calculating the surface area and volume of each geometric shape, we will go through each shape step by step. Let's analyze each one:

---

1. Cube

#### Given:
- Side length \( a = 6 \) cm

#### Surface Area:
The surface area \( A \) of a cube is given by:
\[
A = 6a^2
\]
Substitute \( a = 6 \):
\[
A = 6 \times 6^2 = 6 \times 36 = 216 \, \text{cm}^2
\]

#### Volume:
The volume \( V \) of a cube is given by:
\[
V = a^3
\]
Substitute \( a = 6 \):
\[
V = 6^3 = 216 \, \text{cm}^3
\]

Results for Cube:
\[
\text{Surface Area} = 216 \, \text{cm}^2, \quad \text{Volume} = 216 \, \text{cm}^3
\]

---

2. Sphere

#### Given:
- Diameter \( d = 8 \) cm
- Radius \( r = \frac{d}{2} = \frac{8}{2} = 4 \) cm

#### Surface Area:
The surface area \( A \) of a sphere is given by:
\[
A = 4\pi r^2
\]
Substitute \( r = 4 \):
\[
A = 4\pi (4)^2 = 4\pi \times 16 = 64\pi \, \text{cm}^2
\]

#### Volume:
The volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3}\pi r^3
\]
Substitute \( r = 4 \):
\[
V = \frac{4}{3}\pi (4)^3 = \frac{4}{3}\pi \times 64 = \frac{256}{3}\pi \, \text{cm}^3
\]

Results for Sphere:
\[
\text{Surface Area} = 64\pi \, \text{cm}^2, \quad \text{Volume} = \frac{256}{3}\pi \, \text{cm}^3
\]

---

3. Cylinder

#### Given:
- Radius \( r = 3 \) cm
- Height \( h = 7 \) cm

#### Surface Area:
The total surface area \( A \) of a cylinder is given by:
\[
A = 2\pi r(h + r)
\]
Substitute \( r = 3 \) and \( h = 7 \):
\[
A = 2\pi (3)(7 + 3) = 2\pi (3)(10) = 60\pi \, \text{cm}^2
\]

#### Volume:
The volume \( V \) of a cylinder is given by:
\[
V = \pi r^2 h
\]
Substitute \( r = 3 \) and \( h = 7 \):
\[
V = \pi (3)^2 (7) = \pi \times 9 \times 7 = 63\pi \, \text{cm}^3
\]

Results for Cylinder:
\[
\text{Surface Area} = 60\pi \, \text{cm}^2, \quad \text{Volume} = 63\pi \, \text{cm}^3
\]

---

4. Triangular Prism

#### Given:
- Base triangle sides: \( 5 \) cm, \( 5 \) cm, \( 6 \) cm
- Height of the prism \( h = 10 \) cm

#### Surface Area:
The surface area \( A \) of a triangular prism consists of:
1. The area of the two triangular bases.
2. The area of the three rectangular lateral faces.

##### Step 1: Area of one triangular base
The base is an isosceles triangle with sides \( 5 \), \( 5 \), and \( 6 \). We use Heron's formula to find the area.

- Semi-perimeter \( s \):
\[
s = \frac{5 + 5 + 6}{2} = 8 \, \text{cm}
\]

- Area \( A_{\text{triangle}} \):
\[
A_{\text{triangle}} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{8(8-5)(8-5)(8-6)} = \sqrt{8 \times 3 \times 3 \times 2} = \sqrt{144} = 12 \, \text{cm}^2
\]

##### Step 2: Area of the three rectangular lateral faces
- Two rectangles with dimensions \( 5 \times 10 \):
\[
\text{Area of each rectangle} = 5 \times 10 = 50 \, \text{cm}^2
\]
\[
\text{Total area of two rectangles} = 2 \times 50 = 100 \, \text{cm}^2
\]

- One rectangle with dimensions \( 6 \times 10 \):
\[
\text{Area of the rectangle} = 6 \times 10 = 60 \, \text{cm}^2
\]

##### Step 3: Total surface area
\[
A = 2 \times \text{Area of one triangular base} + \text{Area of the three rectangles}
\]
\[
A = 2 \times 12 + 100 + 60 = 24 + 100 + 60 = 184 \, \text{cm}^2
\]

#### Volume:
The volume \( V \) of a triangular prism is given by:
\[
V = \text{Base Area} \times \text{Height of the prism}
\]
Substitute the base area \( 12 \, \text{cm}^2 \) and height \( 10 \, \text{cm} \):
\[
V = 12 \times 10 = 120 \, \text{cm}^3
\]

Results for Triangular Prism:
\[
\text{Surface Area} = 184 \, \text{cm}^2, \quad \text{Volume} = 120 \, \text{cm}^3
\]

---

5. Rectangular Prism

#### Given:
- Length \( l = 8 \) cm
- Width \( w = 4 \) cm
- Height \( h = 5 \) cm

#### Surface Area:
The surface area \( A \) of a rectangular prism is given by:
\[
A = 2(lw + lh + wh)
\]
Substitute \( l = 8 \), \( w = 4 \), and \( h = 5 \):
\[
A = 2(8 \times 4 + 8 \times 5 + 4 \times 5) = 2(32 + 40 + 20) = 2 \times 92 = 184 \, \text{cm}^2
\]

#### Volume:
The volume \( V \) of a rectangular prism is given by:
\[
V = lwh
\]
Substitute \( l = 8 \), \( w = 4 \), and \( h = 5 \):
\[
V = 8 \times 4 \times 5 = 160 \, \text{cm}^3
\]

Results for Rectangular Prism:
\[
\text{Surface Area} = 184 \, \text{cm}^2, \quad \text{Volume} = 160 \, \text{cm}^3
\]

---

6. Square Pyramid

#### Given:
- Base side \( s = 6 \) cm
- Slant height \( l = 5 \) cm
- Height of the pyramid \( h = 4 \) cm

#### Surface Area:
The surface area \( A \) of a square pyramid consists of:
1. The area of the square base.
2. The area of the four triangular lateral faces.

##### Step 1: Area of the square base
\[
\text{Base Area} = s^2 = 6^2 = 36 \, \text{cm}^2
\]

##### Step 2: Area of one triangular face
Each triangular face is an isosceles triangle with base \( s = 6 \) cm and slant height \( l = 5 \) cm. The area of one triangular face is:
\[
\text{Area of one triangle} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 6 \times 5 = 15 \, \text{cm}^2
\]

##### Step 3: Total area of the four triangular faces
\[
\text{Total area of four triangles} = 4 \times 15 = 60 \, \text{cm}^2
\]

##### Step 4: Total surface area
\[
A = \text{Base Area} + \text{Total area of four triangles}
\]
\[
A = 36 + 60 = 96 \, \text{cm}^2
\]

#### Volume:
The volume \( V \) of a pyramid is given by:
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
Substitute the base area \( 36 \, \text{cm}^2 \) and height \( 4 \, \text{cm} \):
\[
V = \frac{1}{3} \times 36 \times 4 = \frac{1}{3} \times 144 = 48 \, \text{cm}^3
\]

Results for Square Pyramid:
\[
\text{Surface Area} = 96 \, \text{cm}^2, \quad \text{Volume} = 48 \, \text{cm}^3
\]

---

Final Answers:

1. Cube:
\[
\boxed{216 \, \text{cm}^2, 216 \, \text{cm}^3}
\]

2. Sphere:
\[
\boxed{64\pi \, \text{cm}^2, \frac{256}{3}\pi \, \text{cm}^3}
\]

3. Cylinder:
\[
\boxed{60\pi \, \text{cm}^2, 63\pi \, \text{cm}^3}
\]

4. Triangular Prism:
\[
\boxed{184 \, \text{cm}^2, 120 \, \text{cm}^3}
\]

5. Rectangular Prism:
\[
\boxed{184 \, \text{cm}^2, 160 \, \text{cm}^3}
\]

6. Square Pyramid:
\[
\boxed{96 \, \text{cm}^2, 48 \, \text{cm}^3}
\]