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Steps to Solve

1. Identify the correct name for an 11-sided polygon

An 11-sided polygon is called a hendecagon.

2. Determine which statement is true about geometric shapes

Evaluate each statement: - A cube has 8 vertices, 12 edges, and 6 faces, not 8 faces. - A prism is a polyhedron with two congruent faces. This statement is true, these faces are the bases. - Cylinder is a solid figure with two parallel bases. This is true, but not a polyhedron.

3. Calculate the total surface area of a cube

The surface area of one face of the cube is $side^2 = 5^2 = 25 \text{ cm}^2$. A cube has 6 faces, so the total surface area is $6 \times 25 = 150 \text{ cm}^2$.

4. Determine the number of lines of symmetry in a regular pentagon

A regular pentagon has 5 lines of symmetry. Each line of symmetry goes from a vertex to the midpoint of the opposite side.

5. Calculate the amount of wallpaper needed

Jamie needs to cover the lateral surface area of the room. The room has two walls with dimensions 5m x 3m and two walls with dimensions 4m x 3m. The total area is $2 \times (5 \times 3) + 2 \times (4 \times 3) = 2 \times 15 + 2 \times 12 = 30 + 24 = 54 \text{ m}^2$.

6. Determine the number of edges in a triangular prism

A triangular prism has 2 triangular faces and 3 rectangular faces. It has 6 vertices and 9 edges.

7. Calculate the volume of a rectangular prism

The volume of a rectangular prism is $length \times width \times height = 10 \times 5 \times 4 = 200 \text{ cm}^3$.

8. Calculate the area of a circle

The area of a circle is $\pi r^2$, where $r$ is the radius. In this case, the radius is 7 cm, so the area is $\pi \times 7^2 = 49\pi \approx 153.94 \text{ cm}^2$. The closest answer is $154 \text{ cm}^2$.

9. Determine how many faces meet at each vertex of a cube

Three faces meet at each vertex of a cube.

10. Calculate the surface area of a dice

A dice is a cube. The surface area of one face is $2^2 = 4 \text{ cm}^2$. A cube has 6 faces, so the total surface area is $6 \times 4 = 24 \text{ cm}^2$.

11. Problem 1a: area of the triangular base

The sides of the triangle are 6 cm, 8 cm, and 10 cm. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$, this is a right triangle. Thus the area is $(1/2) \times base \times height = (1/2) \times 6 \times 8 = 24 \text{ cm}^2$.

12. Problem 1b: perimeter of the triangular base

The perimeter is the sum of the sides: $6 + 8 + 10 = 24 \text{ cm}$.

13. Problem 1c: lateral area of the prism

The lateral area is the perimeter of the base times the height of the prism: $24 \times 12 = 288 \text{ cm}^2$.

14. Problem 1d: total surface area of the prism

The total surface area is the lateral area plus twice the area of the base: $288 + 2 \times 24 = 288 + 48 = 336 \text{ cm}^2$.

15. Problem 1e: volume of the prism

The volume is the area of the base times the height of the prism: $24 \times 12 = 288 \text{ cm}^3$.

16. Problem 2a: slant height of the cone

The slant height $l$ is given by $l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}$.

17. Problem 2b: lateral area of the cone

The lateral area is $\pi r l = \pi \times 5 \times 13 = 65\pi \approx 204.20 \text{ cm}^2$.

18. Problem 2c: total surface area of the cone

The total surface area is the lateral area plus the area of the base: $\pi r l + \pi r^2 = 65\pi + \pi (5^2) = 65\pi + 25\pi = 90\pi \approx 282.74 \text{ cm}^2$.

19. Problem 2d: volume of the cone

The volume is $(1/3) \pi r^2 h = (1/3) \pi (5^2) (12) = (1/3) \pi (25) (12) = 100\pi \approx 314.16 \text{ cm}^3$.

20. Problem 3a: lateral area of the pyramid

The lateral area of a pyramid is given by $(1/2) \times perimeter \times slant\ height$. The perimeter is $4 \times 8 = 32 \text{ cm}$. The slant height $l$ is given by $l = \sqrt{h^2 + (side/2)^2} = \sqrt{15^2 + (8/2)^2} = \sqrt{225 + 16} = \sqrt{241} \approx 15.52 \text{ cm}$. Thus, the lateral area is $(1/2) \times 32 \times \sqrt{241} = 16\sqrt{241} \approx 248.32 \text{ cm}^2$.

21. Problem 3b: total surface area of the pyramid

The total surface area is the lateral area plus the area of the base: $16\sqrt{241} + 8^2 = 16\sqrt{241} + 64 \approx 248.32 + 64 = 312.32 \text{ cm}^2$.

22. Problem 3c: volume of the pyramid

The volume is $(1/3) \times base\ area \times height = (1/3) \times 8^2 \times 15 = (1/3) \times 64 \times 15 = 320 \text{ cm}^3$.