Geometry and Trigonometry - Nets of pyramids, cones, and compound 3D shapes

A square-based pyramid has a base side length of $6\text{ cm}$ and a slant height of $5\text{ cm}$. Draw the net and calculate its total surface area.

$\text{Area of square base} = 6 \times 6 = 36\text{ cm}^2$ $\text{Area of one triangular face} = \frac{1}{2} \times 6 \times 5 = 15\text{ cm}^2$ $\text{Total Surface Area} = 36 + (4 \times 15) = 36 + 60 = 96\text{ cm}^2$

The net consists of one $6 \times 6$ square and four identical triangles with base $6\text{ cm}$ and height $5\text{ cm}$. We sum the area of the base and the four lateral faces.

A cone has a radius of $r = 3\text{ cm}$ and a vertical height of $h = 4\text{ cm}$. Find the slant height $l$ and the area of the sector required for its net.

$l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}$ $\text{Area of the sector (Curved Surface Area)} = \pi \times 3 \times 5 = 15\pi \approx 47.12\text{ cm}^2$

First, we use the Pythagorean theorem to find the slant height ($l$), which is the radius of the sector in the net. Then, we use the curved surface area formula $\pi r l$ to find the area of that sector.

A compound shape is formed by a cylinder of radius $2\text{ cm}$ and height $5\text{ cm}$, with a hemisphere placed on one end. What is the total surface area of the combined shape? (Leave answer in terms of $\pi$)

$\text{Area of circular base} = \pi(2)^2 = 4\pi$ $\text{Curved area of cylinder} = 2\pi(2)(5) = 20\pi$ $\text{Curved area of hemisphere} = 2\pi(2)^2 = 8\pi$ $\text{Total Surface Area} = 4\pi + 20\pi + 8\pi = 32\pi\text{ cm}^2$

In a compound shape, we only count the exterior surfaces. The net would include one circle (the base of the cylinder), one rectangle (the curved surface of the cylinder), and the curved surface of the hemisphere. The interface where the hemisphere meets the cylinder is hidden and not included.

1. Identify the exposed surfaces:

• The circular base of the cylinder: $\text{Area} = \pi r^2 = \pi (4)^2 = 16\pi \text{ cm}^2$.
• The curved surface of the cylinder: $\text{Area} = 2\pi r h = 2\pi(4)(10) = 80\pi \text{ cm}^2$.
• The curved surface of the cone: $\text{Area} = \pi r l = \pi(4)(7) = 28\pi \text{ cm}^2$.
• (The face where they meet is internal and not counted).

1. Total Surface Area = $16\pi + 80\pi + 28\pi = 124\pi \text{ cm}^2$.

The total surface area of a compound shape only includes the exterior surfaces. We sum the cylinder's base, the cylinder's lateral area, and the cone's lateral area.

1. Area of the base: $8 \times 6 = 48\text{ cm}^2$.
2. Area of the two triangles with base $8\text{ cm}$: $2 \times (\frac{1}{2} \times 8 \times 5) = 40\text{ cm}^2$.
3. Area of the two triangles with base $6\text{ cm}$: $2 \times (\frac{1}{2} \times 6 \times 5.39) = 32.34\text{ cm}^2$.
4. Total Surface Area = $48 + 40 + 32.34 = 120.34\text{ cm}^2$.

For a rectangular pyramid, there are two pairs of congruent triangular faces with different slant heights. We calculate the area of the base and all four triangles and sum them.