Geometry and Measurement - Nets of 3D Shapes

A cube has a side length of $5$ cm. Calculate its total surface area using the concept of its net.

The net of a cube consists of $6$ identical squares. Area of one square face = $s \times s = 5 \text{ cm} \times 5 \text{ cm} = 25 \text{ cm}^2$. Total Surface Area = $6 \times 25 \text{ cm}^2 = 150 \text{ cm}^2$.

Since a cube has 6 faces that are all the same, we find the area of one face and multiply by 6.

A cylinder has a radius of $r = 7$ cm and a height of $h = 10$ cm. Find the area of the rectangular part of its net. (Take $\pi \approx \frac{22}{7}$)

The length of the rectangle in the net is equal to the circumference of the base: $L = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm}$. The width of the rectangle is equal to the height of the cylinder: $W = 10 \text{ cm}$. Area of rectangle = $L \times W = 44 \times 10 = 440 \text{ cm}^2$.

When a cylinder is 'unrolled', its lateral surface forms a rectangle where the length is the circumference of the base.

Calculate the total surface area of a rectangular prism with length $l = 10$ cm, width $w = 4$ cm, and height $h = 5$ cm by summing the areas of its faces.

The net has three pairs of faces: Pair 1 (Base/Top): $2 \times (10 \times 4) = 80 \text{ cm}^2$ Pair 2 (Front/Back): $2 \times (10 \times 5) = 100 \text{ cm}^2$ Pair 3 (Sides): $2 \times (4 \times 5) = 40 \text{ cm}^2$ Total Surface Area calculation: $\begin{array}{r} 80 \\ 100 \\ + 40 \\ \hline 220 \end{array}$ Total Surface Area = $220 \text{ cm}^2$.

We use the sum of the areas of all six rectangular faces represented in the net to find the total surface area.

1. Area of the square base: $A_{base} = s^2 = 6^2 = 36 \text{ cm}^2$
2. Area of one triangular face: $A_{tri} = \frac{1}{2} \times b \times l = \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2$
3. Since there are 4 triangular faces: $A_{total\_tri} = 4 \times 24 = 96 \text{ cm}^2$
4. Total Surface Area: $SA = 36 + 96 = 132 \text{ cm}^2$

The surface area is the sum of the base area and the lateral area (the four triangles). The slant height of the pyramid becomes the height of the triangles in the net.

1. Area of the two circular bases: $2 \times \pi r^2 = 2 \times 3.14 \times 3^2 = 2 \times 3.14 \times 9 = 56.52 \text{ cm}^2$
2. Length of the rectangular face (Circumference): $L = 2\pi r = 2 \times 3.14 \times 3 = 18.84 \text{ cm}$
3. Area of the rectangular face: $A_{rect} = L \times h = 18.84 \times 10 = 188.4 \text{ cm}^2$
4. Total Surface Area: $SA = 56.52 + 188.4 = 244.92 \text{ cm}^2$

The net consists of two circles and one rectangle. The rectangle wraps around the circle, so its length must equal the circle's circumference.