Question: Lesson 2.1: Surface Area of 3-D Objects
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c. Is there a more appropriate SI unit of measure for the surface area of the playground equipment? Explain.
d. What would be a more appropriate imperial unit of measure for the surface area of this playground equipment?
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2. In a robotics competition, a robot must be built with a specific maximum starting height, width, and depth. In order to meet the height requirements, a robot must fit in the judge’s plexi-glass box. The box is a rectangular prism that has a surface area of 1 728 in², a width of 15 inches, and a length of 18 inches. The Mad Hatter robot is 17.75 inches tall in its starting position. Does the Mad Hatter meet the height requirement?
1. **Problem Statement:** Determine if the Mad Hatter robot meets the height requirement to fit inside a plexi-glass box with given dimensions and surface area.
2. **Given:**
- Surface area of box, $$SA = 1728\text{ in}^2$$
- Width, $$w = 15\text{ in}$$
- Length, $$l = 18\text{ in}$$
- Robot height, $$h_{robot} = 17.75\text{ in}$$
3. **Formula for surface area of a rectangular prism:**
$$SA = 2(lw + lh + wh)$$
4. **Find the height $$h$$ of the box:**
$$1728 = 2(18 \times 15 + 18 \times h + 15 \times h)$$
$$1728 = 2(270 + 18h + 15h)$$
$$1728 = 2(270 + 33h)$$
$$1728 = 540 + 66h$$
5. **Isolate $$h$$:**
$$1728 - 540 = 66h$$
$$1188 = 66h$$
$$h = \frac{1188}{66}$$
6. **Simplify the fraction:**
$$h = \frac{\cancel{1188}^{18} \times 66/18}{\cancel{66}^{3} \times 22/3} = \frac{18 \times 66/18}{3 \times 22/3}$$
Actually, better to just divide directly:
$$h = 18$$
7. **Interpretation:**
The box height is $$18\text{ in}$$.
8. **Compare robot height to box height:**
$$17.75\text{ in} < 18\text{ in}$$
9. **Conclusion:**
The Mad Hatter robot meets the height requirement because it is shorter than the box height.
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**Note:** Questions 2c and 2d about appropriate SI and imperial units for surface area are not solved here as per instructions to solve only the first problem.