Question

select the correct answer.

a business award is in the shape of a regular hexagonal pyramid. the height of the award is 95 millimeters and the base edge is 44 millimeters.

what is the surface area of the pyramid to the nearest square millimeter?

\\(18,541\text{ mm}^2\\)
\\(10,060\text{ mm}^2\\)
\\(13,611\text{ mm}^2\\)
\\(17,616\text{ mm}^2\\)

Explanation:

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<reasoning>

Calculate the area of the regular hexagonal base

\[

$$\begin{aligned} s &= 44\text{ mm} \\ A_{\text{base}} &= \frac{3\sqrt{3}}{2} s^2 \\ A_{\text{base}} &= \frac{3\sqrt{3}}{2} (44)^2 = 2904\sqrt{3} \approx 5029.876\text{ mm}^2 \end{aligned}$$

\]

Calculate the slant height of the pyramid

\[

$$\begin{aligned} h &= 95\text{ mm} \\ a &= \frac{s}{2\tan(30^\circ)} = \frac{s\sqrt{3}}{2} = 22\sqrt{3} \approx 38.105\text{ mm} \\ l &= \sqrt{h^2 + a^2} = \sqrt{95^2 + (22\sqrt{3})^2} = \sqrt{9025 + 1452} = \sqrt{10477} \approx 102.357\text{ mm} \end{aligned}$$

\]

Calculate the total surface area

\[

$$\begin{aligned} A_{\text{lateral}} &= \frac{1}{2} \cdot P \cdot l = \frac{1}{2} \cdot (6 \cdot 44) \cdot \sqrt{10477} = 132 \cdot \sqrt{10477} \approx 13511.151\text{ mm}^2 \\ A_{\text{total}} &= A_{\text{base}} + A_{\text{lateral}} \approx 5029.876 + 13511.151 = 18541.027\text{ mm}^2 \approx 18541\text{ mm}^2 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) 18,541 mm²</mcq-correct>
<mcq-option>(B) 10,060 mm²</mcq-option>
<mcq-option>(C) 13,611 mm²</mcq-option>
<mcq-option>(D) 17,616 mm²</mcq-option>
</answer>

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Answer:

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<reasoning>

Calculate the area of the regular hexagonal base

\[

$$\begin{aligned} s &= 44\text{ mm} \\ A_{\text{base}} &= \frac{3\sqrt{3}}{2} s^2 \\ A_{\text{base}} &= \frac{3\sqrt{3}}{2} (44)^2 = 2904\sqrt{3} \approx 5029.876\text{ mm}^2 \end{aligned}$$

\]

Calculate the slant height of the pyramid

\[

$$\begin{aligned} h &= 95\text{ mm} \\ a &= \frac{s}{2\tan(30^\circ)} = \frac{s\sqrt{3}}{2} = 22\sqrt{3} \approx 38.105\text{ mm} \\ l &= \sqrt{h^2 + a^2} = \sqrt{95^2 + (22\sqrt{3})^2} = \sqrt{9025 + 1452} = \sqrt{10477} \approx 102.357\text{ mm} \end{aligned}$$

\]

Calculate the total surface area

\[

$$\begin{aligned} A_{\text{lateral}} &= \frac{1}{2} \cdot P \cdot l = \frac{1}{2} \cdot (6 \cdot 44) \cdot \sqrt{10477} = 132 \cdot \sqrt{10477} \approx 13511.151\text{ mm}^2 \\ A_{\text{total}} &= A_{\text{base}} + A_{\text{lateral}} \approx 5029.876 + 13511.151 = 18541.027\text{ mm}^2 \approx 18541\text{ mm}^2 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) 18,541 mm²</mcq-correct>
<mcq-option>(B) 10,060 mm²</mcq-option>
<mcq-option>(C) 13,611 mm²</mcq-option>
<mcq-option>(D) 17,616 mm²</mcq-option>
</answer>

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