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svlad2 [7]
3 years ago
7

Is this the correct surface area? Explain how you know​

Mathematics
1 answer:
yarga [219]3 years ago
5 0

Answer:

  No. The actual total surface area is double that: 108 in².

Step-by-step explanation:

The area Charlotte has computed is the area of the faces you can see. There are three other faces you cannot see that are the same size. So, the total area is double the amount Charlotte has computed.

  A = 2(LW +H(L+W)) = 2(6·4+3(6+4)) = 2(24+30) = 2(54) = 108 . . . . sq in

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Answer:

23\sqrt{3}\ un^2

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

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Answer:

A. 10.2

B.3 √5
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