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#1 (permalink) |
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Shas'Vre
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you start with a hexagon and number each point starting at the top with one moving clockwise number the rest till you get to 6. Points 1 - 4 draw a line going though the center of the shape this line is 6'. now draw a line from 2 - 6 how long is this line?
KQD
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-Disney: He's the puppet master, both our damnation and salvation. He rewards our successes and punishes us for our hubris (often in subtle but horrible ways). I suggest we do our best not to anger him, he is the greatest asset to our campaign and potentially the most fearsome foe we'll ever face. |
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#2 (permalink) |
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Shas'El
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I make the answer to be 3*SQRT(3). Now, I'll think I'll write everything out in a word document and attach it to this post as it'll be hell to explain without diagrams! Please bear with me...
[hr] Edit, on second thoughts, as this isn't an academic mathematical problem and given Masked Thespian's solution, that probably isn't required. Nicely done anyway, MT.
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I never bluff, TK. |
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#3 (permalink) |
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Shas'El
![]() ![]() ![]() ![]() ![]() Join Date: Jan 2008
Location: Ota, Gunma Prefecture, Japan
Posts: 3,764
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![]() So, based on the above picture, given that the line segment between vertices 1 and 4 is six inches in length, you want to know how long the line segment joining vertices 2 and 6 is. In this case, we can break it down to simple trigonometry plus elementary properties of a regular hexagon. In the case of a regular hexagon, the length of the line segment from vertices 1 to 4 is double the length of one side, such as the line segment between 1 and 2. In addition, the internal angles of a regular hexagon are 120 degrees, therefore half of this is 60 degrees. This gives us the following triangle: ![]() Using SOHCAHTOA, the opposite angle A12 (60 degrees), and the hypotenuse 12 (3"), we have that the length of the line segment A2 is equal to Sine (60) * 3. The full line segment from 6 to 2 is double this and is thus equal to Sine (60) * 6, or 5.20 (3sf) inches. This is equal to Tau-killer's answer of 3*3[sup]1/2[/sup].
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[table][tr][td]Regards, MT.[/td][td] [spoiler=Answer]The One Ring![/spoiler] [hr] |
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#4 (permalink) |
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Shas'Vre
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*sigh I've made another mistake, I was meant to put 6 feet.
I am hoping to build a hexagonal table where points 1 - 4 is 6 feet. along this line will be the split where I'll build the table to fold away. Now the wood I intent to buy is 8" * 4" I want to know if points 2 - 6 is less than 4" so I can save money. KQD
__________________
-Disney: He's the puppet master, both our damnation and salvation. He rewards our successes and punishes us for our hubris (often in subtle but horrible ways). I suggest we do our best not to anger him, he is the greatest asset to our campaign and potentially the most fearsome foe we'll ever face. |
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#5 (permalink) |
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Shas'El
![]() ![]() ![]() ![]() ![]() Join Date: Jan 2008
Location: Ota, Gunma Prefecture, Japan
Posts: 3,764
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Well, 6 inches or 6 feet, the mathematics are identical. You will need approximately five and one fifth feet (or around 62.5 inches) of wood for the width of your table so an 8'x4' (or 96 inches x 48 inches) sheet would not be sufficient.
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[table][tr][td]Regards, MT.[/td][td] [spoiler=Answer]The One Ring![/spoiler] [hr] |
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#6 (permalink) |
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Shas'Vre
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thanks for the maths help anyway it's been bugging me all night, more so because I should know it.
Knowing now that I'll need to buy two sheets should I forgo the 6" table and just make an 8" hexagon table instead? KQD
__________________
-Disney: He's the puppet master, both our damnation and salvation. He rewards our successes and punishes us for our hubris (often in subtle but horrible ways). I suggest we do our best not to anger him, he is the greatest asset to our campaign and potentially the most fearsome foe we'll ever face. |
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