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a problem (maths)
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Old 06 Sep 2009, 13:05   #1 (permalink)
Shas'Vre
 
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Default a problem (maths)

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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Old 06 Sep 2009, 13:26   #2 (permalink)
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Default Re: a problem (maths)

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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Old 06 Sep 2009, 13:40   #3 (permalink)
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Default Re: a problem (maths)



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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Old 06 Sep 2009, 13:51   #4 (permalink)
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Default Re: a problem (maths)

*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

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Old 06 Sep 2009, 13:53   #5 (permalink)
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Default Re: a problem (maths)

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] [[/td][/tr][/table][hr]What's an abelian group with an associative, distributive secondary operator and the power to corrupt mortals?
[spoiler=Answer]The One Ring![/spoiler]
[hr]
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Old 06 Sep 2009, 13:58   #6 (permalink)
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Default Re: a problem (maths)

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

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Old 06 Sep 2009, 14:18   #7 (permalink)
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Default Re: a problem (maths)

Well if you make an 8' hexagon table then the necessary width is roughly 6.9', so you should comfortably have enough material for that with two 8' x 4' sheets.
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