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Math Help - plz help

  1. #1
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    plz help

    I have been looking through my book for an example and I'm not able to find one. May someone help me.

    The question is: What is the area of a regular hexagon whose sides are each 12 inches long? (Round to the nearest square inch.)
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  2. #2
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    Since it is a regular hexagon, you canfind the area using the formula:where l is the side
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alinailiescu View Post
    Since it is a regular hexagon, you canfind the area using the formula:where l is the side
    in case you're wondering how alinailiescu got that formula:

    For a regular polygon, if given the length l of the sides, the area is given by:

    A = [(l^2)*N]/[4tan(pi/N)]

    where l is the length of a side and N is the number of sides.

    For a regular hexagon:
    N = 6

    => A = (6*l^2)/(4tan(pi/6))
    .......= (6*l^2)/(4/sqrt(3))
    .......= (6*sqrt(3)*l^2)/4
    .......= (3*sqrt(3)*l^2)/2
    now replace l with the length of one side

    For more formulas and a cool interactive utility, see Regular polygon area formula
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  4. #4
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    well..
    I think here's a way to work it out without that formula (btw, this might form basis of the forumla I'm not too sure)

    Since it's a regular hexagon each interior angle is 120 deg.
    Split the hexagon into its 4 traingles, find the area of each traingle and x4 for the total area.

    area of any triangle = 1/2 ab Sin c
    we have 12" and 12" for two sides and 120 deg for the other angle (30 deg ea for rest since 12 and 12 are same). This is sufficient information to use this formula.

    so area of this triangle = 62.35382907

    Area of hexagon = 4 x 62.35382907
    = 249.4153163

    is that correct? I may have gone wrong in workings so...
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  5. #5
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    How can you split the hexagon into 4 triangles, each nhaving an angle of 120?
    I would split the hexagon into 6 equilateral triangles, having the side 12.
    So, area of any triangle = 1/2 ab Sin c
    area=6*1/2(12)(12)sin 60=374.1229744
    Since sin60=sqrt(3)/2
    area hexagon=6*(1/2)l^2*sqrt(3)/2
    area hexagon=3l^2sgrt(3)/2
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  6. #6
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    Quote Originally Posted by alinailiescu View Post
    How can you split the hexagon into 4 triangles, each nhaving an angle of 120?
    I would split the hexagon into 6 equilateral triangles, having the side 12.
    So, area of any triangle = 1/2 ab Sin c
    area=6*1/2(12)(12)sin 60=374.1229744
    Since sin60=sqrt(3)/2
    area hexagon=6*(1/2)l^2*sqrt(3)/2
    area hexagon=3l^2sgrt(3)/2
    yup thanks, I can see how it's incorrect now since for some of the 120deg, it combines more than one triangle to make it in some cases. The reason I tried with 4 traingles is because of that (n-2) sides thing, which can give the degrees of each interior angle.

    Thanks for enlighting me
    I thought the answer was a bit dodgy
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