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Math Help - largest possible perimeter

  1. #1
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    largest possible perimeter

    Three non-overlapping regular plane polygons, at least two of which are congruent, all have side length 1. The polygons meet at a point A in such a way that the sum of the three interior angles at A is 360. Thus the three polygons form a new polygon with A as an interior point. What is the largest possible perimeter that this polygon can have?

    a) 12 b) 14 c) 18 d ) 21 e)24

    The answer is d)21 but I think I can make all of them, if not more.
    Could someone please take a look at my attached work and
    show me what I am not understanding?

    Thanks.

    Vicky.
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  2. #2
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    Quote Originally Posted by Vicky1997 View Post
    Three non-overlapping regular plane polygons, at least two of which are congruent, all have side length 1. The polygons meet at a point A in such a way that the sum of the three interior angles at A is 360. Thus the three polygons form a new polygon with A as an interior point. What is the largest possible perimeter that this polygon can have?

    a) 12 b) 14 c) 18 d ) 21 e)24

    The answer is d)21 but I think I can make all of them, if not more.
    Could someone please take a look at my attached work and
    show me what I am not understanding?

    Thanks.

    Vicky.
    1. The formula you used determines the sum of all interior angles of a polygon. The value of one single interior angle is calculated by

    |\alpha| = 180^\circ \cdot \left(\frac{n-2}{n}\right)

    where n is the number of vertices of the polygon.

    2. Let n denote the number of vertices of the congruent polygons and k the number of vertices of the other one then you'll get:

    2 \cdot 180^\circ \cdot \left(\frac{n-2}{n}\right) + 180^\circ \cdot \left(\frac{k-2}{k}\right) = 360^\circ

    Simplify this equation and you'll get

    n = \frac{4k}{k-2}

    3. The perimeter of the 3 polygons is

    p = 2n+k - 6

    4. I'll leave the rest for you.
    Attached Thumbnails Attached Thumbnails largest possible perimeter-polygonperim_winkelsumme.png  
    Last edited by earboth; January 7th 2010 at 10:51 PM.
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  3. #3
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    Quote Originally Posted by earboth View Post
    1. The formula you used determines the sum of all interior angles of a polygon. The value of one single interior angle is calculated by

    |\alpha| = 180^\circ \cdot \left(\frac{n-2}{n}\right)

    where n is the number of vertices of the polygon.

    2. Let n denote the number of vertices of the congruent polygons and k the number of vertices of the other one then you'll get:

    2 \cdot 180^\circ \cdot \left(\frac{n-2}{n}\right) + 180^\circ \cdot \left(\frac{k-2}{k}\right) = 360^\circ

    Simplify this equation and you'll get

    n = \frac{4k}{k-2}

    3. The perimeter of the 3 polygons is

    p = 2n+k - 6

    4. I'll leave the rest for you.

    Thank you!! Thankyou!!

    I think I got the answer.
    Could you please take a look at my work

    Vicky.
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  4. #4
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    Quote Originally Posted by Vicky1997 View Post
    Thank you!! Thankyou!!

    I think I got the answer.
    Could you please take a look at my work

    Vicky.
    To answer your last question:

    Have a look at your own sketch (see below).

    As you can see there are always 2 sides of each polygon which belong to the interior of the complete figur, that means, they don't belong to the perimeter.

    Therefore the perimeter is calculated by:

    p=2(n-2)+(k-2)

    Expand the brackets and you'll get the formula I've posted.
    Attached Thumbnails Attached Thumbnails largest possible perimeter-largestperimeter.png  
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