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Math Help - Area Of Regular Polygons

  1. #1
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    Area Of Regular Polygons

    Ok, I need help with my geometry homework. If anyone can thoroughly explain or something I'd appreciate it much.

    "If the area of a trapezoid is 4x^2+4x+1 and one base is 2x+3 and the height is 2x+1, what is the length of the other base? Answers is in terms of x. "

    Also I know that the area of a regular hexagon with a radius of 7 has an area of 127, but I've gotten down to 147 x sqrt(3) x 1/2. I know for sure that it's 127, I just don't know how to get 127 from there and have it in simplest radical form.

    And... to find the area of an equilateral triangle with the sides of 5 cms.

    If anyone can help.. i'd appreciate it.
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  2. #2
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    Quote Originally Posted by Blayde View Post
    Ok, I need help with my geometry homework. If anyone can thoroughly explain or something I'd appreciate it much.

    "If the area of a trapezoid is 4x^2+4x+1 and one base is 2x+3 and the height is 2x+1, what is the length of the other base? Answers is in terms of x. "
    .
    A=\frac{1}{2}(b_1+b_2)h

    4x^2+4x+1=\frac{1}{2}(2x+3+b)(2x+1)
    8x^2+8x+2=(2x+3+b)(2x+1)
    (2x+1)(4x+2)=(2x+3+b)(2x+1)
    4x+2=2x+3+b
    4x-1=2x+b
    2x-1=b
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  3. #3
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    Quote Originally Posted by Blayde View Post
    Ok, I need help with my geometry homework. ...

    Also I know that the area of a regular hexagon with a radius of 7 has an area of 127, but I've gotten down to 147 x sqrt(3) x 1/2. I know for sure that it's 127, I just don't know how to get 127 from there and have it in simplest radical form.

    ...
    Hello,

    the area of a regular hexagon consists of 6 equilateral triangles. The side of this triangle is equal to r, in your case r = 7 cm.

    A_{6-gon}=6 \cdot \frac{1}{2} \underbrace{\cdot \underbrace{r}_{base} \cdot \underbrace{\frac{1}{2} \cdot r \cdot \sqrt{3}}_{height}}_{of \ triangle}=\frac{3}{2} r^2 \cdot \sqrt{3}

    Plug in r = 7 and you'll get: A = 127.3057...cm²

    From the formula above you see that the area of an equilateral triangle with the side a is:

    A_{triangle}=\frac{1}{2} \cdot a \cdot \frac{1}{2} \cdot a \cdot \sqrt{3}=\frac{1}{4} \cdot a^2 \cdot \sqrt{3}

    With a = 5 cm you'll get: A = 10.825... cm²

    EB
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  4. #4
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    The area of a regular polygon with "radius" r (meaning the distance from the center to a vertex) that has n number of sides is:

    r^2n\left(\sin\left(\frac{360}{2n}\right)\times\co  s\left(\frac{360}{2n}\right)\right)

    At least I think....

    You should check that first.
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  5. #5
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    Quote Originally Posted by earboth View Post
    Hello,

    the area of a regular hexagon consists of 6 equilateral triangles. The side of this triangle is equal to r, in your case r = 7 cm.

    A_{6-gon}=6 \cdot \frac{1}{2} \underbrace{\cdot \underbrace{r}_{base} \cdot \underbrace{\frac{1}{2} \cdot r \cdot \sqrt{3}}_{height}}_{of \ triangle}=\frac{3}{2} r^2 \cdot \sqrt{3}

    Plug in r = 7 and you'll get: A = 127.3057...cm²

    From the formula above you see that the area of an equilateral triangle with the side a is:

    A_{triangle}=\frac{1}{2} \cdot a \cdot \frac{1}{2} \cdot a \cdot \sqrt{3}=\frac{1}{4} \cdot a^2 \cdot \sqrt{3}

    With a = 5 cm you'll get: A = 10.825... cm²

    EB

    Thanks everyone.



    Thank you guys so much.
    Last edited by Blayde; January 7th 2007 at 11:05 AM.
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