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Math Help - Geometry help, please

  1. #1
    Newbie JoanneMac's Avatar
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    Exclamation Geometry help, please

    I just want to make sure I did this problem right or not. Please correct me if I 'm wrong.

    Find the area of a regular hexagon with side 2* sqrt of 3cm and apothem 3cm.
    Here's what I got:
    The formula is A=1/2ap, p(perimeter)=6s=6*2sqrtof3=12*sqrtof3.
    a(apothem)=1/2s*sqrtof3=1/2*2*sqrtof3=sqrtof3.
    A=1/2*sqrtof3*12*sqrtof3=6(sqrtof3)^2=6*3=18.
    I'm I correct?

    thanks for your help.
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  2. #2
    Super Member
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    Quote Originally Posted by JoanneMac View Post
    I just want to make sure I did this problem right or not. Please correct me if I 'm wrong.

    Find the area of a regular hexagon with side 2* sqrt of 3cm and apothem 3cm.
    Here's what I got:
    The formula is A=1/2ap, p(perimeter)=6s=6*2sqrtof3=12*sqrtof3.
    a(apothem)=1/2s*sqrtof3=1/2*2*sqrtof3=sqrtof3.
    A=1/2*sqrtof3*12*sqrtof3=6(sqrtof3)^2=6*3=18.
    I'm I correct?

    thanks for your help.
    There must be a minor mistake (the factor \sqrt{3} is missing)

    The area of a regular hexagon consists of 6 equilateral triangles. The height in an equilateral triangle is (according to Pythagorean theorem)

    h = \frac12 s\cdot \sqrt{3}

    The area of the hexagon is:

    a = 6 \cdot \frac12 \cdot s \cdot \frac12 s\cdot \sqrt{3}

    With s = 2\sqrt{3} you'll get:

    a= 6 \cdot \frac12 \cdot 2\sqrt{3} \cdot \frac12 \cdot 2\sqrt{3} \cdot \sqrt{3} = \boxed{18\sqrt{3}}
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by JoanneMac View Post
    I just want to make sure I did this problem right or not. Please correct me if I 'm wrong.

    Find the area of a regular hexagon with side 2* sqrt of 3cm and apothem 3cm.
    Here's what I got:
    The formula is A=1/2ap, p(perimeter)=6s=6*2sqrtof3=12*sqrtof3.
    a(apothem)=1/2s*sqrtof3=1/2*2*sqrtof3=sqrtof3.
    A=1/2*sqrtof3*12*sqrtof3=6(sqrtof3)^2=6*3=18.
    I'm I correct?

    thanks for your help.
    Hello Joanne,

    A=\frac{1}{2}ap

    The perimeter is 12\sqrt{3} and the apothem is 3.

    A=\frac{1}{2}(3)(12\sqrt{3})=18\sqrt{3}
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