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Area of a Pentagon

  1. Nov 16, 2004 #1
    Hi All,

    Just want you to double check my method and result for a question that I thought of.

    What is the area of a regular pentagon???

    I divided the pentagon into three isosceles. Two triangles, the two on the sides if the base is on the bottom, have two sides which are 200m each and an unknown side with the angles 108°, 36° and 36°. The middle triangle has the base as 200m and the angles as 36°, 72° and 72°.

    So applying the Cosine Rule the two sides will be:

    a² = b² + c² - 2bc cosA
    a² = 200² + 200² - [(2 x 200 x 200)(cos 108°)]
    a² = 4000 + 4000 - (80000 x -0.309)
    a² = 8000 + 24721.35
    a² = 32721.35
    a = (square root)32721.35
    a = 180.89m

    Then work out the perpendicular height of the triangle:

    Sin = Opp/Hyp
    Sin 36° = y/200
    Sin 36° x 200 = y
    y = 117.56m

    Then work out the area of the triangle:

    1/2 (117.56 x 180.89) = 10632.47m²

    The middle triangle:
    The perpendicular height:

    Sin = Opp/Hyp
    Sin 72° = y/180.89
    Sin 72° x 180.89 = y
    y = 172.03m

    Area of triangle:

    1/2 (200 x 172.03) = 17203.7m²

    Total Area: (10632.47 x 2) + 17203.7 = 27836.17m²

    If you have time to check then please do but I just want a double check.


    The Bob (2004 ©)
  2. jcsd
  3. Nov 16, 2004 #2


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    Areas of regular N-Gons

    It looks like the number you're getting is much too small.
    Do you really think that the base of the 108-36-36 triangle should be shorter than the legs? You've got the base at 180 meters, and the legs at 200.

    An alternative method for finding the area:
    Any regular n-gon can be divided into n identical isoscoles triangles where the base of the iscoloes triangle is one of the sides of the polygon, and the angle opposite to the base is [tex]\frac{2\pi}{n}[/tex] (in radians). So, if the side length of the original n-gon is [itex]l[/itex], then the altitude of the triangle will be:
    [tex]\frac{l}{2} \cot(\frac{\pi}{n})[/tex]
    so the area of the triangle will be
    [tex]\frac{l^2}{4} \cot(\frac{\pi}{n})[/tex]
    so the area of the n-gon will be
    [tex]\frac{nl^2 \cot(\frac{\pi}{n})}{4}[/tex]
  4. Nov 17, 2004 #3
    I see. I have recalculated it to equal 323.607m. I made a mistake in my sqauring, of all the things to get wrong but I am interested in the [tex]cot[/tex], what it is and how I can use the equation because I don't really understand it too well, mainly because I do not know what the cot is.

    Cheers for the help, please help a little more. :smile:

    The Bob (2004 ©)
    Last edited: Nov 17, 2004
  5. Nov 17, 2004 #4
    I looked the cot up and I can't get this to work at all. I have tried it with a square, as that is simple to calculate and all I get is a wrong answer.

    The Bob (2004 ©)
  6. Nov 17, 2004 #5


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    I'm using radians so
    Applying the formula for a square ([itex]n=4[/itex]):
    [tex]\frac{l^2n\cot(\frac{\pi}{n})}{4}=\frac{l^2 4 \cot (\frac{\pi}{4})}{4}=l^2[/tex]
    which is the correct result.
    If you're using degrees, you'd need to use [itex]\cot(\frac{180}{n})[/itex] instead.

    Applying the formula for a pentagon:
    [tex]\frac{l^2n\cot(\frac{\pi}{n})}{4}=l^2 \frac{5}{4} \cot(\frac{\pi}{5}) \approx 1.7204 l^2[/tex]
    In your case [itex]l=200[/itex]
    so [itex]1.7204 \times 40000 = 68816[/itex]
    (There's a couple of square meters of rounding error there.)

    Understanding the formula:
    Every regular n-gon has a center which is equidistant from all of the vertices. (Demonstrating that this is true is fairily easy.) That means that it can be disected into n identical isoscoles triangles that all meet at the center. So, if we calculate the area of one of the triangles, and multiply by n we get the area of the entire n-gon.
    The angle opposite the base of the isoscoles triangle is going to be [itex]\frac{2pi}{n}[/itex] (or [itex]\frac{360}{n}[/itex] degrees) and the base length is [itex]l[/itex] - the length of one of the sides of the inital [itex]n[/itex]-gon. From there it's not particularly difficult to work out the area of the triangle (and consequently the n-gon) using trig.
  7. Nov 17, 2004 #6


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    Here's how I would do that problem. Drawing lines from the center of the pentagon to each vertex divides it into 5 congruent triangles. The base angles of the triangles are all the same so there are 10 of them: their total measure is 10θ The vertex angles of all the triangles form a complete circle: measure 360 degrees, so the total of all angles is 10&theta+ 360 which must be 5(180)= 900. 10θ+ 360= 900 gives
    10&theta= 540 so &theta= 54 degrees.

    Taking s as the length of a side of that pentagon, the base of each triangle is s. Let h be the height of each triangle. Then h/(s/2)= 2h/s= tan(54) so h= (s/2)tan(54).
    The area of each triangle is (1/2)hs= (s2/4)tan(54).

    with s= 200 m that is (40000/4)tan(54)= 10000(1.376819)= 13763.82 m2.

    That can be applied to any regular polygon: From the center, draw lines to each vertex. With n vertices we will have n triangles and so 2n angles: 2nθ+ 360= n(180). 2nθ= n(180)- 2(180) so θ = ((n-2)/n)(90).

    Taking s as the length of a side, each triangle has base of length s and height h where h/(s/2)= tan(θ). That is, h= (s/2)tan(&theta). The area of each triangle is (1/2)(s2/2)tan(&theta) and since there are n triangles, the area of the regular n-gon with each side of length s is
    n(s2/4)tan(&theta)= (ns2/4)tan(((n-2)/n)(90) degrees).

    In particular, if n= 4, this is (s2)tan(45 degrees) = s2.

    If n= 3, this is (3/4)s2 tan(30)degrees= (√(3)/4)s2, again, the correct answer.
  8. Nov 17, 2004 #7
    isn't cotx just 1/tanx so cant the equation be nl^2/4tan(pi/n) That would be alot more calculator friendly because i pretty sure most calcs dont have the recipricol trig funtions.
  9. Nov 17, 2004 #8
    My final Pentagon came out as 70402m².

    I do not get how to work out the cot still and Holly's Method makes no sense to me.

    I am now really, really interested in both of your methods so could you please explain them more. I will look over them tomorrow, when I have more time and have had sleep and I will see what I know then but any help, simply in these methods making sense, would be a big help.

    Thanks for all you help and all the help to come.

    The Bob (2004 ©)
  10. Nov 17, 2004 #9
    [(nl^2)/4]x{1/[tan(pi/n)]}. I think that is right.

    The Bob (2004 ©)
  11. Nov 17, 2004 #10


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    Well, using radians:
    [tex]\cot(x)=\tan(\frac{\pi}{4} - x)=\frac{1}{\tan(x)}[/tex]

    So we're all on the same page.
  12. Nov 17, 2004 #11
    I wasn't trying to convince anyone of the fact that cotx = 1/tanx. I just wanted to make sure it was doable in this situation without breaking some simple algebraic rule that I missed or something. so infact the equation I posted previously is valid?
  13. Nov 17, 2004 #12


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    Bob, what Nate and HallsofIvy are saying is that you can split the pentagon (or any polygon) up into a number of equal slices, just as if you were cutting a cake into equal parts. The parts will be identical triangles, in this case there are 5 of them.

    You seeing it yet?
  14. Nov 18, 2004 #13
    I see that because that is what I did to work out a hexagon and the larger even numbered polygons. My problem is that I have worked out 3 numbers for the Pentagon by working it out 2 different ways and so I am interested in how NateTG and Holly work it out. Bascially I would like someone to apply their equations to a simple shape (e.g. a sqaure with sides of 10cm) and then see if they get it right and how they did it so I can apply it, learn how to do it and then apply it myself.


    The Bob (2004 ©)
  15. Nov 18, 2004 #14
    Please people. I need a simple example so I can see how the equations work. :biggrin:

    The Bob (2004 ©)
  16. Nov 18, 2004 #15


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    Ok - working in degrees for you:

    A square with 10 cm sides:
    [tex]A=\frac{1}{4} nl^2 \cot(\frac{180}{n})=\frac{1}{4} 4 (10^2) \cot(\frac{180}{4})=100[/tex]

    A 73-gon with 10 cm sides
    [tex]A=\frac{1}{4}nl^2 \cot(\frac{180}{n})=\frac{1}{4} 73 (10^2) \cot(\frac{180}{73})\approx42380[/tex]
  17. Nov 19, 2004 #16
    Cheers NateTG. :biggrin: That is a very cool equation. I assume it only works for regular polygons because only one length is involved. I normally would be a pain and ask if there was one for irregualr polygons but I don't really need it for the work.

    Thanks so much to everyone for helping. :smile:

    The Bob (2004 ©)

    P.S. I owe you one NateTG and Holly. Cheers. :biggrin:
  18. Nov 21, 2004 #17
    I think someone may have explained this but how did you come to each section and what do they have to do to make the equation work:[tex]A=\frac{1}{4} nl^2 \cot(\frac{180}{n})[/tex].

    Cheers Again.

    The Bob (2004 ©)
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