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Area of a Pentagonby The Bob
Tags: pentagon 
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#1
Nov1604, 02:47 PM

P: 1,116

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. Cheers. The Bob (2004 ©) 


#2
Nov1604, 04:25 PM

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P: 2,538

It looks like the number you're getting is much too small.
Do you really think that the base of the 1083636 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 ngon 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 ngon 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 ngon will be [tex]\frac{nl^2 \cot(\frac{\pi}{n})}{4}[/tex] 


#3
Nov1704, 08:30 AM

P: 1,116

Cheers for the help, please help a little more. The Bob (2004 ©) 


#4
Nov1704, 11:33 AM

P: 1,116

Area of a Pentagon
The Bob (2004 ©) 


#5
Nov1704, 12:21 PM

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P: 2,538

I'm using radians so
[tex]\cot(\frac{\pi}{4})=1[/tex] 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 ngon 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 ngon. 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 ngon) using trig. 


#6
Nov1704, 12:39 PM

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PF Gold
P: 39,339

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= (s^{2}/4)tan(54). with s= 200 m that is (40000/4)tan(54)= 10000(1.376819)= 13763.82 m^{2}. 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 = ((n2)/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)(s^{2}/2)tan(&theta) and since there are n triangles, the area of the regular ngon with each side of length s is n(s^{2}/4)tan(&theta)= (ns^{2}/4)tan(((n2)/n)(90) degrees). In particular, if n= 4, this is (s^{2})tan(45 degrees) = s^{2}. If n= 3, this is (3/4)s^{2} tan(30)degrees= (√(3)/4)s^{2}, again, the correct answer. 


#7
Nov1704, 05:15 PM

P: 94

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.



#8
Nov1704, 06:01 PM

P: 1,116

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 ©) 


#9
Nov1704, 06:03 PM

P: 1,116

The Bob (2004 ©) 


#10
Nov1704, 06:21 PM

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P: 2,538

Well, using radians:
[tex]\cot(x)=\tan(\frac{\pi}{4}  x)=\frac{1}{\tan(x)}[/tex] So we're all on the same page. 


#11
Nov1704, 06:50 PM

P: 94

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?



#12
Nov1704, 06:52 PM

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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? 


#13
Nov1804, 03:08 AM

P: 1,116

Cheers. The Bob (2004 ©) 


#14
Nov1804, 05:36 PM

P: 1,116

Please people. I need a simple example so I can see how the equations work.
The Bob (2004 ©) 


#15
Nov1804, 06:14 PM

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A square with 10 cm sides: n=4 l=10 [tex]A=\frac{1}{4} nl^2 \cot(\frac{180}{n})=\frac{1}{4} 4 (10^2) \cot(\frac{180}{4})=100[/tex] A 73gon with 10 cm sides n=73 l=10 [tex]A=\frac{1}{4}nl^2 \cot(\frac{180}{n})=\frac{1}{4} 73 (10^2) \cot(\frac{180}{73})\approx42380[/tex] 


#16
Nov1904, 02:11 AM

P: 1,116

Thanks so much to everyone for helping. The Bob (2004 ©) P.S. I owe you one NateTG and Holly. Cheers. 


#17
Nov2104, 03:41 PM

P: 1,116

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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