Perimeter = Area?
June 23, 2008 am30 6:26 am
I was thinking of asking for a square with perimeter = area, but that’s either silly or boring.
Give me a square, any square. I’ll bisect a side, then bisect that segment, call the result a unit, and the perimeter is 16 units, the area 16 square units. Works for any square. Boring.
So, readers, can we do something interesting with this? Different shapes? Different regular polygons?
The equilateral triangle has ugly irrational perimeter = area, but some of the other segments…
Concentrate on a side? an altitude? apothem? segment from center to vertex?
An interesting problem, dear readers, can we get an interesting problem out of this?
JD2718 
These are questions I wrote for the Onondaga County MTA exam last year. I know they’re not exactly what you’re looking for… (and, what’s wrong with irrational answers?) (and, have you considered a square with the same perimeter as a circle? that could be interesting…)
The segment joining the midpoints of two adjacent sides of a rectangle is 55 in. Determine the length of the diagonal of the rectangle. (Ans: 110)
ABCDEF is a regular hexagon, and diagonals BF and CE are drawn. If AB = 8, determine the area of rectangle FBCE. (Ans: 64rad3)
The physicist in me objects that since area and perimeter have fundamentally different units, perimeter and area can’t be equal.
Thank you, Kate.
Rachel, in a sense, that’s what makes the initial quesiton ho hum.
Years ago I attempted to construct the square root of a segment. Hard lesson learned.
Only vague recollection here, but I believe the equivalent question for a kite is interesting.
Working it out for the case of the regular polygon, you can show that if n is the number of sides and s is the side length, then P = A whenever
s = 4 tan( pi / n )
Generating a bunch of values for various n, the only nice values between 3 and 100 are for n = 3, 4, 5, 6, 10, 12. Though if you don’t like irrationals I don’t know how you’re going to feel about these…
3 gives s = 4 sqrt( 3 )
4 gives s = 4
5 gives s = 4 sqrt( 5 – 2 sqrt( 5 ) )
6 gives s = 4 / sqrt( 3 )
10 gives s = 4 sqrt( 1 – 2 / sqrt( 5 ) )
12 gives s = 4*( 2 – sqrt( 3 ) )
s does of course limit itself to 0 as n approaches infinity and you get a circle : )
Neat!
Oh, far more interesting, it seems that in the case of regular polygons where P = A, the apothem is always 2.
Last comment. I promise ^^
Simplest proof I can think of – let a be the apothem, s be the side length, n be the number of sides
P = n * s, A = (1/2) a * s * n
Then
P = A
n * s = (1/2) a * s * n
1 = (1/2) * a
a = 2
Fox,
I was hoping that the apothem would show up. And let the number of sides approach infinity (polygon becomes a circle?) and 4π = 4π when r = 2. Like magic.
So, how do I get kiddies to discover this?
And,
The list of polygons with nice values, why so short?
Take any polygon (or more generally a plane figure for which area and perimeter make sense), say it has area a and perimeter p in some choice of units. If you change the units so an old unit is t new units, the area will be at^2 and the perimeter will be pt. These, of course, are equal for t=p/a. So you can always make them equal by changing units…
Oh … I just realized you asked another question. Days late and several dollars short. That’s me.
The reason the list of polygons is so short, I’m not entirely sure why, but I think it has something to do with http://mathworld.wolfram.com/EuclideanNumber.html this.
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