area & perimeter the same = equable shape, can anyone give me the measurments of a triangle/circle?

Question

I know an equable shape is the area and perimeter are the same, ie a square 4x4, A=16 and P=16. I need some triangle and circles that are equable, but am having great difficulty in working these out, can anyone help me.

Answer 1

take a circle:
A=pir^2=C=2pir
r^2=2r
(r^2-2r)=0
>>>>>r=2 or r=0

if the radius is 0,we have a point
which still makes mathematical sense
if the radius is 2,the area of the
circle is equal to its perimeter
in numerical terms

therefore,a circle with a radius
of 2 units has the same value as its
perimeter in numerical terms

take a 3x,4x,5x triangle

A=(3x*4x)/2=6x^2
P=3x+4x+5x=12x
A=P
6x^2=12x
6x^2-12x=0
6x(x-2)=0

giving x=2 or x=0

x=0 is a point and makes maths
sense
x=2,is a triangle with 6,8 and
10 units and has a perimeter and
area of the same numerical value

note:- an equilateral triangle is a
more efficient shape than a
3,4,5 triangle

the problem with a triangle is
if you change the ratio of the sides
of the triangle,you change the area
'efficiency' of the triangle and you
run into all sorts of numerical
problems

here i describe 'most efficient' as
the shape of the perimeter that
encloses the maximum possible
area

the circle is the most efficient
shape in the plane

there is a nice little model
at the site below

http://argyll.epsb.ca/jreed/math9/strand3/triangle_area_per.htm

i hope that this helps

Answer 2

Circle:

A = πr², C = 2πr

πr² = 2πr

Divide both sides by π:

r² = 2r

r² - 2r = 0

r(r - 2) = 0

So an equable circle has a radius of 2 (or 0, but that's not a very interesting circle).

The triangle is a little more interesting.

If you assume it's an equilateral triangle with side length s, then the perimeter is 3s. The area would be 1/2 × s × s√3/2 = s²√3/4.

s²√3/4 = 3s

s²√3 = 12s

s²√3 - 12s = 0

s(s√3 - 12) = 0

So either s = 0 (again a very boring triangle), or s = 12/√3 = 4√3.

I don't think that it has to be a "regular polygon" (like a square or an equilateral triangle) to be equable though.

For instance, if you have a rectangle of length l and width w = kl (i.e. the length times some scaling factor), then you want:

lw = 2l + 2w
kl² = 2l + 2kl = 2l(1 + k)
kl² - 2l(1 + k) = 0
l(kl - 2(1 + k)) = 0

kl = 2(1 + k)
l = 2(1 + k)/k = 2/k + 2
w = kl = k(2/k + 2) = 2 + 2k

What this means is, if you pick any value of k, the rectangle with length 2/k + 2 and width 2 + 2k SHOULD be equable. For instance, try k = 2: 2/2 + 2 = 3, 2 + 2(2) = 6. Area is 3(6) = 18, perimeter is 3(2) + 6(2) = 6 + 12 = 18.

I think you probably could do this for triangles too, but you'd need to use lots of cosines to find a formula... email me if you want me to try.

Answer 3

Dr D has the right idea. Imagine that you have a partially filled soft plastic bottle with no air gap, so that it's somewhat crumpled. You will find that it's not hard to deform that plastic bottle into other shapes, even though it contains the same amount of liquid and has the same surface area. Only when the bottle is filled to maximum capacity that it becomes hard to deform. A circle, for example, as a shape of fixed perimeter and area, cannot be deformed (because the circle has the maximum possible area for a given perimeter), but an oval of fixed perimeter and area can be (because an oval does NOT have the maximum possible area for a given perimeter--i.e., other oval-like shapes have same perimeter and area). I've given this question a star because as simple as this question is, it does have significance in structural mechanics, such as in design of spacecraft. Addendum: Expanding on Zanti3's answer, think of any geometric shape, with a "hole" inside of it, also of any gemoetric shape. Obviously the total perimeter and total area are both constant, but the hole can be moved anywhere inside.

Answer 4

The guys above are OK! Challenge: let x, y and z be integer numbered sides of rectangular triangle; how to construct a triangle with A=k·P, where k is also integer? Sirdon1999!

Answer 5

circle having diameter 2 unit is having area & perimeter the same