Constructing Rectangles?

Question

Is it possible to construct a rectangle in which the ratio of the perimeter to the length of the long side is equal to the ratio of the length of the long side to the length of the short side?

Answer 1

Let the sides of the rectangle be a and b, respectively. Then you wish to have (2a+2b)/a = a/b. Let x = a/b. Then we have:

(2a+2b)/a = x
2 + 2b/a = x
2 + 2/x = x
2x + 2 = x²
x² - 2x = 2
x² - 2x + 1 = 3
(x-1)² = 3
x-1 = ±√3
x = 1±√3

Since 1 - √3 < 0, we must have x = 1 + √3. Check of result: suppose a/b = 1 + √3. Then we have:

(2a + 2b)/a
= (2(1 + √3)b + 2b)/((1+√3)b)
= (4 + 2√3)/(1+√3)
= (4 + 2√3)(1 - √3)/(1-3)
= (4 + 2√3 - 4√3 - 6)/(-2)
= (-2 - 2√3)/(-2)
= 1 + √3 = a/b

As required. So indeed, such a construction is possible.

Answer 2

It is possible, for example when the short side = 1,
and the long side = 1 + sqroot(3)

P = 2(1 + 1 + sqroot(3)) = 4 + 2sqroot(3)
P / L= (4 + 2sqroot(3)) / (1 + sqroot(3)) = 1 + sqroot(3)
L / S= (1 + sqroot(3)) / 1 = 1 + sqroot(3)

Answer 3

State the ratios

(2L + 2W)/L = L/W

2LW + 2W^2 = L^2

L^2 - 2LW - 2W^2 = 0

You can solve this for either L or W

L = [2W +- sqrt(4W^2 + 8)]/2

L = [2W +- 2sqrt(W^2 + 2)]/2

L = W +- sqrt(W^2 + 2)

Since sqrt(W^2 + 2) > W

L = W + sqrt(W^2 + 2)

Given a width, you can calculate a length that satisfies the given condition

Answer 4

p = 2l + 2w
p/l = l/w
pw = l^2
pw = 2lw + 2w^2 = l^2
2w^2 + 2lw - l^2 = 0
w = {-2l ± [(2l)^2 - 4(2)(-l^2)]^(1/2)}/(2*2)
there must be an easier way with derivation or something
well you can plug w back in to find l and then plug in l to find w