# Golden ratio rectangles

I'm designing a layout and I would like to use four golden ratio rectangles. The total width of the layout is 960px. How do I find the height (x)? Below is a diagram of the layout.

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Maybe I am missing something, but wouldn't $x$ just be $$960 \frac{\sqrt{5}-1}{2}$$ – Ron Gordon May 27 '13 at 13:06
Ah, a diagram. I guess you want all four small rectangles to be golden? – Gerry Myerson May 27 '13 at 13:18
@Gerry In the future, please read the question before commenting; the link was there from the start, even before it was replaced by the actual diagram. – Lord_Farin May 27 '13 at 13:19

Finding the required height amounts to expressing $a$ and $c$ in terms of the height $h = a+b$. I will do this below; denote with $\phi$ the golden ratio.

We have that $\dfrac h a =\phi$, i.e. $a = \dfrac h \phi$. Now $d = \dfrac h2$, and $\dfrac c d = \phi$.

Using some trivial algebra, we obtain:

$$c = \phi d = \frac\phi2 h$$

Thus we have reduced to solving the equation:

$$960 = 2a + c = \left(\frac2\phi+\frac\phi 2\right)h$$

In conclusion, $h = \dfrac{960}{\frac2\phi+\frac\phi 2} \approx 469.42$.

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Thank you very much @Lord_Farin! – Coen Hyde May 27 '13 at 18:28

Easiest way to calculate this is with using ratios:

ϕ/(ϕ+2)=x/960

solve for x: 960/(1+2/ϕ) which results in 429.32505

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golden ratio is

(a+b)/a = a/b

so assuming that a is width, and b is height, you have: $$(960 + h)/960 = 960/h \\ 960^2 = 960h + h^2$$ a simple quadratic equation to solve, and the result is $$h = 480*(\sqrt 5-1) \approx 593.3$$

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It is not the whole diagram that is supposed to be golden; it's the four constituent rectangles. – Lord_Farin May 27 '13 at 13:22