MHB Help Me Solve the Golden Ratio Question

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    Golden ratio Ratio
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The discussion centers on solving a problem related to the golden ratio, specifically how to express the lengths of segments in terms of each other. The user seeks help in understanding the relationship between lengths AB (x) and BC (y) within a geometric context involving similar rectangles and a square. The mathematical derivation shows that the ratio of x to y equals the golden ratio, approximately 1.618, leading to the conclusion that x can be expressed as (1 + √5)/2 times y. The problem illustrates the application of proportions in geometry to derive the golden ratio. This solution highlights the importance of understanding geometric relationships in mathematical problems.
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Question is attached.
I don't know how to find the golden ratio

I really did try to solve this one, but i couldnt
please please help me out
 

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wrightarya said:
Question is attached.
I don't know how to find the golden ratio

I really did try to solve this one, but i couldnt
please please help me out

Call length AB = x and BC = y. Since AYXD is a square, that means length YB = x - y. Because the two rectangles are similar, their sides are in proportion, so

$\displaystyle \begin{align*} \frac{x}{y} &= \frac{y}{x - y} \\ x \left( x - y \right) &= y^2 \\ x^2 - x\,y &= y^2 \\ x^2 - x\,y + \left( -\frac{y}{2} \right) ^2 &= y^2 + \left( -\frac{y}{2} \right) ^2 \\ \left( x - \frac{y}{2} \right) ^2 &= \frac{4y^2}{4} + \frac{y^2}{4} \\ \left( x - \frac{y}{2} \right) ^2 &= \frac{ 5y^2}{4} \\ x - \frac{y}{2} &= \frac{\sqrt{5}\,y}{2} \\ x &= \frac{y + \sqrt{5}\,y}{2} \\ x &= \left( \frac{1 + \sqrt{5}}{2} \right) \, y \end{align*}$
 
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