Help Me Solve the Golden Ratio Question

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    Golden ratio Ratio
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SUMMARY

The discussion focuses on solving the golden ratio problem using geometric proportions. The key relationship derived is that if AB = x and BC = y, then the golden ratio can be expressed as x = (1 + √5)/2 * y. This conclusion is reached through the establishment of similar rectangles and the application of algebraic manipulation to derive the relationship between x and y. The final formula clearly defines the golden ratio in terms of y.

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  • Understanding of geometric proportions and similar figures
  • Basic algebraic manipulation skills
  • Familiarity with the concept of the golden ratio
  • Knowledge of square properties and relationships
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  • Study the derivation of the golden ratio in various geometric contexts
  • Explore applications of the golden ratio in design and art
  • Learn about the Fibonacci sequence and its connection to the golden ratio
  • Investigate advanced algebraic techniques for solving ratio problems
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Mathematicians, geometry enthusiasts, students studying algebra, and anyone interested in the applications of the golden ratio in various fields.

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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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