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The golden rectangle with the area of 8cm2  is inscribed in a circle. What's the area of that circle?

 Dec 13, 2019
 #1
avatar+23905 
+4

The golden rectangle with the area of 8cm2  is inscribed in a circle.

What's the area of that circle?

 

\(\begin{array}{|rcll|} \hline \dfrac{x}{y} &=& \varphi \quad | \quad \boxed{\varphi =\dfrac{1+\sqrt{5}}{2}= 1.61803398875 } \\ x = y\varphi &\text{or}& x^2 = y^2\varphi^2 \\ \hline xy &=& 8 \\ y\varphi y &=& 8 \\ y^2 &=& \dfrac{8}{\varphi} \quad | \quad \dfrac{1}{\varphi} = \varphi - 1 \\ \mathbf{y^2} &=& \mathbf{8(\varphi - 1)} \\ \hline x^2 &=& y^2\varphi^2 \quad | \quad y^2=\dfrac{8}{\varphi} \\ x^2 &=&\dfrac{8}{\varphi} \varphi^2 \\ \mathbf{x^2} &=& \mathbf{8\varphi} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (2r)^2 &=& x^2+y^2 \\ (2r)^2 &=& 8\varphi+8(\varphi - 1) \\ 4r^2 &=& 16\varphi-8 \quad | \quad : 4 \\ \mathbf{r^2} &=& \mathbf{4\varphi-2} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \text{area}_{\text{circle}} &=& \pi r^2 \\ &=& (4\varphi-2)\pi \quad | \quad \varphi = 1.61803398875 \\ &=& 4.47213595500\pi \quad | \quad \pi = 3.14159265359 \\ \mathbf{\text{area}_{\text{circle}}} &=& \mathbf{14.0496294621~ \text{cm}^2} \\ \hline \end{array}\)

 

The area of that circle is \(\approx \mathbf{14~ \text{cm}^2}\)

 

laugh

 Dec 13, 2019
edited by heureka  Dec 13, 2019
 #2
avatar+178 
+1

Shortcut:

There's a certain ratio betveen the area of a golden rectangle and the area of a circle in which it's been inscribed.

That ratio is:  ≈ 1 : 1.756

So, we have:    8 x 1.756 = 14.048

smiley

 Dec 14, 2019
edited by Dragan  Dec 14, 2019
edited by Dragan  Dec 14, 2019

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