Answer
See Below
Work Step by Step
(a)
It is known that the golden ration of rectangle \[A\] is $\text{ }\!\!\Phi\!\!\text{ =}\frac{\text{longer side}}{\text{shorter side}}$.
From above diagram, it can be seen that in triangle \[B\] the longer side is 1 and shorter side is $\Phi -1$.
So, the fractional expression that models the golden ratio in rectangle \[B\] is:
$\frac{1}{\Phi -1}$
Hence,fractional expression that models the golden ratio in rectangle \[B\] is $\frac{1}{\Phi -1}$.
(b)
The golden ration of rectangle \[A\] is $\text{ }\!\!\Phi\!\!\text{ }$.
The fractional expression that models the golden ratio in rectangle \[B\] is $\frac{1}{\Phi -1}$.
So, equate both the golden ratios as:
$\text{ }\!\!\Phi\!\!\text{ }=\frac{1}{\Phi -1}$
Solve the above equation as:
$\begin{align}
& \text{ }\!\!\Phi\!\!\text{ }\left( \Phi -1 \right)=1 \\
& {{\text{ }\!\!\Phi\!\!\text{ }}^{2}}-\text{ }\!\!\Phi\!\!\text{ }-\text{1}=0 \\
& \text{ }\!\!\Phi\!\!\text{ }=\frac{1\pm \sqrt{1+4}}{2} \\
& =\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}
\end{align}$
The golden ratio $\text{ }\!\!\Phi\!\!\text{ }$cannot be negative, so neglect it.
Hence, value of $\Phi =\frac{1+\sqrt{5}}{2}$.
(c)
It is known that the golden ration of rectangle \[A\] is $\text{ }\!\!\Phi\!\!\text{ =}\frac{\text{longer side}}{\text{shorter side}}$.
From the calculation in part (b), the golden ratio $\Phi $ is:
$\Phi =\frac{1+\sqrt{5}}{2}$
So, the longer side is $\frac{1+\sqrt{5}}{2}$ and shorter side is 1.
