The ratio of the width, w , of the Temple of Hephaestus to its height, h , is $\frac { w } { h } = \frac { 2 h } { \sqrt { 5 } - 1 }$ . This number is called the golden section. Early Greeks believed that the most aesthetically pleasing rectangle had this ratio. If the height is 5 , find the width. Rationalize the denominator and simplify the expression.
A) $\frac { 5 \sqrt { 5 } } { 2 }$
B) $\frac { \sqrt { 5 } + 5 } { 2 }$
C) $\frac { 5 \sqrt { 5 } - 5 } { 2 }$
D) $\frac { 5 \sqrt { 5 } + 5 } { 4 }$
E) $\frac { 5 \sqrt { 5 } + 5 } { 2 }$

Question

The ratio of the width, w , of the Temple of Hephaestus to its height, h , is $\frac { w } { h } = \frac { 2 h } { \sqrt { 5 } - 1 }$ . This number is called the golden section. Early Greeks believed that the most aesthetically pleasing rectangle had this ratio. If the height is 5 , find the width. Rationalize the denominator and simplify the expression. 
A) $\frac { 5 \sqrt { 5 } } { 2 }$ 
B) $\frac { \sqrt { 5 } + 5 } { 2 }$ 
C) $\frac { 5 \sqrt { 5 } - 5 } { 2 }$ 
D) $\frac { 5 \sqrt { 5 } + 5 } { 4 }$ 
E) $\frac { 5 \sqrt { 5 } + 5 } { 2 }$

Quizplus · Accepted Answer

To find the width, we need to solve for w in the given equation:
\begin{align*}
\frac{w}{h}&=\frac{2h}{\sqrt{5}-1}\
w&=\frac{2h}{\frac{1}{2}(\sqrt{5}-1)}\
w&=\frac{4h}{\sqrt{5}-1}\
w&=\frac{4h}{\sqrt{5}-1}\cdot\frac{\sqrt{5}+1}{\sqrt{5}+1}\
w&=\frac{4h(\sqrt{5}+1)}{4}\
w&=h(\sqrt{5}+1)
\end{align*}
Substituting h = 5, we get:
\begin{align*}
w&=5(\sqrt{5}+1)\
w&=5\sqrt{5}+5
\end{align*}
Therefore, the width of the Temple of Hephaestus is $\boxed{	ext{(E) }\frac{5\sqrt{5}+5}{2}}$, which is the simplified expression for $5\sqrt{5}+5$.

The Ratio of the Width, W , of the Temple wh=2h5−1\frac { w } { h } = \frac { 2 h } { \sqrt { 5 } - 1 }hw​=5​−12h​

The Ratio of the Width, W , of the Temple $\frac { w } { h } = \frac { 2 h } { \sqrt { 5 } - 1 }$