Chapter 8.1, Problem 37WE

To determine

Find x , y , and z .

Answer to Problem 37WE

  $x=\sqrt{2}$

  $y=2$

  $z=\sqrt{2}$

Explanation of Solution

Given:

  

Calculation:

The small right triangle with hypotenuse 2 is similar to the large right triangle with hypotenuse $\sqrt{2}+x$ .

So,

  $\begin{array}{l}\frac{2}{\sqrt{2}+x}=\frac{\sqrt{2}}{2}\\ \Rightarrow 2^2=\sqrt{2}\left(\sqrt{2}+x\right)\mathrm{...................}[\text{multiply each side by 2}\left(\sqrt{2}+x\right)\text{]}\\ \Rightarrow 4=2+\sqrt{2}x\\ \Rightarrow \sqrt{2}x=\mathrm{2.................................}[\text{subtract 2 from each side]}\\ \Rightarrow x=\frac{2}{\sqrt{2}}\mathrm{.....................}[\text{divide each side by }\sqrt{2}\text{]}\\ \Rightarrow x=\sqrt{2}\end{array}$

z is the geometric mean of $\sqrt{2}$ and x :

  $\begin{array}{l}z=\sqrt{\sqrt{2}\cdot x}\\ \Rightarrow z=\sqrt{\sqrt{2}\cdot \sqrt{2}}\\ \Rightarrow z=\sqrt{2}\end{array}$

The small right triangle with hypotenuse y is similar to the large right triangle with hypotenuse $\sqrt{2}+x$ .

So,

  $\begin{array}{l}\frac{y}{\sqrt{2}+x}=\frac{x}{y}\\ \Rightarrow y^2=x\left(\sqrt{2}+x\right)\mathrm{...................}[\text{multiply each side by y}\left(\sqrt{2}+x\right)\text{]}\\ \Rightarrow y^2=\sqrt{2}\left(\sqrt{2}+\sqrt{2}\right)\\ \Rightarrow y^2=\sqrt{2}\left(2\sqrt{2}\right)\\ \Rightarrow y^2=4\\ \Rightarrow y=\mathrm{2................................}[\text{take square root each side]}\end{array}$

So,

  $x=\sqrt{2}$

  $y=2$

  $z=\sqrt{2}$