Chapter 8.1, Problem 39WE

To determine

Find x , y , and z .

Answer to Problem 39WE

  $x=4$

  $y=2\sqrt{5}$

  $z=3\sqrt{5}$

Explanation of Solution

Given:

  

Calculation:

The small right triangle with hypotenuse x + 2 is similar to the large right triangle with hypotenuse 2x + 1 .

So,

  $\begin{array}{l}\frac{x+2}{2x+1}=\frac{x}{x+2}\\ $ \Rightarrow {(x+2)}^2=x\left(2x+1\right)\mathrm{...................}[\text{multiply each side by (x+2)}\left(2x+1\right)\text{]}$\Rightarrow x^2+4+4x=2x^2+x\\ \Rightarrow x^2-3x-4=\mathrm{0...........................}[\text{subtract }{x}^2+4+4x\text{ from each side]}\\ \Rightarrow x=\frac{-(-3)\pm \sqrt{{(-3)}^2-4(1)(-4)}}{2(1)}\\ \Rightarrow x=\frac{3+5}{2}\mathrm{.....................................}[\text{length is positive]}\\ \Rightarrow x=4\end{array}$

y is the geometric mean of x = 4and x + 1= 5:

  $\begin{array}{l}y=\sqrt{4(5)}\\ \Rightarrow y=2\sqrt{5}\end{array}$

The small right triangle with hypotenuse z is similar to the large right triangle with hypotenuse 2x + 1 = 9 .

So,

  $\begin{array}{l}\frac{z}{9}=\frac{x+1}{z}\\ \Rightarrow z^2=9\left(x+1\right)\mathrm{...................}[\text{multiply each side by }9z\text{]}\\ \Rightarrow z=\sqrt{9(x+1)}\mathrm{..................}[\text{take square root each side]}\\ \Rightarrow z=3\sqrt{5}\end{array}$

So,

  $x=4$

  $y=2\sqrt{5}$

  $z=3\sqrt{5}$