Question

Find the values of $x,y,$and $z$.

Short answer

Value of,

$\begin{array}{l}x=4\\ y=2\sqrt{5}\\ z=3\sqrt{5}\end{array}$

Step-by-step solution

Step 1. Given information.

The given figure is,

Step 2. Concept used.

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

So,

role="math" localid="1648873824994" $\begin{array}{c}\frac{x+2}{2x+1}=\frac{x}{x+2}\\ {\left(x+2\right)}^2=x\left(2x+1\right)[\text{multiplyeachsideby(x+2)}\left(2x+1\right)\text{]}\\ x^2+4+4x=2x^2+x\\ x^2-3x-4=0[\text{subtract}{x}^2+4+4x\text{fromeachside]}\\ x=\frac{-(-3)\pm \sqrt{{(-3)}^2-4\left(1\right)(-4)}}{2\left(1\right)}\\ x=\frac{3+5}{2}[\text{lengthispositive]}\\ x=4\end{array}$

Step 3. Use geometric mean concept.

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

$\begin{array}{c}y=\sqrt{4\left(5\right)}\\ =2\sqrt{5}\end{array}$

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

So,

$\begin{array}{c}\frac{z}{9}=\frac{x+1}{z}\\ z^2=9\left(x+1\right)[\text{multiplyeachsideby}9z\text{]}\\ z=\sqrt{9(x+1)}[\text{takesquarerooteachside]}\\ z=3\sqrt{5}\end{array}$

Therefore, the value of,

$\begin{array}{c}x=4\\ y=2\sqrt{5}\\ z=3\sqrt{5}\end{array}$