Question

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

Short answer

The value is, $x=18,y=12\sqrt{2},z=4\sqrt{2}$.

Step-by-step solution

Step 1. Given information.

The given figure is,

Step 2. Concept used.

The small right triangle with hypotenuse $6$ is similar to the large right triangle with hypotenuse $x$.

So,

$\begin{array}{c}\frac{6}{x}=\frac{2}{6}\\ 6^2=2x[\text{multiplyeachsideby6x]}\\ x=\frac{36}{2}[\text{divideeachsideby2]}\\ x=18\end{array}$

Step 3. Find  by Pythagoras Theorem.

$\begin{array}{c}{y}^2+6^2={18}^2\\ y^2=324-36[\text{subtract36fromeachside]}\\ y=\sqrt{288}[\text{takesquarerooteachside]}\\ y=\sqrt{2^2\cdot 6^2\cdot 2}\\ y=12\sqrt{2}\end{array}$

Step 4. Use geometric mean concept.

$z$ is the geometric mean of $2$ and $18-2$,

$\begin{array}{c}z=\sqrt{2\cdot 16}\\ z=4\sqrt{2}\end{array}$

Therefore, value is $x=18,y=12\sqrt{2},z=4\sqrt{2}$.