Question

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

Short answer

The value is, $x=\frac{27}{5},y=\frac{48}{5},z=\frac{36}{5}$.

Step-by-step solution

Step 1. Given information.

The given figure is,

Step 2. Concept used.

The small right triangle with base \[x\] is similar to the large right triangle.

So,

$\begin{array}{c}\frac{9}{15}=\frac{x}{9}\\ 9^2=15x[\text{multiplyeachsideby15}\cdot \text{9]}\\ x=\frac{81}{15}[\text{divideeachsideby15]}\\ x=\frac{27}{5}\end{array}$

$\begin{array}{c}x+y=15\\ y=15-x[\text{subtractxfromeachside]}\\ y=15-\frac{27}{5}\\ y=\frac{48}{5}\end{array}$

Step 3. Use geometric mean concept.

z is the geometric mean of x and y .

$\begin{array}{c}z=\sqrt{x\cdot y}\\ z=\sqrt{\frac{27}{5}\cdot \frac{48}{5}}\\ z=\frac{36}{5}\end{array}$

Therefore, value is $x=\frac{27}{5},y=\frac{48}{5},z=\frac{36}{5}$.