Question

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

Short answer

The value is, $x=\sqrt{\frac{1}{18}},y=\sqrt{\frac{1}{12}},z=\sqrt{\frac{1}{6}}$.

Step-by-step solution

Step 1. Given information.

The given figure is,

Step 2. Use geometric mean concept for x.

$x$is the geometric mean of $\frac{1}{3}$ and $\frac{1}{6}$.

So,

$\begin{array}{c}x=\sqrt{\frac{1}{3}\cdot \frac{1}{6}}\\ =\sqrt{\frac{1}{18}}\end{array}$

Step 3. Now, Use Pythagoras Theorem in right angled triangles to find  y and z  :

$\begin{array}{c}{z}^2={\left(\frac{1}{3}\right)}^2+{\left(\sqrt{\frac{1}{18}}\right)}^2\\ z^2=\frac{1}{9}+\frac{1}{18}\\ z^2=\frac{3}{18}\\ z=\sqrt{\frac{1}{6}}[\text{takesquarerooteachside]}\end{array}$

$\begin{array}{c}{y}^2={\left(\frac{1}{6}\right)}^2+{\left(\sqrt{\frac{1}{18}}\right)}^2\\ y^2=\frac{1}{36}+\frac{1}{18}\\ y^2=\frac{3}{36}\\ y=\sqrt{\frac{1}{12}}[\text{takesquarerooteachside]}\\ y=12\end{array}$

Therefore, the value is, $x=\sqrt{\frac{1}{18}},y=\sqrt{\frac{1}{12}},z=\sqrt{\frac{1}{6}}$.