Question

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

Short answer

The value is $x=3\sqrt{7},y=12,z=4\sqrt{7}$.

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 $9$ and $7$.

So,

$\begin{array}{c}x=\sqrt{9\cdot 7}\\ =\sqrt{3^2\cdot 7}\\ =3\sqrt{7}\end{array}$

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

Substitute the values and calculate.

$\begin{array}{c}{z}^2=7^2+{\left(3\sqrt{7}\right)}^2\\ z^2=49+63\\ z^2=112\\ z=\sqrt{4^2\cdot 7}\mathrm{..............}[\text{takesquarerooteachside]}\\ z=4\sqrt{7}\end{array}$

And,

$\begin{array}{c}{y}^2=9^2+{\left(3\sqrt{7}\right)}^2\\ y^2=81+63\\ y^2=144\\ y=\sqrt{144}\mathrm{..............}[\text{takesquarerooteachside]}\\ y=12\end{array}$

Therefore, the value is, $x=3\sqrt{7},y=12,z=4\sqrt{7}$.