Chapter 8.1, Problem 32WE

To determine

Find x , y , and z .

Answer to Problem 32WE

  $x=3\sqrt{7}$

  $y=12$

  $z=4\sqrt{7}$

Explanation of Solution

Given:

  

Calculation:

x is the geometric mean of 9 and 7.

So, $x=\sqrt{9\cdot 7}=\sqrt{3^2\cdot 7}=3\sqrt{7}$

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

  $\begin{array}{l}{z}^2=7^2+{\left(3\sqrt{7}\right)}^2\\ \Rightarrow z^2=49+63\\ \Rightarrow z^2=112\\ \Rightarrow z=\sqrt{4^2\cdot 7}\mathrm{..............}[\text{take square root each side]}\\ \Rightarrow z=4\sqrt{7}\end{array}$

  $\begin{array}{l}{y}^2=9^2+{\left(3\sqrt{7}\right)}^2\\ \Rightarrow y^2=81+63\\ \Rightarrow y^2=144\\ \Rightarrow y=\sqrt{144}\mathrm{..............}[\text{take square root each side]}\\ \Rightarrow y=12\end{array}$

So,

  $x=3\sqrt{7}$

  $y=12$

  $z=4\sqrt{7}$