Chapter 8.1, Problem 31WE

To determine

Find x , y , and z .

Answer to Problem 31WE

x = 10

  $y=2\sqrt{29}$

  $z=5\sqrt{29}$

Explanation of Solution

Given:

  

Calculation:

x is the geometric mean of 4 and 25.

So, $x=\sqrt{4\cdot 25}=\sqrt{2^2\cdot 5^2}=10$

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

  $\begin{array}{l}{z}^2={25}^2+{10}^2\\ \Rightarrow z^2=625+100\\ \Rightarrow z^2=725\\ \Rightarrow z=\sqrt{25\cdot 29}\mathrm{..............}[\text{take square root each side]}\\ \Rightarrow z=5\sqrt{29}\end{array}$

  $\begin{array}{l}{y}^2=4^2+{10}^2\\ \Rightarrow y^2=16+100\\ \Rightarrow y^2=116\\ \Rightarrow y=\sqrt{4\cdot 29}\mathrm{..............}[\text{take square root each side]}\\ \Rightarrow y=2\sqrt{29}\end{array}$

So,

x = 10

  $y=2\sqrt{29}$

  $z=5\sqrt{29}$