Chapter 10.3, Problem 70HP

To determine

Find $h^4$ in terms of x and y

Answer to Problem 70HP

  $h^4=x^2{y}^2$

Explanation of Solution

Given:

  

Use the fact that h is the geometric mean between x and y in the figure at the right to find $h^4$ in terms of x and y

Concept Used:

  

The Triangle CDB and the triangle ADC are similar.

Therefore, we can write:

  $\begin{array}{l}{\text{AC}}^2={\text{AD}}^2+{\text{CD}}^{\text{2}}\Rightarrow x^2+h^2\\ {\text{BC}}^2={\text{CD}}^{\text{2}}+{\text{BD}}^{\text{2}}\Rightarrow h^2+y^2\end{array}$

Calculation:

From the triangles CBD and ADC we already get the relation:

  $\begin{array}{l}{\text{AC}}^2={\text{AD}}^2+{\text{CD}}^{\text{2}}\Rightarrow x^2+h^2\\ {\text{BC}}^2={\text{CD}}^{\text{2}}+{\text{BD}}^{\text{2}}\Rightarrow h^2+y^2\end{array}$

Now from the right triangle ABC

  $\begin{array}{l}$ {\text{AC}}^{\text{2}}\text{+}{\text{BC}}^{\text{2}}={\text{AB}}^{\text{2}}={(\text{AD+BD})}^2\Rightarrow {(x+y)}^2$$ {(x+y)}^2={\text{AC}}^{\text{2}}\text{+}{\text{BC}}^{\text{2}}=x^2+h^2+h^2+y^2$x^2+2xy+y^2=x^2+2h^2+y^2\\ 2xy=2h^2\\ xy=h^2\\ $ {(h^2)}^2={(xy)}^2$h^4=x^2{y}^2\end{array}$

Thus, $h^4=x^2{y}^2$