Q37. Find the values of x,y,and z.... [FREE SOLUTION]

91影视

Geometry

Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

$Math Studyset 91影视 Explanations$ Math

Student Edition 路 227 pages

ISBN: 9780395977279

Chapter 8: Q37. (page 289)

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

Short Answer

Expert verified

Value of,

$\begin{matrix} x = \sqrt{2} \\ y = 2 \\ z = \sqrt{2} \end{matrix}$

Step by step solution

Step 1. Given information.

The given figure is,

Step 2. Concept used.

The small right triangle with hypotenuse $2$ is similar to the large right triangle with hypotenuse $\sqrt{2} + x$ .

So,

$\begin{matrix} \frac{2}{\sqrt{2} + x} = \frac{\sqrt{2}}{2} \\ 2^{2} = \sqrt{2} (\sqrt{2} + x) [multiplyeachsideby2 (\sqrt{2} + x)] \\ 4 = 2 + \sqrt{2} x \\ \sqrt{2} x = 2 [subtract2fromeachside] \\ x = \frac{2}{\sqrt{2}} [divideeachsideby \sqrt{2}] \\ x = \sqrt{2} \end{matrix}$

Step 3. Use geometric mean concept.

$z$ is the geometric mean of $\sqrt{2}$ and $x$ .

$\begin{matrix} z = \sqrt{\sqrt{2} 鈰� x} \\ z = \sqrt{\sqrt{2} 鈰� \sqrt{2}} \\ z = \sqrt{2} \end{matrix}$

The small right triangle with hypotenuse $y$ is similar to the large right triangle with hypotenuse $\sqrt{2} + x$ .

So,

$\begin{matrix} \frac{y}{\sqrt{2} + x} = \frac{x}{y} \\ y^{2} = x (\sqrt{2} + x) ................... [multiplyeachsidebyy (\sqrt{2} + x)] \\ y^{2} = \sqrt{2} (\sqrt{2} + \sqrt{2}) \\ y^{2} = \sqrt{2} (2 \sqrt{2}) \\ y^{2} = 4 \\ y = 2................................ [takesquarerooteachside] \end{matrix}$