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Geometry

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

$Math Studyset Vaia Explanations$ Math

Student Edition · 227 pages

ISBN: 9780395977279

Chapter 8: Q33. (page 289)

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

Short Answer

Expert verified

The value is, $x = \sqrt{\frac{1}{18}}, y = \sqrt{\frac{1}{12}}, z = \sqrt{\frac{1}{6}}$ .

Step by step solution

01

Step 1. Given information.

The given figure is,

02

Step 2. Use geometric mean concept for x.

$x$ is the geometric mean of $\frac{1}{3}$ and $\frac{1}{6}$ .

So,

$\begin{matrix} x = \sqrt{\frac{1}{3} \cdot \frac{1}{6}} \\ = \sqrt{\frac{1}{18}} \end{matrix}$

03

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

$\begin{matrix} z^{2} = {(\frac{1}{3})}^{2} + {(\sqrt{\frac{1}{18}})}^{2} \\ z^{2} = \frac{1}{9} + \frac{1}{18} \\ z^{2} = \frac{3}{18} \\ z = \sqrt{\frac{1}{6}} [takesquarerooteachside] \end{matrix}$

$\begin{matrix} y^{2} = {(\frac{1}{6})}^{2} + {(\sqrt{\frac{1}{18}})}^{2} \\ y^{2} = \frac{1}{36} + \frac{1}{18} \\ y^{2} = \frac{3}{36} \\ y = \sqrt{\frac{1}{12}} [takesquarerooteachside] \\ y = 12 \end{matrix}$

Therefore, the value is, $x = \sqrt{\frac{1}{18}}, y = \sqrt{\frac{1}{12}}, z = \sqrt{\frac{1}{6}}$ .

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Most popular questions from this chapter

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