Chapter 8.4, Problem 2ST1

a.

To determine

To find: the value of x

Answer to Problem 2ST1

The value of x is 4

Explanation of Solution

Given Information: A triangle with a right angle and a perpendicular directed to the hypotenuse.

Formula used:

For a right triangle NDF with a perpendicular NE being drawn to the hypotenuse DF (as shown below), the relation between NE and the hypotenuse is expressed as,

  

  $\begin{array}{l}N{E}^2=DE\times EF\\ NE=\sqrt{DE\times EF}\end{array}$ ……… (1)

Calculation:

Consider the right triangle given below.

  

Here, applying the relation in (1), the value of x can be obtained as,

  $\begin{array}{c}x=\sqrt{2\times 8}\\ =\sqrt{16}\\ =4\end{array}$

b.

To determine

To find: the value of y

Answer to Problem 2ST1

The value of y is $y=2\sqrt{5}$

Explanation of Solution

Given Information: A triangle with a right angle and a perpendicular directed to the hypotenuse.

Formula used:

For a right triangle DNF with a perpendicular NE being drawn to the hypotenuse DF (as shown below), the relation between the one of the legs of the bigger triangle and the normal to it is expressed as, $N{D}^2=DE\times DF$

  

  $ND=\sqrt{DE\times DF}$ ………… (2)

Calculation:

Consider the right triangle given below. Comparing it with the general figure for right triangle in the above description,

  $\begin{array}{c}DF=DE+EF\\ =2+8\\ =10\end{array}$

  

Here, applying the relation in (2), the value of y can be obtained as,

  $\begin{array}{c}y=\sqrt{2\times 10}\\ =\sqrt{20}\\ y=2\sqrt{5}\end{array}$

c.

To determine

To find: the value of z

Answer to Problem 2ST1

The value of z is $z=4\sqrt{5}$

Explanation of Solution

Given Information: A triangle with a right angle and a perpendicular directed to the hypotenuse.

Formula used:

For a right triangle DNF with a perpendicular NE being drawn to the hypotenuse DF (as shown below), the relation between the one of the legs of the bigger triangle and the normal to it is expressed as,

  

  $N{F}^2=EF\times DF$

  $NF=\sqrt{EF\times DF}$ …………. (3)

Calculation:

The length of hypotenuse DF as compared to the general figure was obtained in the previous part as $DF=10$ .

  

Here, applying the relation in (3), the value of z can be obtained as,

  $\begin{array}{c}z=\sqrt{8\times 10}\\ =\sqrt{80}\\ z=4\sqrt{5}\end{array}$