Chapter 8.1, Problem 4CYP

To determine

To calculate: The height of the rock climbing wall.

Answer to Problem 4CYP

The height of the wall is $10\text{ feet}$

Explanation of Solution

From the given question;

The diagram drawn is:

  

Where,

AB is the height of the wall.

Point E is the position of eye which is 5 feet above the ground and 11 feet far from the wall.

That is,

EG = 5 feet = BF

BG = 11 feet

Draw $\text{EF}\perp \text{AB}$

Now,

In $\Delta \text{ EAF and }\Delta \text{ EBF}$

  $\begin{array}{c}\text{EF = EF }\left(\text{common side}\right)\\ \angle \text{ AEF = }\angle \text{ BEF }\left(\text{ angles from the eye to the top and to the bottom of wall are equal}\right)\\ \angle \text{ EFA = }\angle {\text{ EFB = 90}}^{\circ }\end{array}$

  $\therefore \Delta \text{ EAF }\cong \Delta \text{ EBF }\left(\text{congruent by ASA}\right)$

Thus,

  $\text{AF = BF = 5 feet}$

Hence,

The height of the wall is $\text{AB = AF+BF = 10 feet}$