Chapter 8.4, Problem 8MRPS

Solution

To calculate : The straight-line distance from person's home to the beach.

The straight-line distance from person's home to the beach is $20\text{ mi}$ .

Given information :

To get from home to beach a person drives $8$ miles south, then $16$ miles east and then $4$ miles south.

Calculation :

Make figure on the basis of the given information as below:

  

Here, A is the home of the person and E is the beach.

The straight-line distance from person's home to the beach is AE.

Here, $△ABC\cong △EDC$ as both are right-angled triangles.

Since, vertical angles are congruent, so $\angle ACB\cong \angle ECD$ .

If two angles of a triangle are similar with the corresponding two angles of another triangle then the third angle of both the triangles are also similar as the sum of the angles of a triangle is $180°$ .

So, $\angle BAC\cong \angle DEC$ .

This shows that $△ABC\sim △EDC$ and thus the corresponding sides of both the triangles are proportional. So,

  $\frac{AB}{ED}=\frac{BC}{DC}=\frac{CA}{CE}$   ....... (1)

From equation (1)

  $\begin{array}{c}\frac{AB}{ED}=\frac{BC}{DC}\\ \frac{8}{4}=\frac{BC}{16-BC}\\ 128-8BC=4BC\\ 12BC=128\\ BC=10.67\end{array}$

So, $DC=5.33$

Use Pythagoras Theorem to find $AC\text{ and }CE$ .

In $△ABC$ ,

  $\begin{array}{c}A{B}^2+B{C}^2=A{C}^2\\ A{C}^2=64+113.85\\ =177.85\\ AC=13.34\end{array}$

In $△EDC$ ,

  $\begin{array}{c}D{E}^2+D{C}^2=C{E}^2\\ C{E}^2=16+28.41\\ =44.41\\ CE=6.66\end{array}$

Now,

  $\begin{array}{c}AE=AC+CE\\ =13.34+6.66\\ =20\text{ mi}\end{array}$

Therefore, the straight-line distance from person's home to the beach is $20\text{ mi}$ .