Chapter 8, Problem 5TP

Solution

To calculate: The distance $d$ along the edge of the penalty area between the endpoints of the arc.

The distance $d=16 yards$ .

Given information:

A soccer field has a penalty- kick mark $12yards$ from the goal. From the mark’s center, an arc with a radius of $10yards$ is drawn outside the penalty area.

Calculation:

Let $C$ be the center of the circle, $A$ its orthogonal projection onto the line segment $d$ and $B$ any of the two endpoints of the segment $d$ . Then $ABC$ is a right triangle. The

  $\begin{array}{c} \left|CA\right|=18-12=6\\ \left|CB\right|=10\end{array}$

Because it's the distance from the center to some point on the circle, so it must be equal to the radius of the circle also have that $\left|AB\right|=\frac{d}{2}$ because if $D$ is the other endpoint of $d$ , the triangle $CBD$ is isosceles.

The line segment $\overline{CB}$ is the hypotenuse of the triangle $A B C$ because it is opposite of the right angle $\angle BCA$ . Applying the Pythagorean Theorem, get that

  $|CB{|}^2=|CA{|}^2+|AB{|}^2$

Substituting

  $\begin{array}{c}\left|CB\right|=10\\ \left|CA\right|=6\\ \left|AB\right|=d/2 \end{array}$

The following equation for $d$

  $\begin{array}{c} {10}^2=6^2+\frac{d}{2}\\ 100=36+\frac{d^2}{4}\\ 100-36=\frac{d^2}{4}\\ 64=\frac{d^2}{4}\\ 256=d^2\\ \pm 16=d \end{array}$

Since $d$ is the length of a line segment, consider the positive solution, so $d=16 yards$ .