Chapter 9.4, Problem 12WE

Solution

To find: radius of the circle with the center Q .

  $2\sqrt{17}\text{ cm}$

Given:

Chord $\overline{RS}$ is 16 cm long and is at a distance of 2 cm from of the center Q of the circle.

Concept Used:

Theorem 8-2 The Square of the hypotenuse of right triangle is equal to the sum of the squares of the sides in a right triangle..

Explanation:

Consider the figure representing the given question:

Let $\overline{RS}$ be the required chord and let $\overline{QP}$ be perpendicular to the chord $\overline{RS}$ .

Then, by theorem 9-5, it can be said that $\overline{QP}$ bisects $\overline{RS}$ .

That is,

  $RP=PS$

Now it is given that $RS=16$ , so

  $\begin{array}{c}RS=16\\ RP+PS=16\\ PS+PS=16\end{array}$

  $\begin{array}{c}2PS=16\\ PS=\frac{16}{2}\\ PS=8\end{array}$

Let $QS=x$

Consider the right triangle $△QPS$ .

Now, $\overline{QS}$ is the hypotenuse and has length x and the lengths of the legs are $QP=2$ and $PS=8$ .

So, by theorem 8-2,

  $\begin{array}{c}Q{S}^2=Q{P}^2+P{S}^2\\ x^2=2^2+8^2\\ x^2=4+64\\ x^2=68\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{68}\\ x=2\sqrt{17}\end{array}$

So, $QS=2\sqrt{17}\text{ cm}$

So, the radius of the circle is $2\sqrt{17}\text{ cm}$ .