Chapter 9.4, Problem 13WE

Solution

To find: length of the chord $\overline{AB}$ .

  $AB=\sqrt{21}$

Given:

Chord $\overline{AB}$ is 2 cm from the center P of the circle of radius 5 cm.

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 $\overline{AB}$ be the required chord and let $\overline{PC}$ be perpendicular to the chord $\overline{AB}$ .

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

That is,

  $AC=CB$

Let $CB=x$

Consider the right triangle $△PCB$ .

Now, $\overline{PB}$ is the hypotenuse and has length 5 and the lengths of the legs are $CB=x$ and $PC=2$ .

So, by theorem 8-2,

  $\begin{array}{c}P{B}^2=P{C}^2+C{B}^2\\ 5^2=2^2+x^2\\ 25=4+x^2\end{array}$

  $\begin{array}{c}25-4=x^2\\ 21=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{21}\\ x=\sqrt{21}\end{array}$

So, $CB=\sqrt{21}$

Then,

$\begin{array}{c}AB=AC+CB\\ =CB+CB\\ =\sqrt{21}+\sqrt{21}\\ =2\sqrt{21}\end{array}$ [ $\because AC=CB$ ]

So, the length of the chord $\overline{AB}$ is $\sqrt{21}$ .