Chapter 9.4, Problem 11WE

Solution

To find: distance of the chord $\overline{XY}$ from the center O of the circle.

  $2\sqrt{21}\text{ cm}$

Given:

Chord $\overline{XY}$ is 8 cm long and radius of the circle with center O is 10 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 $\overline{XY}$ be the required chord and let $\overline{ON}$ be perpendicular to the chord $\overline{XY}$ .

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

That is,

  $XN=NY$

Now it is given that $XY=8$ , so

  $\begin{array}{c}XY=8\\ XN+NY=8\\ NY+NY=8\end{array}$

  $\begin{array}{c}2NY=8\\ NY=\frac{8}{2}\\ NY=4\end{array}$

Let $ON=x$

Consider the right triangle $△ONY$ .

Now, $\overline{OY}$ is the hypotenuse and has length 10 and the lengths of the legs are $ON=x$ and $NY=4$ .

So, by theorem 8-2,

  $\begin{array}{c}O{Y}^2=O{N}^2+N{Y}^2\\ {10}^2=x^2+4^2\\ 100=x^2+16\end{array}$

  $\begin{array}{c}100-16=x^2\\ 84=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{84}\\ x=2\sqrt{21}\end{array}$

So, $ON=2\sqrt{21}\text{ cm}$

So, the distance of length of the chord $\overline{XY}$ from the center O of the circle is $2\sqrt{21}\text{ cm}$ .