Chapter 9.4, Problem 5ST1

Solution

To find: length of the chord

  $4\sqrt{10}$

Given:

Chord is 3 cm from the center of the circle of radius 7 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 $\overline{PR}$ be the required chord and let $\overline{OQ}$ be perpendicular to the chord $\overline{PR}$ .

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

That is,

  $PQ=QR$

Let $QR=x$

Consider the right triangle $△OQR$ .

Now, $\overline{OR}$ is the hypotenuse and has length 7 and the lengths of the legs are $QR=x$ and $OQ=3$ .

So, by theorem 8-2,

  $\begin{array}{c}O{R}^2=O{Q}^2+Q{R}^2\\ 7^2=3^2+x^2\\ 49=9+x^2\end{array}$

  $\begin{array}{c}49-9=x^2\\ 40=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{40}\\ x=2\sqrt{10}\end{array}$

So, $QR=2\sqrt{10}$

Then,

$\begin{array}{c}PR=PQ+QR\\ =QR+QR\\ =2\sqrt{10}+2\sqrt{10}\\ =4\sqrt{10}\end{array}$ [ $\because PQ=QR$ ]

So, the length of the chord is $4\sqrt{10}$ .