Chapter 9.2, Problem 2WE

Solution

To completer: the given statement.

If $OT=6$ and $JT=10$ , then $JO=\underset{\_}{11.7}$

Given:

The statement:

If $OT=6$ and $JT=10$ , then $JO=\_\_\_\_$

The figure:

  $JT$ is a tangent to circle with centre O at T .

Concept Used:

Theorem 9-2: tangent of the circle is perpendicular to the radius of the circle.

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 given figure:

It is given that $JT$ is a tangent to circle with centre O at T .

So, by theorem 9-2, it can be said that $\overline{JT}$ is perpendicular to $\overline{OT}$ .

That is,

  $m\angle OTJ=90$

Let $JO=x$

Consider the right triangle $△OTJ$ .

Now, $\overline{JO}$ is the hypotenuse and has length x and the lengths of the legs are $OT=6$ and $JT=10$ .

So, by theorem 8-2,

  $\begin{array}{c}J{O}^2=O{T}^2+J{O}^2\\ x^2=6^2+{10}^2\\ x^2=36+100\\ x^2=136\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{136}\\ x=11.7\end{array}$

So, $JO=11.7$

Hence, the complete statement is:

If $OT=6$ and $JT=10$ , then $JO=\underset{\_}{11.7}$