Chapter 9.2, Problem 1WE

Solution

To complete: the given statement for JT .

If $OT=6$ and $JO=10$ , then $JT=\underset{\_}{8}$

Given:

The statement:

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

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 $JT=x$

Consider the right triangle $△OTJ$ .

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

So, by theorem 8-2,

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

  $\begin{array}{c}100-36=x^2\\ 64=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{64}\\ x=8\end{array}$

So, $JT=6$

Hence, the complete statement is:

If $OT=6$ and $JO=10$ , then $JT=\underset{\_}{8}$