Question

$\overline{JT}$is tangent to $\odot$$O$ at $T$. Complete.

If $OT=6$ and $JT=10$, then $JO=?$

Short answer

If $OT=6$and $JT=10$, then $JO=11.7$.

Step-by-step solution

Step 1. Given information.

The statement here given is,

If $OT=6$and $JT=10$, then $JO=?$

The figure:

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

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

Step 2. Concept Used.

According to theorem $9·2$,

Tangent of the circle is perpendicular to the radius of the circle.

According to 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.

Step 3. 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 $\Delta OTJ$.

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 $8·2$,

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

Step 4. Now, take square root of both sides.

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

So,$JO=11.7$

Therefore, the complete statement is,

If $OT=6$ and $JT=10$, then $JO=11.7$.