Question

$\overline{JK}$ is tangent to $\odot P$ and $\odot Q$.

$JK=?$

Short answer

The given quadrilateral $JPQK$ must be a Trapezium.

Step-by-step solution

Step 1. Given information.

The figure here given is,

$JK$ is tangent to the circle with centers $P$ and $Q$.

Step 2. Draw the construction in figure.

The measures of the sides are,

$\begin{array}{l}PQ=17\\ QK=3\\ PD=3\\ PJ=11\end{array}$

$OT$ is required to find the measures of sides $JD$and $JK$.

Now,

$PJ\perp JK$ And $QK\perp JK$

Joint $PJ$and $QK$and make a line parallel to $PQ$ that is $DK$.

Then, it implies that,

$PQ=DK\left(=17\right)$ ........(1)

To find $DK$,

Step 3. Concept used.

Triangle $JKD$ is a right angle triangle so one can use Pythagoras Theorem,

The measures of sides in this right angle triangle are,

$\begin{array}{l}DK=17\\ JD=11–3\\ =8\end{array}$

Step 4. Use Pythagoras Theorem as follows.

$\begin{array}{l}DK²=JK²+JD²\\ 17²=JK²+8²\\ 289=JK²+64\\ JK²=289–64\\ =225\end{array}$

Take positive square to get,

$JK=15$

So, the required length is $JK=15$.

In quadrilateral $PQJK$, sum one pair of adjacent angle is $180°$.

Therefore, it can be concluded that the opposite sides is parallel.