Q17.

$\bar{J K}$ is tangent to $⊙ P$ and $⊙ Q$ .

$J K = ?$

Verified

Answer

The given quadrilateral $J P Q K$ must be a Trapezium.

1Step 1. Given information.

The figure here given is,

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

2Step 2. Draw the construction in figure.

The measures of the sides are,

$\begin{array}{l} P Q = 17 \\ Q K = 3 \\ P D = 3 \\ P J = 11 \end{array}$

$O T$ is required to find the measures of sides $J D$ and $J K$ .

Now,

$P J ⊥ J K$ And $Q K ⊥ J K$

Joint $P J$ and $Q K$ and make a line parallel to $P Q$ that is $D K$ .

Then, it implies that,

$P Q = D K (= 17)$ ........(1)

To find $D K$ ,

3Step 3. Concept used.

Triangle $J K D$ is a right angle triangle so one can use Pythagoras Theorem,

The measures of sides in this right angle triangle are,

$\begin{array}{l} D K = 17 \\ J D = 11 - 3 \\ = 8 \end{array}$

4Step 4. Use Pythagoras Theorem as follows.

$\begin{array}{l} D K ² = J K ² + J D ² \\ 17 ² = J K ² + 8 ² \\ 289 = J K ² + 64 \\ J K ² = 289 - 64 \\ = 225 \end{array}$

Take positive square to get,

$J K = 15$

So, the required length is $J K = 15$ .

In quadrilateral $P Q J K$ , sum one pair of adjacent angle is $180 °$ .

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