In this exercise we are asked to find the measure of $J Q .$ Let's find this segment on the given diagram.

From the diagram we can see that $\overline{Q K},$ $\overline{Q M},$ and $\overline{Q P}$ are perpendicular to the sides of the triangle. This means that these segments are the distances from $Q$ to the sides of $△ J L N .$ Also, we are told that $Q$ is the incenter of $△ J L N .$ Thus, we can use the Incenter Theorem.

The angle bisectors of a triangle intersect at a point called the incenter that is equidistant from the sides of the triangle.

According to the theorem, the incenter

Q

is equidistant from the sides of the triangle

△ J L N .

This means that distances

Q K,

Q M,

and

Q P

are the same.

Q K = Q M = Q P

We also know that the measure of

Q M

is

9 .

Therefore, the measure of

Q P

is

9

too. Now, let's consider the triangle

△ J Q P .

△ J Q P

is a right triangle, because

\overline{Q P}

is perpendicular to

\overline{J P} .

Thus, we can use the Pythagorean Theorem.

a^{2} + b^{2} = c^{2}

In our case, the legs of the triangle are

\overline{J P}

and

\overline{Q P},

and the hypotenuse is

\overline{J Q},

so this theorem can be written the following way.

J P^{2} + Q P^{2} = J Q^{2}

We know the length of the two triangle's legs. Let's substitute

16.5

for

J P

and

9

for

Q P,

and solve the equation we will get for

J Q .

J P^{2} + Q P^{2} = J Q^{2}

SubstituteII

$J P = 16.5$ , $Q P = 9$

16.5^{2} + 9^{2} = J Q^{2}

CalcPow

Calculate power

272.25 + 81 = J Q^{2}

AddTerms

Add terms

353.25 = J Q^{2}

RearrangeEqn

Rearrange equation

J Q^{2} = 353.25

SqrtEqn

$LHS = RHS$

J Q = 18.79494 \dots

RoundDec

Round to $1$ decimal place(s)

J Q \approx 18.8

The measure of

J Q

is about

18.8 .

Exercise