Question

given that point c is the circumcenter of △lmn, what is the length of qn? √11 5 √5 √17

Explanation:

Step1: Recall circumcenter property

The circum - center is equidistant from the vertices of the triangle. Let $CM = CN=CL$. Consider right - triangle $CMQ$ and $CNR$.
In right - triangle $CMQ$, assume $CM = r$, $MQ = 5$, and in right - triangle $CNR$, $CR=\sqrt{17}$, $RN=\sqrt{19}$.
Let's use the Pythagorean theorem in right - triangle $CMQ$: $CM^{2}=CQ^{2}+MQ^{2}$. In right - triangle $CNR$, $CN^{2}=CR^{2}+RN^{2}$. Since $CM = CN$, we have $CQ^{2}+MQ^{2}=CR^{2}+RN^{2}$.

Step2: Calculate $CQ^{2}$

We know $MQ = 5$, $CR=\sqrt{17}$, $RN=\sqrt{19}$. Let $CQ^{2}=x$. Then $x + 5^{2}=\sqrt{17}^{2}+\sqrt{19}^{2}$.
$x+25 = 17 + 19$.
$x+25=36$.
$x = 11$.

Step3: Find $QN$ in right - triangle $CQN$

In right - triangle $CQN$, since $CQ^{2}=11$ and $CN$ is the circum - radius and $CQ\perp MN$ (a property related to the circum - center and perpendicular bisectors), and using the Pythagorean theorem in right - triangle $CQN$ with $CQ$ and $CN$ (where $CN^{2}=CQ^{2}+QN^{2}$ and considering the right - angled nature of $\triangle CQN$). Since $CQ^{2}=11$ and $CN$ is the circum - radius and we are interested in $QN$, and we know that in right - triangle $CQN$ with right - angle at $Q$, by the Pythagorean theorem, if we assume $CQ$ is one leg and $QN$ is the other leg and $CN$ is the hypotenuse. And since we found $CQ^{2}=11$, and considering the right - triangle formed by $C$, $Q$, and $N$, $QN=\sqrt{CN^{2}-CQ^{2}}$. Also, from the fact that we can consider the right - triangle $CMQ$ and $CQN$ relationships, and since we know the values from the previous steps, in right - triangle $CQN$ with $CQ$ as a leg, $QN=\sqrt{11}$.

Answer:

A. $\sqrt{11}$