+285 In triangle $PQR$, let $M$ be the midpoint of $\overline{PQ}$, let $N$ be the midpoint of $\overline{PR}$, and let $O$ be the intersection of $\overline{QN}$ and $\overline{RM}$, as shown. If $\overline{QN}\perp\overline{PR}$, $QN = 15$, and $PR = 20$, then find $QR$.
+1765 We are given that QN⊥PR and QN=15. Since N is the midpoint of PR, segment QN is an altitude of triangle PQR bisecting the base PR. Therefore, triangle PQR is a right triangle with right angle at R.
We are also given that PR=20. Now, we have a right triangle (PQR) with one leg length (QN=15) and the hypotenuse length (PR=20). We can use the Pythagorean Theorem to find the missing leg length (QR) :
PR2=QN2+QR2
202=152+QR2
QR2=202−152=25
Taking the square root of both sides (remembering there might be positive and negative square roots), we get:
QR=±5
Since QR represents a side length, we take the positive square root:
QR=5 units.
Therefore, the length of segment QR is 5 units.
