In triangle PQR, let M be the midpoint of QR, let N be the midpoint of PR, and let O be the intersection of QN and RM, as shown. If QN perp PR, QN = 12, and PR = 14, then find the area of triangle PQR.
+195 Since QN is perpendicular to PR, we have PN^2 + NQ^2 = PQ^2 by the Pythagorean theorem. Since N is the midpoint of PR, we have PN = NR = (1/2)PR = 7, and since QN = 12, we have NQ^2 = QM^2 = 144. Therefore, we have:
PQ^2 = PN^2 + NQ^2 = 7^2 + 144 = 193.
Since M is the midpoint of QR, we have QM = MR = (1/2)QR, so we can write:
[QOM] = (1/2)(QM)(ON) = (1/2)(1/2)(QR)(ON) = (1/4)[PQR][ON]/[PR]
Similarly, we have:
[RON] = (1/4)[PQR][OM]/[QR]
Adding these two equations, we get:
[QOM] + [RON] = (1/4)[PQR]([ON]/[PR] + [OM]/[QR])
Now, we can use the fact that O is the intersection of QN and RM, so QNOM and RONM are similar triangles. Therefore, we have:
[ON]/[PR] = [OM]/[QR]
Substituting this into the previous equation, we get:
[QOM] + [RON] = (1/4)[PQR]([ON]/[PR] + [OM]/[QR]) = (1/4)[PQR]([OM]/[QR] + [OM]/[QR]) = (1/2)[PQR][OM]/[QR]
Since O is the intersection of the medians RM and QN, we have:
[OM]/[QR] = 1/3
Therefore, we have:
[QOM] + [RON] = (1/2)[PQR]/3
But [QOM] + [RON] = [QNR] = (1/2)(QN)(NR) = (1/2)(12)(7) = 42, so we can write:
42 = (1/2)[PQR]/3
Therefore, we have:
[PQR] = 84.
Therefore, the area of triangle PQR is 84 square units.
