In triangle $PQR$, we have $PQ = PR = 13$ and $QR = 10$. Let $M$ be the midpoint of $\overline{PQ}$ and $N$ be on $\overline{QR}$ such that $\overline{PN}$ is an altitude of triangle $PQR$. If $\overline{PN}$ and $\overline{RM}$ intersect at $X$, then what is $PX$?
Drop the perpendicular from M to QR at Y. Then MY = 7/2, so PX = 3*MY = 3*7/2 = 21/2.
+128407 See the following image :
PQR is isosceles with PQ = PR.....so QN = (1/2) (QR) = (1/2)(10) = 5
By the Pythagorean Theorem
sqrt ( QP^2 - QN^2) = PN
sqrt ( 13^2 - 5^2) = sqrt ( 169 - 25) = sqrt (144) = 12 = PN
Since PQR is isosceles with PQ = PR then, X is a median......so
PX = (2/3) (12) = 8
