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avatar+388 

In triangle $PQR,$ let $X$ be the intersection of the angle bisector of $\angle P$ with side $\overline{QR}$, and let $Y$ be the foot of the perpendicular from $X$ to side $\overline{PR}$.  If $PQ = 28,$ $QR = 35,$ and $PR = 17,$ then compute the length of $\overline{XY}$.

 Jun 4, 2024
 #2
avatar+129399 
+1

                               P

 

             28                     Y     17              

 

Q                     X                      R

                  35

 

Because  PX is a bisector,  then

QX / PQ = RX / PR

PR / PQ = RX / QX

17/28  = RX / QX

So

RX = 17 / ( 17+ 28) * QR  =  (17 / 45) * 35  =  119/9

 

Law of Cosines

PQ^2  = QR^2 + PR^2  - 2(QR * PR) cos (PRQ)

28^2 = 35^2 + 17^2 - 2 (35 * 17) cos (PRQ)

cos (PRQ)  =  [ 28^2 - 35^2 - 17^2 ] /  [-2 * 35 * 17 ] = 73/119

sin (PRQ)  = sqrt [ 119^2 - 73^2 ] / 119  =    8sqrt (138) / 119

 

Law of Sines   (angle XYR = 90 )

 

XY /sin (PRQ)  = XR / sin(XYR)

 

XY / [8sqrt (138) / 119 ] =  (119/9) / sin 90        (sin 90  =  1)

 

XY  =  [ 8sqrt (138) / 119 ] ( [ 119/ 9]   =   (8/9)sqrt (138) ≈  10.44

 

cool cool cool

 Jun 4, 2024

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