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help im stuck

 

Chords $\overline{PQ}$ and $\overline{RS}$ of a circle meet at $X$ inside the circle.  If $RS = 36$, $PX = 8$, and $QX = 24$, then what is the smallest possible value of $RX$?

 Aug 19, 2023
 #1
avatar+128732 
+1

                  R

 

                  m

 

P       8       X     24       Q

 

                 36 - m

                 

 

                   S

 

Intersecting Chords Theorem

 

Let RX =  m     and SX =  36 - m 

 

PX * QX =  RX * SX

 

8  *  24  =   m ( 36 - m)

 

192  =  36m - m^2

 

m^2 - 36m + 192  = 0

 

m^2 - 36m =  -192         complete the square  on m

 

m^2 - 36m + 324  = -192 + 324

 

(m - 18)^2  =  132          take the  positive root

 

m - 18 =  sqrt [ 132]

 

m =  sqrt [132] + 18 =     2sqrt33 + 18  =  2 [ sqrt 33 + 9 ]  =  smallest positive value of RX   

 

 

cool cool cool

 Aug 19, 2023

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