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Let PQRS be a square with side length 10. The points X and Y lie outside the square such that XQ = 6, XP = 8, YS = 6, and YR = 8. Find XY^2.

 May 15, 2020
 #1
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XY^2 = 288.

 May 15, 2020
 #2
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nice explanation

Guest May 15, 2020
 #3
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XY^2 = 392  smiley

Guest May 15, 2020
 #4
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MX = (8*6)/10 = 4.8

 

MQ = sqrt(QX² - MX²) = 3.6          MQ = NL

 

NX = 2*MX + 10 = 19.6

 

LY = PQ - MQ = 6.4

 

NY = LY - MQ = 2.8

 

XY^2 = NX^2 + NY^2 = 392   indecision

 

 May 20, 2020
edited by Dragan  May 20, 2020
edited by Dragan  May 20, 2020

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