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Shown below is rectangle EFGH. Its diagonals meet at Y. Let X be the foot of in altitude is dropped from E to FH. If EX = 24 and GY = \(26\), find the perimeter of rectangle .

 

 Apr 22, 2022
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The diagonals bisect each other.....therefore  EY = GY = 26  and EG =  52  = HF

 

And because  triangle   EYX  is right then    XY  = sqrt ( EY^2 - EX^2) =  sqrt (26^2 - 24^2) = sqrt 100   = 10 

 

But the diagonals of  a  rectangle are congruent  so   EY = FY = GY =  HY

 

And   HX = HY - XY  =  26   - 10    =   16

 

And since triangle EHX is right, then   EH   = sqrt ( HX^2 + EX^2) =  sqrt ( 16^2 + 24^2 ) =   sqrt ( 832)

 

And triangle HEF is also right so EF  =sqrt  ( HF^2 - EH^2 )  =  sqrt (  52^2 - 832)  = sqrt (1872)

 

So.....the perimeter  of   EFGH  = 2 [ EH + EF ]  =    2 [ sqrt 832  + sqrt 1872 ]   ≈  144.22

 

cool cool cool

 Apr 22, 2022

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