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In rectangle ABCD, shown here, CE is perpendicular to BD. If BC = 3 and DC = 8, what is CE?

 

 Jun 10, 2022
 #1
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DBC is a right triangle.....so....

 

BD =   sqrt  [ DC^2 + BC^2 ] =  sqrt [  8^2 + 3^2 ]    =  sqrt (73)

 

Let  EB = x    and DE = sqrt (73)  - x

 

And  EBC  and   EDC  are also right triangles

 

So

 

CE^2   =     BC^2 - EB^2  =   3^2  - x^2  =   9 - x^2         (1)

and

CE^2  =  DC^2  - DE^2  =   8^2  - (sqrt (73)  - x )^2  =   64 - x^2 + 2xsqrt (73) - 73  = 

-x^2 + 2xsqrt (73) - 9         (2)

 

Set  (1)  = (2)

 

9 - x^2  =  -x^2 + 2 sqrt (73) x - 9      simplify

 

18 =  2  sqrt (73) x 

 

9 =  sqrt (73) x

 

x =  9  / sqrt (73)  =  EB

 

 

So

 

CE =  sqrt   [  BC^2 - EB^2  ]  =  sqrt [ 3^2  - (9/sqrt (73))^2 ]   =  sqrt  [ 9 - (81 / 73 ) ]  ≈  2.8

 

 

cool cool cool

 Jun 10, 2022

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