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Let point P be inside rectangle ABCD such that PA = 1, PB = 7 , and PC = 8. Find PD .

Guest Oct 14, 2017
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First....let A be located at (0,0)....construct a circle with radius 1 and centered at A

And this circle wil pass through P  and have the equation

x^2 + y^2  =  1  →  y^2  =  1 - x^2   (1)

 

And let B be located at (c,0).........construct a circle with radius 7 and centered at B

And this circle will pass through P and have the equation

(x - c)^2 + y^2 = 49          sub (1) in  for y^2  

(x -c)^2  + [1 - x^2 ]  = 49  

(x - c)^2  = 48 + x^2   (2)

 

Next......let C  be located at (c,d).........construct a circle with radius 8 and centered at C

And this circle will pass through P and have the equation

(x - c)^2 + (y - d)^2  = 64        sub (2) in for  (x - c)^2  

[48 + x^2]  +  (y - d)^2  = 64  

(y - d)^2 = 16 - x^2      (3)

 

Finally......

Let D  be located at (0,d).........construct a circle with radius DP and centered at D

And this circle will pass through P and have the equation

x^2 + (y - d)^2  = DP^2       sub (3) in for (y - d)^2 

x^2 + [ 16  - x^2]  = DP^2    .......  simplify..........

16 = DP^2        take the positive root

4  = DP =  PD

 

 

cool cool cool

CPhill  Oct 15, 2017

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