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

Guest Oct 14, 2017

#1**+1 **

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

CPhill
Oct 15, 2017