Question 636584: Let P be a point inside a square S so that the distances from P to the four vertices, in order, are 7, 35, 49, and x. What is x?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 636584
Let P be a point inside a square S so that the distances from P to the four vertices, in order, are 7, 35, 49, and x. What is x?
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Make the square's vertices at (0,0), (0,s), (s,s) and (s,0)
The square is s by s units
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P is 7 units from (0,0), 35 units from (0,s) and 49 units from (s,s)
P is the intersection of 3 circles with the vertices as centers and the radii as given.
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The circles are:
x^2 + y^2 = 49
(x-s)^2 + y^2 = 1225
(x-s)^2 + (y-s)^2 = 2401
Solve for s
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(x-s)^2 + y^2 = 1225
(x-s)^2 + (y-s)^2 = 2401
--------------------------- Subtract
(y-s)^2 - y^2 = 1176
-2sy + s^2 = 1176
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x^2 + y^2 = 49
(x-s)^2 + y^2 = 1225
--------------------------- Subtract
x^2 - (x-s)^2 = - 1176
2sx - s^2 = -1176
s^2 - 2sx = 1176
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==================
x^2 + y^2 = 49
(x-s)^2 + (y-s)^2 = 2401
--------------------------- Subtract
-2sx + s^2 - 2sy + s^2 = 2352
s^2 - sx - sy = 1176
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