Question

ABCD is a square. P is a point within ABCD such that PA=1,PB=29 and PC=41. What is area of square ABCD ?

Answer 1

I started by placing a square of sides k so that one vertex is at B(0,0), and letting sides fall on the x and y axes
so the other vertices are C(k,0), D(k,k) and A(0,k)
I let point P be (x,y)

PO = 29 ----> x^2 + y^2 = 29^2
PA = 1 -----> x^2 + (y-k)^2 = 1^2
PC = 41 ----> (x-k)^2 + y^2 = 41^2

so I had 3 messy equations in 3 unknowns.
After a page of frustrating algebraic manipulations,
(which I really don't feel like typing here)

I finally found
x = 1/√2 , y = 41/√2 , and k = 21√2
but it was really the k value we are after,
since the area of the square is simply k^2

if k = 21√2, then
k^2 = (21√2)^2 = 882
So the area is 882 units^2

btw, you can test the distances PO, PA, and PC with your calculator, they work

Answer 2

Hmmm. I don't buy it. You can't have

x = 1/√2 , y = 41/√2 , and k = 21√2

and

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

Maybe I'll visit the figure.

Answer 3

x^2 + (y-k)^2
= (1/√2)^2 + (41/√2 - 21√2)^2 ,,,,,, 21√2 = 42/√2
= 1/2 + (-1/√2)^2
= 1/2+1/2
= 1

Answer 4

My bad. I misread it. Even after looking at it several times, I still saw 21/√2 rather than 21√2.