Consider the diagram below:



(r - 10)^2 + (r - 20)^2 = r^2;
r = 10 or r = 50.
r cannot be 10, so it's 50.

Answer: D.


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A) 10
B) 40
C) 30
D) 50
E) None of the above



Notice that, once Bunuel creates the equation (r - 10)² + (r - 20)² = r², he doesn't try to solve it (a fair bit of work).
Instead, he starts plugging in the answer choices.
As he's doing that, he's looking for variations of some common Pythagorean triples. These are possible INTEGERS lengths that are possible in a right triangles.
The 4 most common Pythagorean triples are:
3-4-5
5-12-13
8-15-17
7-24-25
You'll see that each of these triples satisfies the Pythagorean Theorem (a² + b² = c²)
For example, 3² + 4² = 5² AND 5² + 12² = 13², etc)

So, when he plugs answer choice D (r = 50) into his equation (r - 10)² + (r - 20)² = r² he gets: (50 - 10)² + (50 - 20)² = 50²
Simplify to get: 40² + 30² = 50²
PERFECT, this is a variation of 3² + 4² = 5², but in this instance, each side is 10 times the length of the 3-4-5 right triangle.
As such, Bunuel automatically know that r = 50 is a solution to his equation.
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we can easily eliminate A.
since the sides of the rectangular figure are 10x20, and since 2r = side of the big square, it must be true that 10 can't be the radius... otherwise, one side would be exactly half of the side of the big square...
method given by bunuel is a good approach...
r^2 = (r-10)^2 + (r-20)^2
r^2 = r^2 -20r +100 + r^2 -40r + 400
r^2 - 60r +500 = 0
(r-10)(r-50)=0
r=50

Need help.! Can we apply PA*PB=PT^2 if PB is the diameter of circle?

Bunuel : Can you please explain me where I went wrong in my approach ?

I somehow ended up imagining the figure on a coordinate plane and used equation of circle and distance formula. However, I got stuck after that.

I understood your solution posted above. However, I wish to know where I went wrong in my approach so as to avoid mistakes in future.

Appreciate help.

Thankyou in advance.

Regards,
Saakhi

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