JeffTargetTestPrepw wrote:

Bunuel wrote:

In the figure, the circles with centers A and B are tangent to each other at C, and are tangents to the lines k and m at D and F respectively. If the radius of each circle is √2 and k is

parallel to m, what is the length of EF?

(A) 2√2 + 1

(B) 2√2 + 2

(C) 2√2 + 3

(D) 3√2

(E) 3√2 + 1

Attachment:

2017-11-21_1032_003.png

We see that EF consists of the radius of the top circle, a leg of the triangle, and the radius of the bottom circle.

We see that the two radii combined are √2 + √2 = 2√2.

We also see that BA = 2√2 and that the triangle is a 45-45-90 right triangle, so each leg is 2.

Thus, EF = 2√2 + 2.

Answer: B

JeffTargetTestPrep , or

anyone who can explain, I have a question about this problem: How do we know the triangle is a right

isosceles triangle?

The sides are equal right up to the point they cross each circumference (because radii are equal). How do we know the distance of each leg's small segment outside each circle -- from where the leg crosses each circumference to where it intercepts line EF -- is equal?

I have been doodling around with external tangent lines to see whether or not that might help to prove that the legs of the right triangle are equal in length. So far, nothing.

And if it is isosceles because its two smaller angles = 45°, I don't understand that, either.

The two answers assert that the sides are equal. Would someone please explain why?

_________________

The only thing more dangerous than ignorance is arrogance.

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