Key property: A line tangent to a circle is perpendicular to the circle's radius at the point of intersection

First, if we add the centers of each triangle as as well as the point E, we get the following:



At this point, if we draw lines from point E to each of the three vertices, look at the following:



At this point we may recognize that: (area of ∆AEB) + (area of ∆CEA) + (area of ∆CEB) = area of ∆ACB

Area of triangle = (base)(height)/2

So, we get: (40)(r)/2 + (30)(5r)/2 + (50)(r)/2 = (40)(30)/2
Simplify to get: 20r + 75r + 25r = 600
Simplify: 120r = 600
Solve: r = 600/120 = 5

Answer: A

Cheers,
Brent

That's a great idea, and you did arrive at the correct answer, but this was a bit of a coincidence. Here's why....

If we want to compare the total area of the 3 circles with the area of triangle ABC, we can't write: (pi)(r^2) = 600, because this assumes that we have just one circle.
If r = the radius of ONE circle, then (pi)(r^2) = the area of one circle, which means (3)(pi)(r^2) = the total area of THREE circles.
So, we need to write: (3)(pi)(r^2) = 600
From here we get: (pi)(r^2) = 200
And: r^2 ≈ 64 (after dividing both sides by pi)
Solve: r ≈ 8

Why is this number so much greater than the correct answer? Because, we are equating the total area of the circles with the area of triangle ABC, when we can see that the 3 circles definitely don't take up the entire area of the triangle.

Cheers,
Brent

wow