Question 1175957: A right triangle with legs of lengths 5 and 12 units has circles centered at each vertex so that the circles are externally tangent to each other. What is the sum of the numbers of square units in the area of the circles? Express in terms of π.
Found 2 solutions by CubeyThePenguin, greenestamps:
Answer by CubeyThePenguin(3113)   (Show Source):
You can put this solution on YOUR website! In order of increasing radius, the circles' radii will be a, b, and c.
a + b = 5
a + c = 12
b + c = 13
Add all of the equations together:
a + b + c = 30/2 = 15
So, we get a = 2, b = 3, and c = 10.
The total area of the circles is 
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Let the vertices of the triangle be A(0,0), B(12,0), and C(0,5).
Let a, b, and c be the radii of the circles centered at A, B, and C respectively.
Side AB has length 12 and side AC has length 5; the hypotenuse BC has length 13.
Then, since the circles are externally tangent to each other,
a+b=12
a+c=5
b+c=13
This system of equations is easy to solve by first adding all three equations together to get
2a+2b+2x=30
a+b+c=15
Then comparing that equation to the earlier three equations gives us a=2, b=10, and c=3.
Use the formula for the area of a circle to complete the problem.
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