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25 Mar 2019, 01:16
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (01:59) correct
26%
(02:41)
wrong
based on 34
sessions
History
Date
Time
Result
Not Attempted Yet
The vertices of a 3-4-5 right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?
(A) \(12\pi\)
(B) \(\frac{25\pi}{2}\)
(C) \(13\pi\)
(D) \(\frac{27\pi}{2}\)
(E) \(14\pi\)
Attachment:
2006_AMC_12A_Problem_13.gif [ 9.14 KiB | Viewed 6121 times ]
eabhgoy
Joined: 12 Apr 2011
Last visit: 14 Jan 2021
Posts: 112
Given Kudos: 85
Location: United Arab Emirates
Concentration: Strategy, Marketing
Schools: CBS '21 Yale '21 INSEAD Jan'21 Intake IMD '21 (A)
GMAT 1: 670 Q50 V31
GMAT 2: 720 Q50 V37
GPA: 3.2
WE:Marketing (Telecommunications)
25 Mar 2019, 02:01
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Bunuel
Let us consider the radius of the circles to be:
Smallest circle = x (bottom left)
Medium = y (top)
Biggest = z (bottom right)
Since this is a 3-4-5 circle, then we can deduce the sides to be:
x + y = 3
x + z = 4
y + z = 5
Adding all three: 2(x + y + z) = 3 + 4 + 5 = 12
=> x + y + z = 6
=> x = 1, y = 2, z = 3
Area of 3 circles = pie (x^2 + y^2 + z^2) = pie (3^3 + 2^2 + 1^2) = 14pie
Hence E is the correct answer.
Archit3110
Major Poster
25 Mar 2019, 03:04
Kudos
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Bunuel
we can solve <30 sec
visualize the given image
the largest length = 5
smallest = 3
and third side = 4
the value of radius can be 3,2,1 ; square = 9pi+4pi+1pi = 14 pi
IMO E
