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09 Aug 2017, 23:36
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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51% (03:02) correct
49%
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wrong
based on 195
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In the figure above, line segment AD is the diameter of circle O, line segment AO is the diameter of circle B, line segment OD is the diameter of circle C, and circle E is tangent to each of the other circles. If the radius of circle O is 4, what is the radius of circle E?
A. 2/3
B. 3/4
C. 1
D. 4/3
E. 3/2
Attachment:
FourCircles.png [ 11.47 KiB | Viewed 8815 times ]
15 Aug 2017, 10:58
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Bunuel
Since the radius of circle O is 4 (i.e., AO = 4), BO = 2. We can form a right triangle by connecting B, E, and O. If we let x = radius of circle E, the legs of right triangle BOE are BO = 2 and OE = 4 - x, and its hypotenuse is BE = 2 + x. Now, using the Pythagorean theorem, we have:
2^2 + (4 - x)^2 = (2 + x)^2
4 + 16 - 8x + x^2 = 4 + 4x + x^2
16 - 8x = 4x
16 = 12x
x = 16/12 = 4/3
Answer: D
General Discussion
10 Aug 2017, 00:09
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Radius OA=4
therefore OB=OC=2
Now let radius of circle E =x
BE=EC=2+x
As BE=EC and OB=OC, implies EO is perpendicular to BC
Now in triangle EOB :
OB=2
EB=2+x
EO =4-x (radius of circle O - radius of circle E)
solving Pythagoras : OB^2+OE^2=BE^2
=> x=4/3
Answer : D
therefore OB=OC=2
Now let radius of circle E =x
BE=EC=2+x
As BE=EC and OB=OC, implies EO is perpendicular to BC
Now in triangle EOB :
OB=2
EB=2+x
EO =4-x (radius of circle O - radius of circle E)
solving Pythagoras : OB^2+OE^2=BE^2
=> x=4/3
Answer : D
