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In the figure above, line segment AD is the diameter of circle O, line

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Bunuel
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In the figure above, line segment AD is the diameter of circle O, line [#permalink] New post   09 Aug 2017, 23:36
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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

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ScottTargetTestPrep
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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink] New post   15 Aug 2017, 10:58
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Bunuel wrote:

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

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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
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In the figure above, line segment AD is the diameter of circle O, line [#permalink] New post   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
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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink] New post   25 May 2020, 03:54
Bunuel chetan2u ScottTargetTestPrep -
Can you let me know where I'm going wrong?

BC = 4
BE = EC = 2+x = say Y
So BC^2 = Y^2+Y^2
16 = 2Y^2
8 = Y^2
Y = 2Root2
since 2+x= 2Root2
x = 2root2-2
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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink] New post   06 Jun 2020, 04:41
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TarPhi wrote:
Bunuel chetan2u ScottTargetTestPrep -
Can you let me know where I'm going wrong?

BC = 4
BE = EC = 2+x = say Y
So BC^2 = Y^2+Y^2
16 = 2Y^2
8 = Y^2
Y = 2Root2
since 2+x= 2Root2
x = 2root2-2


Response:

Your solution assumes that BC^2 should equal BE^2 + EC^2, which is not necessarily true. That is the Pythagorean theorem, and it holds only for right triangles. If the angle BEC were a right angle, your approach would be correct; however, we have no such information.
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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink] New post   07 Aug 2021, 19:29
D it is..

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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink]
07 Aug 2021, 19:29
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