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

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Joined: 02 Sep 2009
Posts: 58320
In the figure above, line segment AD is the diameter of circle O, line  [#permalink]

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09 Aug 2017, 23:36
00:00

Difficulty:

95% (hard)

Question Stats:

48% (03:12) correct 52% (03:03) wrong based on 63 sessions

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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 2641 times ]

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In the figure above, line segment AD is the diameter of circle O, line  [#permalink]

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10 Aug 2017, 00:09
1
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

solving Pythagoras : OB^2+OE^2=BE^2
=> x=4/3

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Re: In the figure above, line segment AD is the diameter of circle O, line  [#permalink]

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15 Aug 2017, 10:58
2
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

Attachment:
FourCircles.png

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

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Re: In the figure above, line segment AD is the diameter of circle O, line  [#permalink]

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20 Feb 2019, 00:17
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Re: In the figure above, line segment AD is the diameter of circle O, line   [#permalink] 20 Feb 2019, 00:17
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