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

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 amFollowed by: 2 members

by Gmat_mission » Wed Jun 24, 2020 8:17 am

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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. $3/2$

[spoiler]OA=D[/spoiler]

Source: Veritas Prep

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Source: Beat The GMAT — Problem Solving | Wed Jun 24, 2020 8:17 am

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Re: In the figure above, line segment $AD$ is the diameter of circle $O,$ line segment $AO$ is the diameter of cir

by Scott@TargetTestPrep » Mon Jun 29, 2020 5:51 am

Gmat_mission wrote: ↑
Wed Jun 24, 2020 8:17 am
FourCircles.png

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. $3/2$

[spoiler]OA=D[/spoiler]

Solution:

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

Source: Veritas Prep

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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