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# In the diagram above, point B is the center of Circle #1 and point D i

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Math Expert
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In the diagram above, point B is the center of Circle #1 and point D i  [#permalink]

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10 Mar 2015, 06:00
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35% (medium)

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73% (01:52) correct 27% (02:16) wrong based on 172 sessions

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gpp_img7.png [ 15.74 KiB | Viewed 3346 times ]
In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?

A. 1.5
B. $$\sqrt{3}$$
C. 3
D. $$\sqrt{6}$$
E. 6

Kudos for a correct solution.

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Re: In the diagram above, point B is the center of Circle #1 and point D i  [#permalink]

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10 Mar 2015, 06:22
Hi

let's say R1 is the radius of smaller circle and R2 is of the bigger circle

According to question
A1/A2 = 3/2
[pi(R1)^2 ] / [pi(R2)^2] = 3/2
R1/R2 = root3 / root 2

Now CE= 2[R2] & BC =R1

therefore CE/BC = [2(root3)] / root2 = root6
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Re: In the diagram above, point B is the center of Circle #1 and point D i  [#permalink]

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10 Mar 2015, 19:01
1
Assume radius of circle 1 = R & radius of circle 2 = r;
given, Area of Circle 1/Area of Circle 2 = 3/2 = R^2/r^2; therefore R/r = Root 3/Root 2
Required, CE:BC = 2*R: r = 2 * Root 3/Root 2; simplify (multiply numerator and denominator by root 2), and answer = Root 6
Therefore answer choice D Root 6
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Posts: 58327
Re: In the diagram above, point B is the center of Circle #1 and point D i  [#permalink]

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15 Mar 2015, 21:04
Bunuel wrote:

In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?

A. 1.5
B. $$\sqrt{3}$$
C. 3
D. $$\sqrt{6}$$
E. 6

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

This is tricky. We are given the ratio of areas, and we want to know a ratio involving the radii. Let R be the radius of the bigger circle, and r be the radius of the smaller circle.
$$ratio \ of \ areas = \frac{3}{2} = \frac{\pi{R^2}}{\pi{r^2}}=\frac{\pi{R^2}}{\pi{r^2}}$$.

Take the square root and rationalize the denominator:
$$\frac{R}{r}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}$$

Well, CE = 2R, and BC = r, so CE/BC = 2R/r, which is twice this ratio $$\frac{CE}{BC}=\sqrt{6}$$.

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Re: In the diagram above, point B is the center of Circle #1 and point D i  [#permalink]

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20 Oct 2015, 09:52
[quote="Bunuel"]
Attachment:
gpp_img7.png
In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?

A. 1.5
B. $$\sqrt{3}$$
C. 3
D. $$\sqrt{6}$$
E. 6

Solution :

Lets assume radius of circle 1 - r, and circle 2 as - R

Area of circle 1 - pi r*2
Area of circle 2 - pi R*2

Therfore pi R*2/pi r*2 = 3/2
R*2/r*2 = 3/2
R/r = root 3/ root 2

We have to find out the ratio of the diameter of circle 2/ radius of circle 1, which can be written mathematically as 2R/r

2R/r = 2* root 3/ root 2
= root 2 * root 3
= root 6

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Re: In the diagram above, point B is the center of Circle #1 and point D i  [#permalink]

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06 Apr 2018, 02:36
Bunuel wrote:
Attachment:
gpp_img7.png
In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?

A. 1.5
B. $$\sqrt{3}$$
C. 3
D. $$\sqrt{6}$$
E. 6

Kudos for a correct solution.

Area(C2) : Area (C1) = 3 :2
Diameter(C2) : Diameter(C1) = $$\sqrt{(3/2)}$$
CE: AB = $$\sqrt{(3/2)}$$
CE: 2BC = $$\sqrt{(3/2)}$$
CE:BC = 2*$$\sqrt{(3/2)}$$ = $$\sqrt{4*(3/2)}$$ = $$\sqrt{6}$$

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Re: In the diagram above, point B is the center of Circle #1 and point D i   [#permalink] 06 Apr 2018, 02:36
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