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# In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.

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Math Expert
Joined: 02 Sep 2009
Posts: 59628
In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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06 Jun 2017, 11:13
1
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Difficulty:

65% (hard)

Question Stats:

61% (02:52) correct 39% (03:15) wrong based on 75 sessions

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In the circle pictured above, AB=2√5, CD=1, and AC is a diameter. What is the circumference of the circle?

A. 5π
B. 6π
C. 7π
D. 8π
E. 9π

Attachment:

ABCD_diameter.PNG [ 11.3 KiB | Viewed 2302 times ]

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Joined: 23 May 2017
Posts: 231
Concentration: Finance, Accounting
WE: Programming (Energy and Utilities)
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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06 Jun 2017, 11:30
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Attachment:

FullSizeRender (15).jpg [ 64.54 KiB | Viewed 2070 times ]

Ans:A
##### General Discussion
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Joined: 24 Apr 2016
Posts: 316
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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06 Jun 2017, 12:12
Let's take the Right Angled Triangle ADB

$$AB^2$$ = $$BD^2$$ + $$AD^2$$

$$(2√5)^2$$ = $$BD^2$$ +$$(r+r-1)^2$$

20 = $$BD^2$$ + 4$$r^2$$+1-4r --------> Equation 1

Let's take the Right Angled Triangle ABC

$$AC^2$$ = $$AB^2$$+ $$BC^2$$

4$$r^2$$ = $$(2√5)^2$$ + $$BC^2$$
4$$r^2$$ = 20 + $$BC^2$$ --------> Equation 2

Let's take the Right Angled Triangle BDC

$$BC^2$$ = $$BD^2$$ + $$DC^2$$

$$BC^2$$= $$BD^2$$+ 1--------> Equation 3

Replace $$BC^2$$ from equation 3 to Equation 2

4$$r^2$$ = 20 + $$BD^2$$+ 1
4$$r^2$$ = 21 +$$BD^2$$
$$BD^2$$ = 4$$r^2$$ - 21 --------> Equation 4

Replace $$BD^2$$ from equation 4 to Equation 1

20 = 4$$r^2$$ - 21 + 4$$r^2$$ + 1 - 4r

8$$r^2$$ - 4r - 40 = 0

2$$r^2$$ - r - 10 = 0

Solving for r we get

r = -2 or 5/2

Circumference of the circle = 2 * π *$$\frac{5}{2}$$ = 5π

Answer is A
Attachments

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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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06 Jun 2017, 12:42
Bunuel wrote:

In the circle pictured above, AB=2√5, CD=1, and AC is a diameter. What is the circumference of the circle?

A. 5π
B. 6π
C. 7π
D. 8π
E. 9π

My approach is a bit different. I used similar triangles. Please refer to the figure below:

triangle ABD and triangle BCD are similar because:
1. angle ADC=angle BDC=90
2. Side BD common
3. Angle BAD= Angle CBD;

Therefore:
AB/BD=BC/CD
=> 2 root5=BC * BD
AND,
AB/AD=BC/BD
=>AD=BD^2

The problem is pretty simple now,
By pythogorous theorem in traingle ABD we have,
20=AD^2 + BD^2
=> 20=AD^2 + AD
Solving quad eq, we have AD=-5 and 4; But side cannot be negative, therefore AD=4

This means, Diameter=5
Hence circumfrence=5pie

Solution => A
Attachments

approach.PNG [ 11.49 KiB | Viewed 1975 times ]

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Joined: 21 Mar 2016
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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08 Jun 2017, 10:57
gmatexam439 wrote:
Bunuel wrote:

In the circle pictured above, AB=2√5, CD=1, and AC is a diameter. What is the circumference of the circle?

A. 5π
B. 6π
C. 7π
D. 8π
E. 9π

My approach is a bit different. I used similar triangles. Please refer to the figure below:

triangle ABD and triangle BCD are similar because:
1. angle ADC=angle BDC=90
2. Side BD common
3. Angle BAD= Angle CBD;

Therefore:
AB/BD=BC/CD
=> 2 root5=BC * BD
AND,
AB/AD=BC/BD
=>AD=BD^2

The problem is pretty simple now,
By pythogorous theorem in traingle ABD we have,
20=AD^2 + BD^2
=> 20=AD^2 + AD
Solving quad eq, we have AD=-5 and 4; But side cannot be negative, therefore AD=4

This means, Diameter=5
Hence circumfrence=5pie

Solution => A

how did you get the highlighted part???
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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08 Jun 2017, 11:11
mohshu wrote:

how did you get the highlighted part???

A perpendicular from the angle opposite to hypot on to the hypot always divides the traingle into 2 similar traingles which in turn are similar to the parent triangle.
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Joined: 02 Jan 2017
Posts: 43
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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22 Jun 2017, 16:28
My opinion is this.
Comparing AB to AC, we see that AB is slightly lesser than AC. ie trying to make AB horizontal.
Also, √5 is slightly more than √4. so 2√5 is slightly more than 4' so circumference is πd = 5π. This is the closest answer more than 4'π

Sent from my D6503 using GMAT Club Forum mobile app
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Joined: 07 Jan 2017
Posts: 5
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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25 Jun 2017, 20:12
Hi mohshu,

Could you please explain me how did you get the below equation?
Therefore:
AB/BD=BC/CD
=> 2 root5=BC * BD
AND,
AB/AD=BC/BD
=>AD=BD^2
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Posts: 13733
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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25 May 2019, 07:31
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.   [#permalink] 25 May 2019, 07:31
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# In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.

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