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Math Expert V
Joined: 02 Sep 2009
Posts: 59674
In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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1
10 00:00

Difficulty:   65% (hard)

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

### HideShow timer Statistics 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 2303 times ]

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Manager  S
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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3
2
Attachment: FullSizeRender (15).jpg [ 64.54 KiB | Viewed 2071 times ]

Ans:A
##### General Discussion
Senior Manager  G
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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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

Attachments Solution.PNG [ 10.28 KiB | Viewed 2011 times ]

Retired Moderator V
Joined: 28 Mar 2017
Posts: 1192
Location: India
GMAT 1: 730 Q49 V41 GPA: 4
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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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:
2. Side BD common

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

The problem is pretty simple now,
By pythogorous theorem in traingle ABD we have,

This means, Diameter=5
Hence circumfrence=5pie

Solution => A
Attachments approach.PNG [ 11.49 KiB | Viewed 1976 times ]

Senior Manager  S
Joined: 21 Mar 2016
Posts: 494
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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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:
2. Side BD common

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

The problem is pretty simple now,
By pythogorous theorem in traingle ABD we have,

This means, Diameter=5
Hence circumfrence=5pie

Solution => A

how did you get the highlighted part???
Retired Moderator V
Joined: 28 Mar 2017
Posts: 1192
Location: India
GMAT 1: 730 Q49 V41 GPA: 4
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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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.
Intern  B
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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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
Intern  B
Joined: 07 Jan 2017
Posts: 4
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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Hi mohshu,

Could you please explain me how did you get the below equation?
Therefore:
AB/BD=BC/CD
=> 2 root5=BC * BD
AND,
Non-Human User Joined: 09 Sep 2013
Posts: 13742
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

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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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