GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 11 Dec 2019, 14:34

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

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

### Show Tags

06 Jun 2017, 11:13
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 ]

_________________
Manager
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]

### Show Tags

06 Jun 2017, 11:30
3
2
Attachment:

FullSizeRender (15).jpg [ 64.54 KiB | Viewed 2071 times ]

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

### Show Tags

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

Attachments

Solution.PNG [ 10.28 KiB | Viewed 2011 times ]

Retired Moderator
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]

### Show Tags

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

### Show Tags

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:
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
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]

### Show Tags

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.
Intern
Joined: 02 Jan 2017
Posts: 43
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.  [#permalink]

### Show Tags

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

### Show Tags

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,
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]

### Show Tags

25 May 2019, 07:31
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter.   [#permalink] 25 May 2019, 07:31
Display posts from previous: Sort by