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

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Bunuel
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In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter. [#permalink] New post   06 Jun 2017, 11:13
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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


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Leo8
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter. [#permalink] New post   06 Jun 2017, 11:30
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gmatexam439
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter. [#permalink] New post   06 Jun 2017, 12:42
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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:
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
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter. [#permalink] New post   06 Jun 2017, 12:12
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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

Answer is A
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Re: In the circle pictured above, AB=2 5 , CD=1, and AC is a diameter. [#permalink] New post   01 Jun 2022, 12:44
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To be honest, yes we can use the conventional approach and arrive at the answer but for time saving purposes there is a very fast and simple method here based on elimination of answer choices

ABD is a right triangle with hypotenuse AB as 2 √ 5 so AB^2 = 20

In Triangle ABD, AB^2 = AD^2 + BD^2

Now if you look at the options the circumference is provided
2*Pie*r or Pie*D and AD = Diameter of Circle -1

Now if you have a look at the options, in all options except the first one, AD when squared is returning a value larger than AB^2 which is not possible since AB is the hypotenuse of right triangle ABD

So we can eliminate all except A which is the answer
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter. [#permalink] New post   08 Jun 2017, 10:57
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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:
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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gmatexam439
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter. [#permalink] New post   08 Jun 2017, 11:11
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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.
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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter. [#permalink] New post   22 Jun 2017, 16:28
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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

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Re: In the circle pictured above, AB=2 √ 5 , CD=1, and AC is a diameter. [#permalink] New post   25 Jun 2017, 20:12
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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,
AB/AD=BC/BD
=>AD=BD^2
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Re: In the circle pictured above, AB=2 5 , CD=1, and AC is a diameter. [#permalink] New post   16 Jun 2022, 01:42
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I thought for 10 mins, but actually this problem be solved in 20 secs.

Circumference is 2Pr, so PD, If Diameter is 6, than AD 6-1=5, which is greater than the hypotenuse (since square of 5 is 25, which is greater than square of hypotenuse, which is 20). Therefore, only D=5 meets this criteria.
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Re: In the circle pictured above, AB=2 5 , CD=1, and AC is a diameter. [#permalink]
16 Jun 2022, 01:42
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