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Bunuel
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Bunuel

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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gmatexam439
Bunuel

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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mohshu

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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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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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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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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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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