Circle
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- eagleeye
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Hi PGMAT:
The correct answer should be A. Let me explain:
We are asked if Circumference of the circle is greater than 2(AB+CD).
First since it is a circle we know that:
a. Circumference = 2*pi*r. (where pi~3.14 and r is the radius of circle.
b. Since O is the center of the circle, AB is the diameter, so AB= 2r.
So our question reduces to whether 2*pi*r > 2(2r+CD) which is pi*r > 2r+CD
1. First statement CD=1/2*AB. Since AB = 2r, and CD is half of that CD=r.
Then we have 2r+CD = 2r+r = 3r. Clearly pi*r > 3r since pi>3.
SUFFICIENT.
2. Next we are given that x is 90. Note that depending on where CD is located between O and B, it can be vary from 0(close to B) to 2r (close to O).
Now for x=90 degrees, the smallest value of CD is close to 0. Let's use CD=0 In that case, pi*r is definitely greater than 2r+0.
However, for the largest value of CD, which is close to the diameter, Let's use CD=diameter=2r.
Then 2r+CD is approx. 2r+2r = 4r. Now pi*r <4*r, since pi is less than 4. So, depending on where CD is located on the x axis, we get different answers. INSUFFICIENT.
Hence the final answer is A.
Let me know if this helps
The correct answer should be A. Let me explain:
We are asked if Circumference of the circle is greater than 2(AB+CD).
First since it is a circle we know that:
a. Circumference = 2*pi*r. (where pi~3.14 and r is the radius of circle.
b. Since O is the center of the circle, AB is the diameter, so AB= 2r.
So our question reduces to whether 2*pi*r > 2(2r+CD) which is pi*r > 2r+CD
1. First statement CD=1/2*AB. Since AB = 2r, and CD is half of that CD=r.
Then we have 2r+CD = 2r+r = 3r. Clearly pi*r > 3r since pi>3.
SUFFICIENT.
2. Next we are given that x is 90. Note that depending on where CD is located between O and B, it can be vary from 0(close to B) to 2r (close to O).
Now for x=90 degrees, the smallest value of CD is close to 0. Let's use CD=0 In that case, pi*r is definitely greater than 2r+0.
However, for the largest value of CD, which is close to the diameter, Let's use CD=diameter=2r.
Then 2r+CD is approx. 2r+2r = 4r. Now pi*r <4*r, since pi is less than 4. So, depending on where CD is located on the x axis, we get different answers. INSUFFICIENT.
Hence the final answer is A.
Let me know if this helps
- ronnie1985
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Perimeter = 2*pi*r = 6.28*r (approx)
S1. AB = 2r (dia) and CD = r
2(AB+CD) = 6*r<6.28*r
Suff
x = 90 gives no info about relation between dia AB and CD = > Not Suff
(A)
S1. AB = 2r (dia) and CD = r
2(AB+CD) = 6*r<6.28*r
Suff
x = 90 gives no info about relation between dia AB and CD = > Not Suff
(A)
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