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If 1/a^2 + a^2 represents the diameter of circle O and 1/a +
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Updated on: 21 Oct 2013, 08:19
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If 1/a^2 + a^2 represents the diameter of circle O and 1/a + a = 3, which of the following best approximates the circumference of circle O? A. 28 B. 22 C. 20 D. 16 E. 12
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Originally posted by somethingbetter on 14 Sep 2007, 10:33.
Last edited by Bunuel on 21 Oct 2013, 08:19, edited 3 times in total.
Renamed the topic and edited the question.




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Re: looks pretty simple , huh!!
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14 Sep 2007, 10:58
somethingbetter wrote: If (1/a^2)+a^2 represents the diameter of circle O and (1/a)+a =3 , which of the following best approximates the circumference of circle O?
a. 28 b. 22 c. 20 d. 16 e. 12
Given that (1/a)+a =3
Square both sides of the equation:
We get,
[(1/a)^2 + a^2 + 2*(1/a)*a] = 9
=> (1/a)^2 + a^2 + 2 = 9
=> (1/a)^2 + a^2 = 7  (1)
Diameter D = (1/a)^2 + a^2
= 7 (From (1))
So Radius = D/2 = 7/2
Circumference = 2*Pi*r
= 2*(22/7)*(7/2)
= 22
So the answer should be B.
 Brajesh




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I get 22
(1/a + a) =3
(1/a + a) (1/a +a)=1/a^2 +a^2 +2
92=7
7*pi=22



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Re: looks pretty simple , huh!!
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21 Oct 2013, 06:42
b14kumar wrote: somethingbetter wrote: If (1/a^2)+a^2 represents the diameter of circle O and (1/a)+a =3 , which of the following best approximates the circumference of circle O?
a. 28 b. 22 c. 20 d. 16 e. 12 Given that (1/a)+a =3 Square both sides of the equation: We get,[(1/a)^2 + a^2 + 2*(1/a)*a] = 9=> (1/a)^2 + a^2 + 2 = 9 => (1/a)^2 + a^2 = 7  (1) Diameter D = (1/a)^2 + a^2 = 7 (From (1)) So Radius = D/2 = 7/2 Circumference = 2*Pi*r = 2*(22/7)*(7/2) = 22So the answer should be B.  Brajesh Hi, bumping an old question, but why would you square both sides of the equation...I see that it works, but I don't understand how one would know to do that.



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Re: looks pretty simple , huh!!
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21 Oct 2013, 08:22
AccipiterQ wrote: b14kumar wrote: somethingbetter wrote: If 1/a^2 + a^2 represents the diameter of circle O and 1/a + a = 3, which of the following best approximates the circumference of circle O?
A. 28 B. 22 C. 20 D. 16 E. 12 Given that (1/a)+a =3 Square both sides of the equation: We get,[(1/a)^2 + a^2 + 2*(1/a)*a] = 9=> (1/a)^2 + a^2 + 2 = 9 => (1/a)^2 + a^2 = 7  (1) Diameter D = (1/a)^2 + a^2 = 7 (From (1)) So Radius = D/2 = 7/2 Circumference = 2*Pi*r = 2*(22/7)*(7/2) = 22So the answer should be B.  Brajesh Hi, bumping an old question, but why would you square both sides of the equation...I see that it works, but I don't understand how one would know to do that. We need to find the value of 1/a^2 + a^2, which is a second degree equation. 1/a + a = 3 is a first degree equation. Squaring seems very natural thing to do.
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Re: If 1/a^2 + a^2 represents the diameter of circle O and 1/a +
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12 Apr 2016, 08:25
somethingbetter wrote: If 1/a^2 + a^2 represents the diameter of circle O and 1/a + a = 3, which of the following best approximates the circumference of circle O?
A. 28 B. 22 C. 20 D. 16 E. 12 circumference = 2*pi*r we are given that 1/a^2 + a^2 is diameter. 1/a + a = 3  take square 1/a^2 + a^2 = 7 hence r = 7/2 circumference = 22
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If 1/a^2 + a^2 represents the diameter of circle O and 1/a +
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30 Jan 2017, 03:16
somethingbetter wrote: If \(\frac{1}{a^2} + a^2\) represents the diameter of circle O and \(\frac{1}{a} + a = 3\), which of the following best approximates the circumference of circle O?
A. 28 B. 22 C. 20 D. 16 E. 12 \(Circumference = 2πr\) \(( \frac{1}{a} + a )( \frac{1}{a}+ a ) = \frac{1}{a^2} + 1 + 1 + a^2\) \((3)(3) = \frac{1}{a^2}+ a^2 + 2\) \(\frac{1}{a^2} + a^2 = 7 = Diameter\) \(Circumference = 2πr = 2π ( \frac{7}{2} ) = 7π = 21.99 ≈ 22\)
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Re: If 1/a^2 + a^2 represents the diameter of circle O and 1/a +
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21 Feb 2018, 13:15
somethingbetter wrote: If 1/a^2 + a^2 represents the diameter of circle O and 1/a + a = 3, which of the following best approximates the circumference of circle O?
A. 28 B. 22 C. 20 D. 16 E. 12 Squaring 1/a + a = 3, we have: (1/a + a)^2 = 3^2 1/a^2 + a^2 + 2 = 9 1/a^2 + a^2 = 7 Since 1/a^2 + a^2 = 7 represents the diameter, the circumference is 7π ≈ 22 Answer: B
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