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The area of the circle is 16/r. [#permalink]
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ThisIsWater wrote:
Hi Carcass,

Can you help me in this?

Question never mentioned figure drawn to scale. Also, how can we take "O" as the center?

I have <4pi area with X>90 and also X<90 by my assumptions....Am I correct?


I went through three versions of the same question in different books and in one O is the center is not mentioned and in the others yes. :roll:

If x were equal to 90, the shaded region would have an area equal to 1/ 4 that of the entire circle (since 90 is 1/ 4 of 360). If the angle were equal to 90, the shaded region would have an area of \(4 \pi\) (1/ 4 of the entire circle’s area). Since the area of the shaded region is actually less than \(4 \pi\), x must be less than 90.

B is correct
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The area of the circle is 16/r. [#permalink]
2
Area of Circle =\(π\)\(r^2\)

\(16π\) = \(π\)\(r^2\)

\(r= 4\)

Now, Area of shaded region < \(4π\)

\(\frac{x}{360}\)\(π\)\(r^2\) < 4\(π\)

\(\frac{x}{360}\)\( π\)16 < 4\(π\)

\(\frac{x}{90}\) < 1

x < 90

Hence, B
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Re: The area of the circle is 16/r. [#permalink]
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Carcass wrote:
Can you help me in this?

Question never mentioned figure drawn to scale. Also, how can we take "O" as the center?

I have <4pi area with X>90 and also X<90 by my assumptions....Am I correct?


I went through three versions of the same question in different books and in one O is the center is not mentioned and in the others yes. :roll:

If x were equal to 90, the shaded region would have an area equal to 1/ 4 that of the entire circle (since 90 is 1/ 4 of 360). If the angle were equal to 90, the shaded region would have an area of \(4 \pi\) (1/ 4 of the entire circle’s area). Since the area of the shaded region is actually less than \(4 \pi\), x must be less than 90.

B is correct[/quote]

Thank you Carcass
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Re: The area of the circle is 16/r. [#permalink]
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