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KillerSquirrel
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choices are : (A) 8pi/3 (B) 8pi (C) 4pi/3 (D) 8 (E) 16 I'm 19 Apr 2007, 11:09
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choices are :

(A) 8pi/3

(B) 8pi

(C) 4pi/3

(D) 8

(E) 16

I'm looking for the shortest way possible

Thanks



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choices are : (A) 8pi/3 (B) 8pi (C) 4pi/3 (D) 8 (E) 16 I'm 19 Apr 2007, 11:23
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KillerSquirrel
choices are :

(A) 8pi/3

(B) 8pi

(C) 4pi/3

(D) 8

(E) 16

I'm looking for the shortest way possible

Thanks
Show more


Area of the sector = (60/360)*Area of the circle
= 16pi/6 =8pi/3
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KillerSquirrel
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choices are : (A) 8pi/3 (B) 8pi (C) 4pi/3 (D) 8 (E) 16 I'm 19 Apr 2007, 12:52
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goalsnr
KillerSquirrel
choices are :

(A) 8pi/3

(B) 8pi

(C) 4pi/3

(D) 8

(E) 16

I'm looking for the shortest way possible

Thanks

Area of the sector = (60/360)*Area of the circle
= 16pi/6 =8pi/3
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Nice thanks !

Thats OA
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salr15
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choices are : (A) 8pi/3 (B) 8pi (C) 4pi/3 (D) 8 (E) 16 I'm 19 Apr 2007, 15:02
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A.

radius = 4
Area of circle = 4^2 pi = 16 pi

triangle is an equalateral traingle therefore all angles are 60 degrees.


therefore, 60/360 = x/16 pi

1/6 = x/16 pi
6x = 16 pi

x = 8pi/3
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techjanson
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choices are : (A) 8pi/3 (B) 8pi (C) 4pi/3 (D) 8 (E) 16 I'm 19 Apr 2007, 20:20
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KillerSquirrel
choices are :

(A) 8pi/3

(B) 8pi

(C) 4pi/3

(D) 8

(E) 16

I'm looking for the shortest way possible

Thanks
Show more


Since all sides are equal, its an equilateral triangle with each angle 60 degrees.
So area of region is 60/360 x pi x 16 = 8pi/3

A
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Mishari
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It is an equilateral triangle with all sides equal the radius --> all angles measure 60 degrees. Now using the proportion of angle theory:

60/360 = area of shaded region / total area of circle
1/6 = x / 16 pi [ area of circle = pi x r^2 = pi 4^2 = 16 pi ]

x = 16 pi/ 6 = 8 pi /3

Answer: A



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!