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co-ordinate gprep

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Suyash
 Post subject: co-ordinate gprep
 Post Posted: Thu Mar 27, 2008 7:24 am 
6)Equilateral triangle ABC is inscribed in a circle.If length of the arc ABC is 24,what is approximate diameter of the circle?

a 5 b 8 c 11 d 15 e19.
Oa 11

How to do this Ron?
Top 
shrenik
 Post subject: Re: co-ordinate gprep
 Post Posted: Thu Mar 27, 2008 10:06 am 
here we have to consider two arcs,one is arc BAO and other is BOC (assuming centre of circle O)

then equation is 2*(120/360)*2*(22/7)*r=24

where r=radius of circle and 120 is the angle subtended by the arc at centre.

after calculation 2r=11.45which is approx 11

Suyash wrote:
6)Equilateral triangle ABC is inscribed in a circle.If length of the arc ABC is 24,what is approximate diameter of the circle?

a 5 b 8 c 11 d 15 e19.
Oa 11

How to do this Ron?
    Top 
    RonPurewal
     Post subject: Re: co-ordinate gprep
     Post Posted: Sat Mar 29, 2008 12:44 am 
    Offline
    ManhattanGMAT Staff


    Posts: 7146
    Suyash wrote:
    6)Equilateral triangle ABC is inscribed in a circle.If length of the arc ABC is 24,what is approximate diameter of the circle?

    a 5 b 8 c 11 d 15 e19.
    Oa 11

    How to do this Ron?


    ok.

    first comment: this problem is absolutely perfect for estimation. the answer choices are extremely spread out, so, if you don't clearly know how to solve the problem, just draw a decent picture and take your best guess.

    in your diagram, the arc ABC should be two-thirds of the circle's circumference. therefore, one-third of the circumference (arc AB, or arc BC) is 12. if you've drawn a good picture, then, if you imagine straightening out that 12, it's awfully close to the diameter of the circle. therefore, you'd go with choice c.

    --

    the real way to do the problem:
    two-thirds of the circle's circumference is 24.
    therefore, the circle's circumference is 36.
    36 = pi * diameter
    diameter = 36 / pi
    = 36 / (a lil more than 3)
    = a lil less than 12
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    rajibgmat
     Post subject: In the figure above, equilateral triangle ABC is inscribed
     Post Posted: Sun Nov 23, 2008 8:47 pm 
    [img]
    Image
    [/img]
    Top 
    RonPurewal
     Post subject: Re: In the figure above, equilateral triangle ABC is inscrib
     Post Posted: Sat Nov 29, 2008 7:55 am 
    Offline
    ManhattanGMAT Staff


    Posts: 7146
    rajibgmat wrote:
    [img]
    Image
    [/img]


    that works too.

    in your diagram, you should probably trace out the arcs in color, rather than pointing to them with arrows; especially for larger arcs, it's unclear exactly where the arrow is pointing. (you can't really point at two-thirds of a whole circle - i.e., a 240° arc - with a single arrow, cool colors notwithstanding.)[/list]
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    zarak_khan
     Post subject: Re: co-ordinate gprep
     Post Posted: Mon May 17, 2010 2:24 am 
    Offline
    Course Students


    Posts: 25
    Hi Ron,

    This is how I solved this problem:

    < ABC = 60 since this is an equilateral triangle
    Inscribed angle at center of circle must be twice < ABC = 120

    Now, 120 deg corresponds to 24 and 360 deg corresponds to 2*pi*r where r = radius

    --> 120 / 360 = 24 / (2*Pi*r)
    --> 1/3 = 12/(pi*r)
    --> pi*r = 36
    --> r = 36/pi
    --> 2r = diameter = 72/pi = approx. 22

    Can you please see where my mistake is?

    Thanks!!
    Top 
    RonPurewal
     Post subject: Re: co-ordinate gprep
     Post Posted: Sun May 23, 2010 6:14 am 
    Offline
    ManhattanGMAT Staff


    Posts: 7146
    zarak

    your mistake is here:

    zarak_khan wrote:
    120 deg corresponds to 24


    nope.

    120 degrees would be any of these arcs:
    AB
    AC
    BC

    ... but, in this problem, we are given arc ABC. that is two-thirds of the way around the circle (draw a picture if you don't see why).
    so 240 degrees, not 120 degrees, corresponds to an arc length of 24.
    Top 
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