## Re: [ascent CAT] easy solution wanted

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• hi.. i m not sure if this could be called a solution..but it is rather a quicker way of approximation...tell me if it makes sense the fig consists of a hexagon
Message 1 of 4 , Sep 2, 2003
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hi..
i m not sure if this could be called a solution..but
it is rather a quicker way of approximation...tell me
if it makes sense

the fig consists of a hexagon inscribed in a
circle...the angles subtended at the centre by the
sides will be proportional to the length of the
sides...
so 2x+2x+7x+7x+11x+11x=360
x=9
so the angles are 18,18,63,63,99 and 99

the angle subtended at the centre by the 7cm side is
63..
so in the corr triangle formed by that side the other
2 angles have to be something around 60 each ( since
sides..r..r..and 7)
so the triangle is equilateral and hence r=7

--- bhakta kishore panigrahi
<bhaktakishore@...> wrote: > Hi Amit,
> I am answering 2nd question.
> let angles subtended at centre by sides
> 2,2,7,7,11,11 be
> x1,x1,x2,x2,x3,x3.
> now x1+x2+x3 = 180 degree or x1/2 + x2/2 + x3/2 = 90
> degree.
> or x1/2 + x2/2 = 90 - x3/2
> or sin ( x1/2 + x2/2 ) = cos(x3/2)
> or sin(x1/2)*cos(x2/2) + cos(x1/2)*sin(x2/2) =
> cos(x3/2)
>
> sin(x1/2) = 1/r where r = radius of the circle
> cos(x1/2) = (sqrt(1-r^2))/r
> sin(x2/2) = 7/(2r)
> cos(x2/2) = ( sqrt(4r^2 - 49) )/(2r)
> cos(x3/2) = ( sqrt(4r^2 - 121))/(2r)
>
> putting these sine and cosine values we get the
> following equation
>
> sqrt(4r^2 - 49 ) + 7* sqrt(r^2 - 1) = r* sqrt(4r^2 -
> 121)
>
> if you solve this equation conventionally it is time
> consuming.
> hence check with the options given. 7 comes out to
> be the value for r.
>
> *** anybody having a simpler solution is requested
> to help us.***
>
> Bhakta
>
>
>
>
>
>
>
> --- a_mit97 <a_mit97@...> wrote:
> > Hi all,
> CAT approach.
> >
> > 1. An urn contains a number of colored balls,
> with equal numbers of
> > each color. Adding 20 balls of a new color to the
> urn would not
> > change the probability of drawing (without
> replacement) two balls of
> > the same color.
> >
> > How many balls are in the urn? (Before the extra
> >
> > a) 210
> > b) 95
> > c) 190
> > d) Data Insufficient
> >
> > 2. A hexagon with consecutive sides of lengths 2,
> 2, 7, 7, 11, and
> > 11 is inscribed in a circle. Find the radius of
> the circle.
> >
> > a) 2
> > b) 7
> > c) 11
> > d) None of these
> >
> >
> > Arnab
> >
> >
> >
> >
> >
> >
>
>
> __________________________________
> Do you Yahoo!?
> Yahoo! SiteBuilder - Free, easy-to-use web site
> design software
> http://sitebuilder.yahoo.com
>
>

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• Hi, I congratulate you for method to arrive at the soln in an easier and quicker way. However I have certain points regarding this method which I am describing
Message 2 of 4 , Sep 10, 2003
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Hi,

I congratulate you for method to arrive at the soln in
an easier and quicker way.
However I have certain points regarding this method
which I am describing below.

1. the angles subtended at the centre by different
arcs are in the same proportion as to the proportion
of length of respective arcs.
But you have applied this principle to chords not
arcs.

For further clarification let us take a small example.

In a circle of radius 2 cm a chord of length
2*sqrt(2) cm will subtend 90 degree at the centre.
For a chord of length sqrt(2) cm , according to you
will subtend 45 degree at the centre which is not the
case.
In fact a chord of length 8 - 4 * sqrt(2) cm will
subtend 45 degree at the centre.

2. if none of the angles found by your method comes
closer to 60 degree then how to find the radius.

3. If choices are too closer then how to guess the

Thanks

Bhakta
subtend
--- disha j <girlfromvenus1981@...> wrote:
> hi..
> i m not sure if this could be called a solution..but
> it is rather a quicker way of approximation...tell
> me
> if it makes sense
>
> the fig consists of a hexagon inscribed in a
> circle...the angles subtended at the centre by the
> sides will be proportional to the length of the
> sides...
> so 2x+2x+7x+7x+11x+11x=360
> x=9
> so the angles are 18,18,63,63,99 and 99
>
> the angle subtended at the centre by the 7cm side is
> 63..
> so in the corr triangle formed by that side the
> other
> 2 angles have to be something around 60 each ( since
> sides..r..r..and 7)
> so the triangle is equilateral and hence r=7
>
>
>
>
> --- bhakta kishore panigrahi
> <bhaktakishore@...> wrote: > Hi Amit,
> > I am answering 2nd question.
> > let angles subtended at centre by sides
> > 2,2,7,7,11,11 be
> > x1,x1,x2,x2,x3,x3.
> > now x1+x2+x3 = 180 degree or x1/2 + x2/2 + x3/2 =
> 90
> > degree.
> > or x1/2 + x2/2 = 90 - x3/2
> > or sin ( x1/2 + x2/2 ) = cos(x3/2)
> > or sin(x1/2)*cos(x2/2) + cos(x1/2)*sin(x2/2) =
> > cos(x3/2)
> >
> > sin(x1/2) = 1/r where r = radius of the circle
> > cos(x1/2) = (sqrt(1-r^2))/r
> > sin(x2/2) = 7/(2r)
> > cos(x2/2) = ( sqrt(4r^2 - 49) )/(2r)
> > cos(x3/2) = ( sqrt(4r^2 - 121))/(2r)
> >
> > putting these sine and cosine values we get the
> > following equation
> >
> > sqrt(4r^2 - 49 ) + 7* sqrt(r^2 - 1) = r* sqrt(4r^2
> -
> > 121)
> >
> > if you solve this equation conventionally it is
> time
> > consuming.
> > hence check with the options given. 7 comes out
> to
> > be the value for r.
> >
> > *** anybody having a simpler solution is
> requested
> > to help us.***
> >
> > Bhakta
> >
> >
> >
> >
> >
> >
> >
> > --- a_mit97 <a_mit97@...> wrote:
> > > Hi all,
> in
> > CAT approach.
> > >
> > > 1. An urn contains a number of colored balls,
> > with equal numbers of
> > > each color. Adding 20 balls of a new color to
> the
> > urn would not
> > > change the probability of drawing (without
> > replacement) two balls of
> > > the same color.
> > >
> > > How many balls are in the urn? (Before the
> extra
> > >
> > > a) 210
> > > b) 95
> > > c) 190
> > > d) Data Insufficient
> > >
> > > 2. A hexagon with consecutive sides of lengths
> 2,
> > 2, 7, 7, 11, and
> > > 11 is inscribed in a circle. Find the radius of
> > the circle.
> > >
> > > a) 2
> > > b) 7
> > > c) 11
> > > d) None of these
> > >
> > >
> > > Arnab
> > >
> > >
> > >
> > >
> > >
> > >
> >
> >
> > __________________________________
> > Do you Yahoo!?
> > Yahoo! SiteBuilder - Free, easy-to-use web site
> > design software
> > http://sitebuilder.yahoo.com
> >
> >
>
>
________________________________________________________________________
> Want to chat instantly with your online friends?
> Get the FREE Yahoo!
> Messenger http://uk.messenger.yahoo.com/
>
>

__________________________________
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