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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
Could someone post an explanation to this question? Would be very appreciated!
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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
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Awesome.

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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
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Can we explain the problem this way:

As the length of arc AC=BD, So, Chord AB is parallel to CD. Arc drawn on the chord CD is proportionate to the arc drawn on the chord AB. Since, arc AB<BD or AC, hence AB/CD= arc Y (say)/arc AC+BD+Y. so, the ration falls far short of 1/2. Hence, B
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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
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As a starting point, one might want to look at answer B and consider when would that chord be half of the diameter. We know that a radius would be half the diameter. Given that, we can start looking in that direction. I think that's another start, if someone is stuck, onto the path Greenlight laid out.
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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
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answer is B
that is because the length of AB = 2 CD/2sin (Q/2) Q is the angle of AB
as will as AB = CD/2 so Q =60
AS will as the arc CA= BD so their inner angles is
therefore 180 -Q = 2CA angle
this will lead to CA angle = 60 and this can't be as the question stated that Arce AB is smaller than CD not equal
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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
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Another way that is more intuitive for me:

2 conditions:
- BD > AB
- BC = AC

Therefore BD = (CD - AB)/2 and since BD > AB, (CD - AB)/2 > AB. Simplifying CD - AB > 2AB -> CD > 3AB.
Sub CD > 3AB into quantity A which is AB/CD -> 1AB/GT3AB -> simplifies to 1/MT3, which is smaller than quantity B.
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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
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chord AB=z
chord AC=BD=Y
chord CD=x
let, z be maximum by making, z=y
then,
x=3y

length of chord ab/length of chord cd
= y/x
=y/3y
=1/3
=0.33
at max quantity A is 0.33 which is less than 0.5
so quantity B is larger
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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
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we want to AB/CD also AB=CD-2AC
so substituting this we will get CD-2ac/CD
=> cd/cd-2ac/cd
lets think what will be the fraction of 2ac in cd, from the given conditions it is clear it is greater than 2/3
so lets look at possible maximum value and minimum value
=>1-2.1/3 or 1-2.9/3 value is only decrrasing and is lessa than 1 /2 so B is greater
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Re: In the semicircle above, the length of arc AC is equal to t [#permalink]
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