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#GREpracticequestion In the semicircle above, the length of arc AC is equal to the length of arc BD.jpg [ 7.64 KiB | Viewed 28693 times ]
In the semicircle above, the length of arc AC is equal to the length of arc BD, and the length of arc AB is less than the length of arc BD.
Quantity A
Quantity B
\(\frac{\text{the length of chord AB}}{\text{the length of chord CD}}\)
\(\frac{1}{2}\)
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
In the semicircle above, the length of arc AC is equal to the length of arc BD, and the length of arc AB is less than the length of arc BD.
Quantity A
Quantity B
\(\frac{\text{the length of chord AB}}{\text{the length of chord CD}}\)
\(\frac{1}{2}\)
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
Okay, so here's what we have so far...
As you can see, I've added the entire circle AND I've added the circle's center
Now the question tells us that the length of arc AB is less than the length of arc BD. However, at this point, I want to investigate what would happen if it were the case that the length of arc AB is equal to the length of arc BD We'd get something like this...
Notice that all 3 arcs (AC, AB, and BC) all have the SAME length.
This means that each CENTRAL ANGLE that "holds" these 3 equal arcs must also be equal...
Since all three angles are on the same line (the diameter to be exact), they must add to 180°, which means each angle must be 60°
Since OA and OB are the radii of the circle, we can conclude that ∠OAB and ∠OBA must both be equal, which means they both equal 60°
So, ∆OAB is an EQUILATERAL TRIANGLE, which means all 3 sides have equal length. In fact all 3 sides are equal to the radius of the circle.
Since OC and OD are also radii, we can see that we have a BUNCH of line segments that are all the same length.
At this point, we can see that: (length of chord AB)/(length of chord CD) = 1/2 [since AB = the length of 1 radius, and CD = the length of 2 radii]
So, if it were the case that the length of arc AB is equal to the length of arc BD, then Quantities A and B would be EQUAL.
However, the original question tells us that the length of arc AB is less than the length of arc BD. From this, we can conclude that chord AB is LESS THAN the radius of the circle.
This means (length of chord AB)/(length of chord CD) < 1/2
Re: In the semicircle above, the length of arc AC is equal to t
[#permalink]
13 Mar 2019, 08:37
1
1
Bookmarks
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
Re: In the semicircle above, the length of arc AC is equal to t
[#permalink]
10 Sep 2019, 09:16
1
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.
Re: In the semicircle above, the length of arc AC is equal to t
[#permalink]
17 Sep 2021, 13:11
1
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
Re: In the semicircle above, the length of arc AC is equal to t
[#permalink]
20 Dec 2021, 22:30
2
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.