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Circle with Center O

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Circle with Center O [#permalink] New post 25 Aug 2018, 00:14
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Circle with center O

Quantity A
Quantity B
The length of chord PQ
The length of chord XY


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.

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Re: Circle with Center O [#permalink] New post 25 Aug 2018, 05:14
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Note that both triangles OPQ and OXY are both isosceles triangles (OP=OQ=OX=OY=radius of the circle) => the height drawn from O to either bases cut the chord PQ and XY in equal halt => by Pythagorean theorem: radius^2 = (5.9)^2+(PQ/2)^2 = (5.8)^2+(XY/2)^2. Since (5.9)^2>(5.8)^2 => (PQ/2)^2<(XY/2)^2 => PQ<XY (PQ and XY are positive of course) => B is answer.
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Re: Circle with Center O [#permalink] New post 28 Aug 2018, 10:38
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There is an alternate solution to it that does not need any calculation.
Note that the diameter is the biggest chord in the circle and as the perpendicular distance of chord increases from center, the chord becomes shorter. Therefore, chord XY is bigger as it has lesser perpendicular distance as compared to chord PQ.
Hence answer is B.
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Re: Circle with Center O [#permalink] New post 28 Aug 2018, 10:54
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You could go on full length to prove this but there's just one thing to remember here: The closer to the center, the longer the chord (here we talk about the distance between the center and the point on the chord that's closest to the center)

When it's the furthest from the center it's 0 and when it's closest the length becomes the diameter.

XY is closer to the center so XY is longer.
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Re: Circle with Center O [#permalink] New post 18 Jan 2019, 19:45
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Nearer chord to diameter is larger.
Answer B
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Re: Circle with Center O [#permalink] New post 16 Apr 2020, 03:29
Simon wrote:
Note that both triangles OPQ and OXY are both isosceles triangles (OP=OQ=OX=OY=radius of the circle) => the height drawn from O to either bases cut the chord PQ and XY in equal halt => by Pythagorean theorem: radius^2 = (5.9)^2+(PQ/2)^2 = (5.8)^2+(XY/2)^2. Since (5.9)^2>(5.8)^2 => (PQ/2)^2<(XY/2)^2 => PQ<XY (PQ and XY are positive of course) => B is answer.

how do we infer that the perpendicular from center of circle to PQ or XY cuts it into half?
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Re: Circle with Center O [#permalink] New post 16 Apr 2020, 08:34
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prototypevenom wrote:
Simon wrote:
Note that both triangles OPQ and OXY are both isosceles triangles (OP=OQ=OX=OY=radius of the circle) => the height drawn from O to either bases cut the chord PQ and XY in equal halt => by Pythagorean theorem: radius^2 = (5.9)^2+(PQ/2)^2 = (5.8)^2+(XY/2)^2. Since (5.9)^2>(5.8)^2 => (PQ/2)^2<(XY/2)^2 => PQ<XY (PQ and XY are positive of course) => B is answer.

how do we infer that the perpendicular from center of circle to PQ or XY cuts it into half?


There's a nice/important triangle property that says: The altitude of an isosceles (or an equilateral) triangle bisects the base

Image
(screenshot from my course)

Since OX and OY are both radii of the same circle, we can conclude that OX = OY.
So, following the above property, we know that the altitude cuts XY in half (i.e., bisects XY)

Cheers,
Brent
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Re: Circle with Center O   [#permalink] 16 Apr 2020, 08:34
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