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# In the figure above, point B lies on line segment AC

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Joined: 03 Feb 2013
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In the figure above, point B lies on line segment AC  [#permalink]

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05 Nov 2014, 08:46
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65% (hard)

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In the figure below, point B lies on line segment AC. If the two circular arcs are two semicircles whose diameters are AB and BC, what is the total length of the two arcs?

(1) AB=BC

(2) AC=20
Director
Joined: 03 Feb 2013
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Concentration: Operations, Strategy
GMAT 1: 760 Q49 V44
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WE: Engineering (Computer Software)
In the figure above, point B lies on line segment AC  [#permalink]

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Updated on: 06 Nov 2014, 06:11
Length of one semicircular arc is $$2*\pi*r$$
So the total length of the arc = $$\pi * AB/2 + \pi * BC/2 = \pi * (AB+BC)/2 = \pi * AC/2$$

So statement 2 is sufficient.

Originally posted by kinjiGC on 05 Nov 2014, 08:48.
Last edited by kinjiGC on 06 Nov 2014, 06:11, edited 3 times in total.
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Re: In the figure above, point B lies on line segment AC  [#permalink]

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05 Nov 2014, 08:54
1
kinjiGC wrote:
Length of one semicircular arc is $$2pir$$
So the total length of the arc = $$2 * pi * AB + 2 * pi * BC = 2 * pi * (AB+BC) = 2 * pi * AC$$

So statement 2 is sufficient.

How to write $$\pi$$: mark \pi and press M button to get $$\pi$$
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Joined: 31 Jul 2014
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Re: In the figure above, point B lies on line segment AC  [#permalink]

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06 Nov 2014, 06:08
1
1
Bunuel wrote:
kinjiGC wrote:
Length of one semicircular arc is $$2pir$$
So the total length of the arc = $$2 * pi * AB + 2 * pi * BC = 2 * pi * (AB+BC) = 2 * pi * AC$$

So statement 2 is sufficient.

How to write $$\pi$$: mark \pi and press M button to get $$\pi$$

sorry but, AB is diameter so length of semicircle is 0.5*pi * AB rt?
Director
Joined: 03 Feb 2013
Posts: 835
Location: India
Concentration: Operations, Strategy
GMAT 1: 760 Q49 V44
GPA: 3.88
WE: Engineering (Computer Software)
Re: In the figure above, point B lies on line segment AC  [#permalink]

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06 Nov 2014, 06:10
Bunuel wrote:
kinjiGC wrote:
Length of one semicircular arc is $$2pir$$
So the total length of the arc = $$2 * pi * AB + 2 * pi * BC = 2 * pi * (AB+BC) = 2 * pi * AC$$

So statement 2 is sufficient.

How to write $$\pi$$: mark \pi and press M button to get $$\pi$$

sorry but, AB is diameter so length of semicircle is 0.5*pi * AB rt?

Good catch. Corrected the error.
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Re: In the figure above, point B lies on line segment AC  [#permalink]

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04 Oct 2018, 21:38
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Re: In the figure above, point B lies on line segment AC   [#permalink] 04 Oct 2018, 21:38
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# In the figure above, point B lies on line segment AC

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