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In the circle above, AC and BD are the diameters of length 10. The sum

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In the circle above, AC and BD are the diameters of length 10. The sum  [#permalink]

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New post 23 Nov 2017, 00:09
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

74% (01:44) correct 26% (01:25) wrong based on 54 sessions

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In the circle above, AC and BD are the diameters of length 10. The sum of the lengths of the arcs AXD and BYC is

(A) 4π
(B) 8π
(C) 9π
(D) 10π
(E) 16π


Attachment:
2017-11-22_1023_001.png
2017-11-22_1023_001.png [ 8.41 KiB | Viewed 862 times ]

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Re: In the circle above, AC and BD are the diameters of length 10. The sum  [#permalink]

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New post 23 Nov 2017, 02:46
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Image

AXD and BYC are both angles which are supplementary to 36 degree, Therefore, AXD = BYC = 144
The diameter AC = BD = 10, which makes the radius(r) 5.

Given data: r=5 and θ=144

The length of the arc AXD is \((\frac{θ}{360}) × 2π × r\)
Therefore, the length of the arc AXD is \(\frac{144}{360} * 2π * 5= \frac{4}{10} * 2π * 5= 4π\)

Since both the arcs have the exact same length , the total length of the arcs is 2*4π = 8π(Option B)
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Re: In the circle above, AC and BD are the diameters of length 10. The sum  [#permalink]

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New post 23 Nov 2017, 07:05
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Bunuel wrote:
Image
In the circle above, AC and BD are the diameters of length 10. The sum of the lengths of the arcs AXD and BYC is

(A) 4π
(B) 8π
(C) 9π
(D) 10π
(E) 16π


Attachment:
2017-11-22_1023_001.png

The key is to find the areas of sectors AXD and BYC, combined, as a fraction of the circle.

That fraction will hold for arc lengths and circumference. Arc lengths of arcs AXD and BYC will be the same fraction of the circumference as the sectors are of the circle's area.

Central angles of arcs AXD and BYC

The smaller sectors' central angles are 36° each (they are vertical angles, hence equal).

Each small sector's central angle lies on a straight line with one larger sector's central angle. Larger sector AXD's central angle, e.g., therefore is

180° - 36° = 144°

Sectors AXD and BYC have vertical central angles. Both sectors' central angles hence = 144°. Together, their central angles total 288°.

Central angles determine fractional area. In other words, the central angles, summed, determine what fraction of the circle the sectors AXD and BYC are.

Sectors AXD and BYC make up what fraction of the circle?

\(\frac{Sectors'Angles}{360°}=\frac{Sectors'Area}{CircleArea}\)

\(\frac{288}{360}=\frac{4}{5}=\frac{Part}{Whole}\)


Sectors AXD and BYC make up \(\frac{4}{5}\) of the circle. Their arc lengths will be \(\frac{4}{5}\) of the circumference.

Radius of circle with diameter of 10:
\(10 = 2r\), \(r = 5\)

Circumference: \(2πr = 10π\)

Arc lengths AXD and BYC are \(\frac{4}{5}\) of circumference:

\(\frac{4}{5} * 10π = 8π\)

Answer B
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Re: In the circle above, AC and BD are the diameters of length 10. The sum  [#permalink]

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New post 01 Dec 2017, 11:32
1
I got B as well. Can someone pls explain the answer?

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Re: In the circle above, AC and BD are the diameters of length 10. The sum  [#permalink]

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New post 01 Dec 2017, 11:41
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Re: In the circle above, AC and BD are the diameters of length 10. The sum   [#permalink] 01 Dec 2017, 11:41
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