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# In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete

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Math Expert
Joined: 02 Sep 2009
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In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete  [#permalink]

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17 Jan 2016, 12:08
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55% (hard)

Question Stats:

71% (03:18) correct 29% (03:07) wrong based on 94 sessions

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In the rhombus, BC = 6, AE = 4, and angle DAE = 45°. AD is the diameter of the circle. If a man started at C and followed around the outer edge of this figure to D, F, A, G, E, B, and back to C, approximately how far did he travel?

A. 14 + 27/4*π
B. 14 + 6π
C. 12 + 6π
D. 14 + 9/2*π
E. 12 + 9/2*π

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2016-01-17_2306.png [ 7.02 KiB | Viewed 1497 times ]

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Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete  [#permalink]

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17 Jan 2016, 21:33
Bunuel wrote:

In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diameter of the circle. If a man started at C and followed around the outer edge of this figure to D, F, A, G, E, B, and back to C, approximately how far did he travel?

A. 14 + 27/4*π
B. 14 + 6π
C. 12 + 6π
D. 14 + 9/2*π
E. 12 + 9/2*π

Attachment:
2016-01-17_2306.png

Hi Bunuel,
again a good Q but there is a typo AE , 4, should be AE=4..

I'll solve it in two ways ..
first by POE and second by proper algebra..

1)POE:-
first lets look at the perimeter of rhombus one has to walk= 6+6+(6-4)=14..
our answer will be 14 + something 'π'..
A, B and D remain..
now lets look at the circle..
the radius = side of rhombus/2=6/2=3..
so perimeter of circle= 6 π...
our answer has to be less than 14+6π..

lets see A,B, and D..

A. 14 + 27/4*π= 14+6 1/4*π>14+6π.. not possible

B. 14 + 6π= 14+6π.., but our ansewr has to be less than that.. so eliminate

D. 14 + 9/2*π = 14 + 4 1/2*π <14+6π.. what we were looking for..
ans D..

2) algebric way..
GEOMETRY RULE :- if an arc(DE here) makes an angle x(45) from a point on circumference(A here), the angle from center to this arc will be 2x, 90 degree here..

Sice the arc makes a 90 angle at center, the circumference will be 2 pi *r*90/360... = 1/4 circumference..
thus he walks on 3/4 circumference..
circumference=2pi*r=6pi..
he walks 6pi*3/4=9/2pi on circle..
on rhombus he walks 6=6+2=14..
total=14+9pi/2....
D
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Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete  [#permalink]

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25 Mar 2016, 22:38
DFAG=Total Perimeter (2pi*r -2pi*r/4) ...............(1)

6+6+2=14................(2)

So 14+4.5pi
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Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete  [#permalink]

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04 Apr 2016, 14:48
Here is my approach => Total distance covered on the rhombus => 14
On the circle => 6 * pie - (90/360 * 6*pie) =>4.5 *pie => 9/2 *pie
adding them => 14+ 9/2 * pie
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Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete  [#permalink]

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16 Apr 2018, 10:53
Hi Bunuel,

I have one question, the value of AE must be $$3 \sqrt{2}$$, given DAE is 45 degrees and
DOE is 90 degrees (by central angle theorem), so triangle AOE must be right angled isoceles triangle in the ratio $$1:1:\sqrt{2}$$ and AE must be $$3 \sqrt{2}$$
Please correct me if i am wrong.

Thanks
Re: In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete &nbs [#permalink] 16 Apr 2018, 10:53
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# In the rhombus, BC = 6, AE , 4, and angle DAE = 45°. AD is the diamete

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