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Length of an arc

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by uptowngirl92 » Sun Sep 13, 2009 9:15 pm
AE // BD. AD is also the diameter.If the length of arc BCD is 4pi then what is the area of the circle??

My method:
Join BA. Angle ABD is 90deg. Therefore BAD =60
Now, the formula is:
Length of an arc= circumference x angle presented by the arc/360
Hence, 4pi=2.pi.r.60/360
r=12
Area=144pi.

Now could somebody please point out where I am going wrong?
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by Nermal » Mon Sep 14, 2009 6:20 am
The formula you were applying refers to the angle at the origin of the circle.

As you can see in the graphic attached, the angle has to be 120°.

Then you can plug in: 2pi*r*120/360=4pi
r=6
pi*r^2=pi*6^2=36pi
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by alexchow » Mon Sep 14, 2009 6:22 am
Hi,

I think where you went wrong is you confused a central angle (vertex on center) with an inscribed angle (vertex on circle). Your formula is correct for a central angle but BAD is an inscribed angle.

However, a central angle is always twice the inscribed angle on the same arc, so you could do:
4pi=2*pi*r*120/360 that is r=6 and area=36*pi.
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by grockit_jake » Mon Sep 14, 2009 5:50 pm
If you have 2 points on the outside of a circle, the angle created with the center is 2x the angle created with any other point on the circle.
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by acenikk » Tue Sep 15, 2009 6:37 am
Hi Nirmal, Can you tell me how you got 30 degree for the angle OBD ?

@Uptowngirl, what is the answer ?

For me, the area comes out to be 64Pi. I guess ABDE forms a square. If yes, then the radius can be calculated as 2*Pi*r = 4*4Pi => r = 8.
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by Nermal » Tue Sep 15, 2009 7:17 am
@acenikk:
OBD is 30° because OBD is an isoceles triangles with two sides of radius r. If the sides are of the same length the angles have to be of the same measure.
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by acenikk » Tue Sep 15, 2009 8:02 am
@Nirmal: you are right. thanks.

BTW you have any idea if there exists any relation between the side of a square inscribed in the circle and the length of the arc formed by that side of the square?
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by uptowngirl92 » Tue Sep 15, 2009 2:48 pm
Nermal n' alexchow got it bang on!
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