Problem Solving

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?

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

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.

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.

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.

@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.

@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?

Nermal n' alexchow got it bang on!