How do you divide the area of a circle into parallel identical areas?

1) If it is divided in half, it is cut along the diameter.

2) If it is a third.

is the area of three closed curves.

Let the uppermost curve form a fan-shaped AOB with an angle of 2, the radius of the circle is the upper point of the arc, and O is the center of the circle.

Then in aob, ab=2rsin, o to ab distance d=rcos s aob=r square sin cos

Sector AOB area = (2 2) * R square = R squared.

So the area of the chord ab division = sector AOB area - S aob r squared - r squared sin cos = r squared ( -sin cos ) because the third is divided into circles.

So the other string CD is the same with AB

So chord cd partition area = r squared (-sin cos) so abcd area = circle area - chord cd partition area - chord ab partitioned area.

r-squared-2r-squared (-sin cos) because of the thirds.

So ABCD area = area divided by chord AB.

So r squared -2r squared ( -sin cos ) = r squared ( -sin cos )

So -3 =3 2sin2

Solve , we can get the distance of the chord ab and cd from the center of the circle, and d gives the area of the third circle.

Nothing but the diameter.

Or the title is wrong.

Because the area of the parallelogram is equal to the base height; The base of the parallelogram is half the circumference of the circle, and the height is the radius of the circle, so the area formula of the circle is half of the circumference of the radius; R r is represented by letters, which is r.

1) If it is divided in half, it is cut along the diameter.

2) If it is a third.

is the area of three closed curves.

Let the uppermost curve form a fan to sell the code shape AOB with an angle of 2, the radius of the circle is the upper point of the arc, and O is the center of the circle.

Then in aob, ab=2rsin, o to ab distance d=rcos s aob=r square sin cos

Sector AOB area = (2 2) * R square = R squared.

So the area of the chord ab division = sector AOB area - S aob r squared - r squared sin cos = r squared ( -sin cos ) because the third is divided into circles.

So the other string CD is the same with AB

So chord cd partition area = r squared (-sin cos) so abcd area = circle area - chord cd partition area - chord ab partitioned area.

r-squared-2r-squared (-sin cos) because of the thirds.

So ABCD area = area divided by chord AB.

So r squared -2r squared ( -sin cos ) = r squared ( -sin cos )

So -3 =3 2sin2

Solve , you can get the distance between the chord ab and cd from the center of the bright circle, and d will get the area of the circle of the third-class signal.