6.12 Circles  Part 3:
6.12.1:
Arcs of a circle
Two arcs of two different circles having same radii are said to
be ‘congruent’ if their central angles
are same.
Arc ASB = Arc CTD if _{}AOB = _{}CO’D
6.12.1 Theorem 1: If two arcs are congruent then their chords are equal
To prove: AB=CD
Proof:
1. OA = O’C, 2.
_{}AOB = _{}CO’D (it is given that arcs are congruent) Hence
by SAS Postulate on congruence _{}AOB _{}_{}CO’D Hence AB = CD 

6.12.1 Theorem 2: If two chords of circles having same radii are same,
then their arcs are congruent.
Note: This is converse of the previous theorem.
Use SSS postulate to show that _{}AOB = _{}CO’D
6.12.1:
Areas of sectors/segments of circle
If ‘r’ is the radius of a
circle, we know that the circumference and area of the circle are given by Circumference of the circle = 2_{}r, Area
of the circle = _{}r^{2}, Where _{} is a constant whose
approximate value we use for our calculations is 22/7 (3.1428). If _{} (where _{} is in degrees) is
the angle at center (_{}COD) formed by the arc CSD then 1. Length of the arc CSD = (_{}/180) *_{}r 2. Area of the sector CSDO (shaded portion in the adjoining figure) =
(_{}/360) *_{}r^{2} = (_{}/180) *(_{}r*r)/2 = {(_{}/180) *_{}r}*(r/2) = Length of the arc*(radius/2) Note: _{} radians = 180^{0}
and x^{0} = (x*_{})/180 radians 

Let _{}AOB =_{} in the adjoining figure with AB as chord We note that Area of triangle ABO =
(1/2)*base*height = (1/2)*BO*AM = (1/2) *r*rsin_{}= (1/2) r^{2}*sin_{} (AM = rsin_{} : Refer to section 7.1 for definition of sin of an
angle) From the figure we notice that Area of Sector ASBO = Area of
triangle ABO + Area of segment ASB _{}Area of segment ASB = Area of Sector ASBO  Area of
triangle ABO = (_{}/360) *_{}r^{2 } (1/2) r^{2}*sin_{} = r^{2 }{(_{}*_{}/360)  (sin_{}/2)} Note: For all the above
calculations _{} must be in degrees. 

6.12 Summary of learning
No 
Points to remember 
1 
Congruency
of arcs 
2 
Formula
for length of an arc, area of an arc, Area of a segment 