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 : 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). Let _{} (where _{} is in degrees) be
the angle at center (_{}COD) formed by the arc CSD. Since 360^{0} at
the central angle gives us 2_{}r as the perimeter of the circle then for _{} degree at the
center, length of the arc made by it is 1.
Length of the arc CSD = (_{}) *_{}r (unitary method) Since 360^{0} at
the central angle gives us _{}r^{2} as the area of the circle then for
_{} degree at the
center, area of the sector made by _{} degrees: 2.
Area of the sector CSDO (shaded portion in the
adjoining figure) = (_{}) *_{}r^{2} = (_{}) * (_{})= [(_{}) *_{}r]*_{} =
Length of the arc*_{}(_{}) Note: _{} radians = 180^{0}
and x^{0} = (_{}) radians 

Let
_{}COD =_{} in the adjoining figure with CD as chord We
note that Area
of triangle CDO = _{}*base*height = _{}*DO*CM
= _{}*r*rsin_{}= _{}r^{2}*sin_{} (CM = rsin_{} : Refer to section 7.1 for definition of sin of an
angle) From
the figure we notice that Area
of Sector CSDO = Area of triangle CDO + Area of segment CSD _{}Area of segment CSD = Area of Sector CSDO  Area of _{} CDO = (_{}) *_{}r^{2 } (_{}) r^{2}*sin_{} = r^{2 }{(_{}*_{})  (_{})} Note: For all the above calculations _{} must be in degrees. 

6.12 Problem 1: AB and CD are respectively
arcs of two concentric circles of radii 21Cm and 7Cm with center as O. If_{}AOB= 30, find the
area of the shaded portion
Area
of the shaded portion CABD= area of OCABDOarea of OCDO = (_{}) *_{}21^{2 }  (_{}) *_{}7^{2 }(_{} 21^{2}=7^{2}_{ }*3^{2})_{} _{ } =
_{}*_{}*7*7*(3*31) =_{}=_{} 

6.12
Problem 2: In the
figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn
with BC as diameter. Find the area of the shaded region
Area
of shaded region BPCQB= Area of semi circle BCQB Area of sector BCPB But
Area of sector BCPB = Area of quarter of circle ACPB Area of _{} ABC _{} Area of shaded
region BPCQB = Area of semi circle
BCQB – ( Area of quarter of circle ACPB Area of _{} ABC) = Area of semi circle BCQB – Area
of quarter of circle ACPB + Area of _{} ABC Note
AC=AB=14 and _{}BAC=90 By
Pythagoras theorem BC^{2} = AB^{2}
+ AC^{2} _{}BC(diameter) =_{} =_{}. _{} Radius of semi circle BCQB= _{} Area of
semi circle BCQB = _{}_{}_{}*_{}= _{}*_{}*392= 154 Area of quarter of circle ACPB
=
_{}_{}14^{2 }(_{} one fourth area of circle of radius 14cm)= _{}*_{}*14*14=154 Area of _{} ABC = _{}*14*14 = 98 (_{} Base and height of the triangle are same ) _{} Area of shaded region BPCQB= 154 154 + 98 =98_{} 

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