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, OB = O’D (Radii) 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 = r2,                                  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  3600  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  3600  at  the central angle  gives us r2 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) = ( ) * r2 = ( ) * ( )= [( ) * r]* = Length of the arc* ( ) Note: radians = 1800 and x0 = ( ) radians Let COD = in the adjoining figure with CD as chord We note that Area of triangle CDO = *base*height = *DO*CM = *r*rsin = r2*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 = ( ) * r2 - ( ) r2*sin = r2 {( * ) - ( )} 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 OCABDO-area of OCDO =  ( ) * 212  - ( ) * 72 ( 212=72 *32)   = * *7*7*(3*3-1) = =  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 BC2 =  AB2 + AC2 BC(diameter) = = . Radius  of semi circle BCQB= Area of semi circle BCQB =   * = * *392= 154 Area of quarter of circle ACPB =  142 ( 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