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.
Hence
by SAS Postulate on congruence Hence AB = CD |
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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 Area of the circle = Where
Let Since 3600 at
the central angle gives us 2 for 1.
Length of the arc CSD = ( Since 3600 at
the central angle gives us for
2.
Area of the sector CSDO (shaded portion in the
adjoining figure) = ( = ( =
Length of the arc* Note: |
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Let
We
note that Area
of triangle CDO = (CM = rsin From
the figure we notice that Area
of Sector CSDO = Area of triangle CDO + Area of segment CSD
= ( = r2 {( Note: For all the above calculations |
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6.12 Problem 1: AB and CD are respectively
arcs of two concentric circles of radii 21Cm and 7Cm with center as O. IfAOB= 30, find the
area of the shaded portion
Area
of the shaded portion CABD= area of OCABDO-area of OCDO = ( =
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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
= Area of semi circle BCQB – Area
of quarter of circle ACPB + Area of Note
AC=AB=14 and By
Pythagoras theorem BC2 = AB2
+ AC2
Area of
semi circle BCQB = Area of quarter of circle ACPB
=
Area of
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6.12 Summary of learning
No |
Points to
remember |
1 |
Congruency
of arcs |
2 |
Formula
for length of an arc, Area of a segment |