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Maths
presentation
circle
s
MOHIT PATIL
MAUSAM
MANISH
KIRTI
MAYURI
NAJIB
KEWAL
MAYANK
CONTENTS
๏ƒ˜Introduction
๏ƒ˜Arc of circle
๏ƒ˜ chord diameter of circle
๏ƒ˜Area of circle
๏ƒ˜Circles in real life application
INTRODUCTION
โ€ข CIRCLE
โ€ข A circle is a collection of all points in a plane that arc at a given fixed
point in th The distance between any of the points and the centre is
called the radius. It can also be defined as the locus of a point
equidistant from a fixed point. e plane
โ€ข . A circle is a simple closed curve which divides the plane into two
regions: an interior and an exterior. In everyday use, the term "circle"
may be used interchangeably to refer to either the boundary of the
figure, or to the whole figure including its interior; in strict technical
usage, the circle is the former and the latter is called a disk.
Arc of circle
A continuous piece of circle is called arc of a
circle
Minor arc
A collection of those points of the circle
That lie on and also inside a central angle
Major arc
major arc of a circle is the collection of points
of the circle that lie on and outside a central
angle
Chord, diameter,semicircle
A line segment joining any two points on a
circle is called a chord of circle
A chord passing through centre of circle is
known as diameter of circle
A diameter of circle divides divides it into two
equal parts which are arcs each of this two
arcs is called semicircle
Area of circle
โ€ข The ratio of a circle's circumference to its
diameter is ฯ€ (pi), an irrational constant
approximately equal to 3.141592654. Thus
the length of the circumference C is related
to the radius r and diameter d by:
Circles in real life application
This is a clock. It has different shapes. Your clock can be a
square- shape. But have you notice whatโ€™s inside? I have
once open a destroyed one. It has a gear inside. Do you
know what a gear is? What is its shape? Yes itโ€™s like a
circle. Have you seen a gear that has a shape other than
circle? Difinitely, you can not see a gear that has a shape
other than circle
Theorems
of circle
THEOREM 1
If two arcs of circle are congruent then
corresponding chords are equal
Sol:
โ€ข Given,
โ€ข Arc pq =arc ab and also congruent
โ€ข TO PROVE :- PQ=AB
โ€ข PROOF
โ€ข In OPQ andOAB
โ€ข OP=OQ=Oโ€™R=Oโ€™B (Equal radii of two circles )
โ€ข POQ= ROโ€™S(ARC PQ =arc AB =>m{arc PQ }=m {AB}
โ€ข So by sas criterion of congruence we have POQ CONGRUENT
โ€ข AOB SO PQ=AB {C.P.C.T}
โ€ข HENCE PROVED
THEOREM 2
The perpendicular from center
of circle bisect the chord
SOI:
โ€ข Given
โ€ข A chord PQ of a circle
โ€ข Cons = join OP to OQ
โ€ข Proof in triangle PLO and QLO we have
โ€ข OP =OQ=r {radii of same circle}
โ€ข OL=OL (commom side )
โ€ข OLP= OLQ (each angle right
angle)
โ€ข So by RHS criterion of congruence we
have PLO= QLO
THE LINE JOINING THE
CENTRE OF CIRCLE TO THE
MID POINT OF CHORD IS
PERPEN DI CULAR TO
CHORD
SOL:-
โ€ข Given:- A chord PQ of a circle with mid point m
โ€ข Cons:-join OP to OQ
โ€ข PROOF :- in triangle OPM and OQM
โ€ข OP=OQ =r (radii of same circle)
โ€ข PM =MQ (M is mid point )
โ€ข OM=OM (COMMON SIDE}
โ€ข OPM= OQM
โ€ข OMP= OMP (C.P.C.T)
โ€ข OMP +OMQ=180 (LINEAR PAIR )
โ€ข 2 OMP =180
โ€ข OMP = 90
โ€ข OMP=OMQ =90
โ€ข OM PQ
THERE ARE ONE AND ONLY
ONE CIRCLE PASSING
THROUGH THREE NON CO
LINEAR POINTS
J
K
SOL :
โ€ข GIVEN ,three non collinear point P,Q,R AND O IS CENTRE OF
CIRCLE
TO PROVE THERE IS ONLY ONE CIRCLE PASS THROUGH THREE
NON COLLINEAR POINT
CONS= JOIN P TO Q TO R DRAW PERPENDICULAR BISECTOR AL
AND BM OF PQ AND RQ ALSO WHERE BOTH PERPENDICULAR
BISECTOR MEET NAME IT O;JOIN OR,OQ,OP
PROOF = IN TRIANGLE POJ AND QOJ
OJ=OJ (COMMON SIDE)
OJP=OJQ (EACH 90 )
PJ=JQ (PERPENDICULAR BISECTOR BISECT SIDE )
SO TRIANGLE OPJ = OJQ BY SAS CONGRUENCE SO
OP=OQ (C.P.C.T EQ 1 )
SIMARLY IN OQK AND OKR WE HAVE
OQ =OR ( C.P.C.T) EQ2
NOW FROM EQ 1 AND 2 WE GET
OP=OQ=OR
IF RADIUS ARE EQUAL
โ€ข SO THEN THE CIRCLE WILL PASS
THROUGH THREE NON COLLINEAR
POINTS
โ€ข HENCE PROVED
THE ANGLE SUBTENDED BY
ARC OF A CIRCLE AT
CENTRE IS DOUBLE THE
ANGLE SUBTENDED BY IT AT
ANY POINT ON THE
REMAINING PART OF CIRCLE
SOL=
GIVEN , AN MINOR PQ OF A CIRCLE AND
A POINT R ON REMAINING PART OF
CIRCLE i.e ARC QP
TO PROVE POQ=2 PRQ
CONS=JOIN RO PRODUCE
PROOF =
IN TRIANGLE ROP
PO=RO (RADII OF CIRCLE )
SO POR IS ISOSCELES TRIANGLE
SO RPO =PRO =x (IF SIDES ARE
EQUAL ANGLES ARE EQUAL )
โ€ข OQ= OR (RADII OF CIRCLE )
โ€ข SO ROQ IS ISOSCELES TRIANGLE
โ€ข ORQ = OQR (IF SIDES ARE
EQUAL ANGLES ARE EQUAL)
โ€ข POM IS EXTERIOR ANGLE OF
RPO SO RPO + PRO = POM
โ€ข x+x = POM (exterior angle of
triangle is equal to sum of two interior
angles) 2x = POM( EQ1)
In ROQ
Q
โ€ข NOW IN POQ
โ€ข ORQ +RQO = QOM (exterior angle
of triangle is equal to sum of two interior
angles) 2Y =QOM(EQ 2)
โ€ข FROM EQ 1 AND2 WE GET
โ€ข 2X+2Y =POM +QOM
โ€ข 2(X+Y)= POQ
โ€ข PRQ=1/2 POQ
โ€ข HENCE PROVED
THANK YOU

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mathematics

  • 3. CONTENTS ๏ƒ˜Introduction ๏ƒ˜Arc of circle ๏ƒ˜ chord diameter of circle ๏ƒ˜Area of circle ๏ƒ˜Circles in real life application
  • 4. INTRODUCTION โ€ข CIRCLE โ€ข A circle is a collection of all points in a plane that arc at a given fixed point in th The distance between any of the points and the centre is called the radius. It can also be defined as the locus of a point equidistant from a fixed point. e plane โ€ข . A circle is a simple closed curve which divides the plane into two regions: an interior and an exterior. In everyday use, the term "circle" may be used interchangeably to refer to either the boundary of the figure, or to the whole figure including its interior; in strict technical usage, the circle is the former and the latter is called a disk.
  • 5. Arc of circle A continuous piece of circle is called arc of a circle Minor arc A collection of those points of the circle That lie on and also inside a central angle Major arc major arc of a circle is the collection of points of the circle that lie on and outside a central angle
  • 6. Chord, diameter,semicircle A line segment joining any two points on a circle is called a chord of circle A chord passing through centre of circle is known as diameter of circle A diameter of circle divides divides it into two equal parts which are arcs each of this two arcs is called semicircle
  • 7. Area of circle โ€ข The ratio of a circle's circumference to its diameter is ฯ€ (pi), an irrational constant approximately equal to 3.141592654. Thus the length of the circumference C is related to the radius r and diameter d by:
  • 8. Circles in real life application This is a clock. It has different shapes. Your clock can be a square- shape. But have you notice whatโ€™s inside? I have once open a destroyed one. It has a gear inside. Do you know what a gear is? What is its shape? Yes itโ€™s like a circle. Have you seen a gear that has a shape other than circle? Difinitely, you can not see a gear that has a shape other than circle
  • 10. THEOREM 1 If two arcs of circle are congruent then corresponding chords are equal
  • 11. Sol: โ€ข Given, โ€ข Arc pq =arc ab and also congruent โ€ข TO PROVE :- PQ=AB โ€ข PROOF โ€ข In OPQ andOAB โ€ข OP=OQ=Oโ€™R=Oโ€™B (Equal radii of two circles ) โ€ข POQ= ROโ€™S(ARC PQ =arc AB =>m{arc PQ }=m {AB} โ€ข So by sas criterion of congruence we have POQ CONGRUENT โ€ข AOB SO PQ=AB {C.P.C.T} โ€ข HENCE PROVED
  • 12. THEOREM 2 The perpendicular from center of circle bisect the chord
  • 13. SOI: โ€ข Given โ€ข A chord PQ of a circle โ€ข Cons = join OP to OQ โ€ข Proof in triangle PLO and QLO we have โ€ข OP =OQ=r {radii of same circle} โ€ข OL=OL (commom side ) โ€ข OLP= OLQ (each angle right angle) โ€ข So by RHS criterion of congruence we have PLO= QLO
  • 14. THE LINE JOINING THE CENTRE OF CIRCLE TO THE MID POINT OF CHORD IS PERPEN DI CULAR TO CHORD
  • 15. SOL:- โ€ข Given:- A chord PQ of a circle with mid point m โ€ข Cons:-join OP to OQ โ€ข PROOF :- in triangle OPM and OQM โ€ข OP=OQ =r (radii of same circle) โ€ข PM =MQ (M is mid point ) โ€ข OM=OM (COMMON SIDE} โ€ข OPM= OQM โ€ข OMP= OMP (C.P.C.T) โ€ข OMP +OMQ=180 (LINEAR PAIR ) โ€ข 2 OMP =180 โ€ข OMP = 90 โ€ข OMP=OMQ =90 โ€ข OM PQ
  • 16. THERE ARE ONE AND ONLY ONE CIRCLE PASSING THROUGH THREE NON CO LINEAR POINTS J K
  • 17. SOL : โ€ข GIVEN ,three non collinear point P,Q,R AND O IS CENTRE OF CIRCLE TO PROVE THERE IS ONLY ONE CIRCLE PASS THROUGH THREE NON COLLINEAR POINT CONS= JOIN P TO Q TO R DRAW PERPENDICULAR BISECTOR AL AND BM OF PQ AND RQ ALSO WHERE BOTH PERPENDICULAR BISECTOR MEET NAME IT O;JOIN OR,OQ,OP PROOF = IN TRIANGLE POJ AND QOJ OJ=OJ (COMMON SIDE) OJP=OJQ (EACH 90 ) PJ=JQ (PERPENDICULAR BISECTOR BISECT SIDE ) SO TRIANGLE OPJ = OJQ BY SAS CONGRUENCE SO OP=OQ (C.P.C.T EQ 1 ) SIMARLY IN OQK AND OKR WE HAVE OQ =OR ( C.P.C.T) EQ2 NOW FROM EQ 1 AND 2 WE GET OP=OQ=OR
  • 18. IF RADIUS ARE EQUAL โ€ข SO THEN THE CIRCLE WILL PASS THROUGH THREE NON COLLINEAR POINTS โ€ข HENCE PROVED
  • 19. THE ANGLE SUBTENDED BY ARC OF A CIRCLE AT CENTRE IS DOUBLE THE ANGLE SUBTENDED BY IT AT ANY POINT ON THE REMAINING PART OF CIRCLE
  • 20. SOL= GIVEN , AN MINOR PQ OF A CIRCLE AND A POINT R ON REMAINING PART OF CIRCLE i.e ARC QP TO PROVE POQ=2 PRQ CONS=JOIN RO PRODUCE PROOF = IN TRIANGLE ROP PO=RO (RADII OF CIRCLE ) SO POR IS ISOSCELES TRIANGLE SO RPO =PRO =x (IF SIDES ARE EQUAL ANGLES ARE EQUAL )
  • 21. โ€ข OQ= OR (RADII OF CIRCLE ) โ€ข SO ROQ IS ISOSCELES TRIANGLE โ€ข ORQ = OQR (IF SIDES ARE EQUAL ANGLES ARE EQUAL) โ€ข POM IS EXTERIOR ANGLE OF RPO SO RPO + PRO = POM โ€ข x+x = POM (exterior angle of triangle is equal to sum of two interior angles) 2x = POM( EQ1) In ROQ
  • 22. Q โ€ข NOW IN POQ โ€ข ORQ +RQO = QOM (exterior angle of triangle is equal to sum of two interior angles) 2Y =QOM(EQ 2) โ€ข FROM EQ 1 AND2 WE GET โ€ข 2X+2Y =POM +QOM โ€ข 2(X+Y)= POQ โ€ข PRQ=1/2 POQ โ€ข HENCE PROVED