Aryan's Quadrilaterals

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WELCOME TO OUR PRESENTATION
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Creane memorial high school
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FA-3
GROUP Thales
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GROUP MEMBERS
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S. No. Name Roll No.
01 Aryan Sinha 07
02 Gaurav Vats 14
03 Hritik Kumar 15
04 Samyak Jain 37
05 Shiraz Ali 40
06 Shwetank Kumar 45
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Preface

• Our project provides complete information on the
  topic quadrilaterals and maximum efforts have
  been taken to make it more comprehensive and
  lucid to understand. The Project uses most simple
  words so that it is easy to understand and
  expressive. The success of this effort depends on
  the steps taken by each member to encourage by
  reflecting their own learning and to pursue
  imaginative work.
• We hope this project will be liked by you due to
  keen devotion of all our members and guidelines
  given by you.

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ACKNOWLEDGEMENT

• First of all we would like to express our deepest
  thank to our principal who gave us this
  wonderful project to work on. Then we would like
  to give special gratitude to our teacher Sir Ravi
  who encouraged and stimulated us suggestions to
  do the best. We would also thank each of us for
  excellent coordination. Then we would thank our
  parents who supported us in all the ways they
  could. At last we would thank to all others who
  helped us in one or other manner to finish the
  project.

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quadrilaterals
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QUADRILATERALS
The word quadrilateral comes from two words
quad- four and lateral- sides. Thus a closed
plane figure made up of four line segments is
called a quadrilateral. A quadrilateral has 4
sides, 4 vertices, 4 angles and 2 diagonals.
D
C
O
A
B
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Sum of angles of a quadrilateral is 360
C
D

• Given A quad. ABCD.
• To prove?A?B?C?D 360 .
• Construction Join A to C.
• Proof In ? ABC,
• ?CAB?ACB?CBA 180 .
  (A.S.P).1
• In ? ACD ,
• ?ADC?DCA?CAB 180 .
  (A.S.P)....2
• Adding 1 and 2
• ?CAB?ACB?CBA?ADC?DCA?CAD1801
  80.
• (?CAB?BAC)?ABC(?BCA?ACD)?ADC36
  0 .
• ??A?B?C?D 360 .

A
B
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TYPES OF QUADRILATERALS
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(No Transcript)
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TRAPEZIUM

• A quadrilateral in which one pair of opposite
  sides are parallel and other pair is
  non-parallel, is called a trapezium.

• Isosceles Trapezium- A trapezium in which non
  parallel sides are equal is called isosceles
  trapezium.
• Properties of isosceles trapezium-
• Diagonals are equal.
• Base angles are equal.

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PARALLELOGRAM

• A quadrilateral in which both pairs of opposite
  side are parallel is known as parallelogram.

• Properties of parallelogram-
• Opposite sides are equal.
• Opposite angles are equal.
• Diagonals bisect the parallelogram into two
  congruent triangles.
• Diagonals bisect each other.
• Adjacent angles are supplementary.

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rectangle

• A parallelogram in which one of the angles is
  right angle is known as rectangle.

• Properties of rectangle-
• All the properties of parallelogram.
• Diagonals are equal.
• All angles are right angle.

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rhombus

• A parallelogram in which all sides are equal is
  known as rectangle.

• Properties of rhombus-
• All the properties of parallelogram.
• Diagonals bisect each other at right angles.
• Diagonals bisect the
• All angles are right angle.

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square

• A rectangle in which all sides are equal is known
  as square.

• Properties of a square-
• The diagonals of a square are equal to each
  other.
• The diagonals of a square are perpendicular to
  each other.
• All properties of parallelogram

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KITE

• A quadrilateral in which two pairs of adjacent
  sides are equal is known as kite.

• Properties of a kite-
• Opposite angles are equal.
• The diagonals bisect each other at right angles
• All properties of rectangle.

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theorems
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• The sum of all the angles of a quadrilateral is
  360 degrees.

• A diagonal of a parallelogram divides it into two
  congruent triangles.
• In a parallelogram opposite sides are equal.
• If each pair of opposite sides of a quadrilateral
  is equal then it is a parallelogram.
• In a parallelogram opposite angles are equal.
• If in a quadrilateral each pair of opposite
  angles are equal, then it is a parallelogram.

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• The diagonals of a parallelogram bisect each
  other.
• If the diagonals of a quadrilateral bisect each
  other, then it is a parallelogram.
• A quadrilateral is a parallelogram if a pair of
  opposite sides is equal and parallel.
• The line segment joining the mid-points of two
  sides of a triangle is parallel to the third
  side.
• The line drawn through the mid-point of one side
  of a triangle parallel to another side bisects
  the third side.

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The diagonals of a parallelogram divides it into
two congruent triangles
C
D

• Given A parallelogram ABCD.
• To prove ? ABC ? ? ADC
• Construction Joined AC and BD.
• Proof In ? ABC and ? ADC,
• AB is parallel to CD and AC is
  transversal.
• ?BAC ?DCA (alternate angles)
• ?BCA ?DAC (alternate angles)
• AC AC
  (common side)
• ? ? ABC ? ? CDA (ASA rule)
• Similarly, ? ABD ? ? CDB
• Hence Proved.

A
B
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IN A PARALLELOGRAM OPPOSITE SIDES ARE EQUAL
C
D

• Given A parallelogram ABCD.
• To Prove AB DC and AD BC
• Construction Join A to C
• Proof In ? ABC and ? ADC,
• AB is a parallelogram to CD and AC is the
  transversal.
• ?BAC ?DCA (Alternate angles)
• ? BCA ? DAC (Alternate angles)
• AC AC (Common
  side)
• ? ? ABC ? ? CDA (ASA rule)
• Now AB DC and AD BC (C.P.C.T)
• Hence, proved.

A
B
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IF OPPOSITE SIDES OF A QUADRILATERAL ARE EQUAL
THEN IT IS A PARALLELOGRAM
C
D

• Given A quad ABCD in which ABCD ADBC
• To Prove ABCD is a parallelogram.
• Construction Joined AC.
• Proof In ? ABC and ? CDA,
• AB CD (Given)
• AD BC (Given)
• AC AC (Common side)
• ? ? ABC ? ? CDA (SSS Rule)
• ? ? BAC ? DCA (C.P.C.T)
• But, these are alternate interior ?s.
• ? ABCD
• Similarly, AD BC
• ? ABCD is a parallelogram.

A
B
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In a parallelogram opposite angles are equal.
IN A PARALLELOGRAM OPPOSITE ANGLES ARE EQUAL
C
D

• Given A parallelogram ABCD.
• To Prove ?A ?C ?B?D
• Proof In the parallelogram ABCD,
• ?AB is a parallel to CD AD is a transversal
• ??A?D 180 (co-interior angles)1
• Similarly
• ?A?B180 ..2
• From 1 and 2,
• ?A?D ?A?B.
• ??D ?B.
• Similarly ,?A ?C.

A
B
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A QUADRILATERAL IS SAID TO BE A PARALLELOGRAM IF
EACH PAIR OF OPPOSITE ANGLES IS EQUAL
IF OPPOSITE ANGLES OF A QUADRILATERAL ARE EQUAL
THEN IT IS A PARALLELOGRAM
C
D

• Given In a quadrilateral ABCD ?A?C
• ?B ?D.
• To prove It is a parallelogram.
• Proof ?A?B?C?D360
• 2?A 2?B360 (? ?A?C and?B?D)
• 2(?A?B)360
• ?A?B 180. (co-interior angles.)
• ?AD BC
• Similarly we can prove AB CD.
• This implies that ABCD is a parallelogram.

A
B
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THE DIALGONALS OF A PARALLELOGRAM BISECT EACH
OTHER
C
D

• Given A parallelogram ABCD
• To Prove AC and BD bisects each other.
• Proof ? AD BC BD is transversal.
• ? ?CBD?ADB (alternate interior angles)
• ? In ? BOC and ? AOD,
• ? CBD ? ADB
• ? DAC ? ACB
• BCAD (opposite sides of gm)
• ? ? BOC ? ? AOD (by ASA rule)
• ? AOOC BOOD (C.P.C.T)
• This implies that diagonals of a parallelogram
  bisect each other.

O
B
A
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If the diagonals of a quadrilateral bisect each
other then it is a parallelogram.
IF DIAGONALS OF A QUADRILATERAL BISECT EACH OTHER
THEN IT IS A PARALLELOGRAM
C
D

• Given In a quadrilateral ABCD,
• AO OC BO OD
• To prove ABCD is a parallelogram.
• Proof In ? AOD ? BOC
• AO OC (given)
• BO OD (given)
• ?AOD ?BOC (V.O.As)
• ?, ? BOC ? ? AOD (by SAS rule)
• ? ?ADB?CBD ?DAC?ACB (C.P.C.T)
• Since alternate angles are equal, AD is parallel
  to BC.
• Similarly, we can prove AB is parallel to CD.
• This implies that ABCD is a parallelogram.

O
B
A
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IF ONE PAIR OF OPPOSITE SIDES OF QUADRILATERAL
ARE EQUAL AND PARALLEL THEN IT IS PARALLELOGRAM
C
D

• Given In a quadrilateral ABCD,
• AB CD ABCD
• To prove ABCD is a parallelogram .
• Construction Join A to C.
• Proof In ? ABC ? ADC,
• ABCD (given)
• ?BAC ?DCA (alternate angles) AC AC
  (common)
• ? ? ABC ? ? ADC (by SAS rule)
• ? ?ACB ?DAC and AD BC (c.p.c.t)
• ? AD is parallel to BC and AD BC
• ? ABCD is a parallelogram.

A
B
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THE LINE JOINING THE MID POINTS OF TWO SIDES OF A
TRIANGLE IS PARALLEL AND HALF TO THIRD SIDE

• Given S T are mid points of PQ PR
  respectively.
• To prove ST QR ST1/2QR
• Construction Drawn UR SQ and
• Proof SQUR SQUR
• ? SQRU is a gm.
• ? SU QR SU QR
• ST QR
• In ? PST ? RUT
• ?P ?R (alt. int angles)
• ?T ?T (V.O.As)
• PT TR (given)
• ? ? PST ? ? RUT (ASA rule)
• ? STTU (C.P.C.T)
• ? ST ½SU ?ST ½QR

P
---------------
---------------
---------------
U
T
S
R
Q
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THE LINE DRAWN FROM MID POINT OF ONE SIDE OF A
TRIANGLE PARALLEL TO OTHER BISECTS THE THIRD SIDE
P

• Given ST QR and QS PS
• To Prove PT TR
• Construction Drawn RUQS and ST is extended to
  form SU.
• Proof ST QR
• ? SU QR
• QSRU ? QRUS is a gm
• ? QS RU
• but QS PS
• PS RU
• In ?PST and ?RUT
• PS RU
• ?T ?T (V. O. ?s)
• ?P ?R (Alt. Int. ?s)
• ? ?PST ?RUT (AAS)
• PT TR (CPCT)

---------------
---------------
---------------
U
T
S
R
Q
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THE DIAGONALS OF A RECTANGLE ARE EQUAL
C
D

• Given A rectangle ABCD in which
  AC and BD are its diagonals.
• To prove ACBD
• Proof In ?ABD and? BAC,
• ABBA (common)
• ?A?B (each equal to 90 degree)
• ADBC (opposite sides of gm)
• ??ABD ? ?BAC (SAS-criteria).
• Hence, BDAC

A
B
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IF THE DIAGONALS OF A PARALLELOGRAM ARE EQUAL
THEN IT IS A RECTANGLR
C
D

• Given- A gm ABCD in which AC BD.
• To prove- ABCD is a rectangle.
• Proof- In ? ABC and ? DCB
• AB DC (OPP SIDES OF A gm)
• BC CB (common)
• AC DB (given)
• ? ABC ? ? DCB
• But, DC AB and CB cuts them.
• ?ABC ?DCB 180 co.int.?s
• ?ABC ?DCB 90 Using (1)
• Thus, ABCD is a gm, one of whose angles is 90
• HENCE, ABCD is a rectangle.

O
A
B
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THE DIAGONALS OF A RHOMBUS ARE PERPENDICULAR
BISECTORS OF EACH OTHER

• Given- A rhombus ABCD whose diagonals
  
  AC and BD intersect at a point O.
• To prove- ?BOC ?DOC ?AOB ? DOA 90
• Proof- Clearly ABCD is agm, in which
  ABBCCDDA.
• ? OAOC and OB OD (ABCD Is a gm)
• Now, In triangle BOC and DOC,
• OB OD,BC DC and OC OC
• So , triangle BOC ? DOC
• ? BOC ?DOC (c.p.c.t).
• But, ?BOC ?DOC 180 (linear pair)
• ?BOC ?DOC 90
• Hence , the diagonals of a rhombus bisect each
  other at a right angle.

D
C
O
A
B
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If THE DIAGONALS of a parallelogram are
perpendicular to each other, it is a rhombus

• Given- A gm ABCD whose diagonals AC and BD
• intersect at O such that AC is perpendicular
• to BD.
• To prove- ABCD is a rhombus.
• Proof- ? The diagonals of a gm bisect each
  other
• ? OA OC and OB OD
• In ? AOD and ? COD
• OA OC ?AOD ?COD 90 and OD is common
• ? AOD ? ? COD
• AD CD ( c.p.c.t.).
• Now , AB CD and AD BC ( opp. sides of a gm
  )
• and AD CD AD BC
• Hence , ABCD is a rhombus.

D
C
O
B
A
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Conclusion

• The project on quadrilaterals was very helpful,
  fun and educational too. It gave us a lot of
  information about quadrilateral and enhanced our
  knowledge to a great extent. It also helped us to
  learn how to work in group and coordinate with
  each other as a team. We had a great fun while
  doing this project.

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THE END