Quadrilaterals
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Quadrilaterals

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Quadrilaterals Quadrilaterals Presentation Transcript

  • QUADRILATERAL :A quadrilateral is a geometrical figure whichhas four sides, four angles, four vertices, andtwo diagonals. The sum of all angles of aquadrilateral is 360 B A A B C D C D
  • There are actually six types of quadrilaterals.They are as follows: TRApEzIUm pARALLELogRAm REcTAngLE RhombUs sQUARE KITE
  • A BTRApEzIUm : D CIf in a quadrilateral one pair of oppositesides are equal then the given quadrilateralis called a TRApEzIUm.E.g. : In the above figure if AB isparallel to CD then the figure is aquadrilateral.
  • A B PARALLELOGRA M:If in a quadrilateral both thepairs of opposite sides areparallel , then the given C Dquadrilateral is aPARALLELOGRAM. E.G.- In the above figure if AB is parallel to CD and AD is parallel to BC then the figure is a parallelogram.
  • REcTAngLE:If in a quadrilateral one of its D Cangles is a right angle then thequadrilateral is aREcTAngLE. A BE.G. : If in the above figureangle A is a right angle then thefigure is a rectangle .
  • D CRhombUs:If in a parallelogram all sides areequal, then the parallelogram is aRhombUs. A BE.G. : If in the above figureAB=BC=CD=DA, then it isrhombus.
  • sQUARE: D CA parallelogram whose one angle isa right angle and all the sides areequal, then it is called asQUARE. A BE.G.: If in the above figureAB=BC=CD=DA, and angle B isa right angle, then the given figureis a square.
  • AKITE:In a quadrilateral if two pairs of B Dadjacent sides are equal. Then it isnot a parallelogram. It is called aKITE. CE.G. : If in the above figure AB =AD and BC = CD, then it is not aparallelogram. It is a kite.
  • A square is a rectangle and also arhombus. A parallelogram is a trapezium. A kite is not a parallelogram. A trapezium is not a parallelogram. A rectangle or a rhombus is not asquare .
  •  The sum of angles of a quadrilateral is 360 degrees. A diagonal of a parallelogram divides it into twocongruent triangles. In a parallelogram opposite sides are equal. If each pair of opposite sides of a quadrilateral isequal, then it is a parallelogram. In a parallelogram opposite angles are equal.If in a quadrilateral each pair of opposite angles isequal, then it is a parallelogram.
  • 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 ofopposite sides is equal and parallel.The line segment joining the mid-points of two sidesof a triangle is parallel to the third side. The line drawn through the mid-point of one side ofa triangle parallel to another side bisects the third side.
  • ThEoREm : Sum of angles of a quadrilateralD Cis 360Given: A quadrilateral ABCD.To prove: angles A + B+ C+ D= 360.Construction: Join A to C. A BProof: In triangle ABC, angle CAB + angle ACB + angle CBA = 180. (A.S.P) – 1 In triangle ACD, angle ADC + angle DCA + angle CAB = 180 (A.S.P) -2 Adding 1 and 2 angles CAB+ACB+CBA+ADC+DCA+CAD=180+180 angles (CAB+BAC)+ABC+(BCA+ACD)+ADC= 360. Therefore, angles A+B+C+D=360.
  • ThEoREm: The diagonal of a parallelogram divides it into two congruent triangles. D CGiven: A parallelogram ABCD and its diagonal AC.To prove: Triangle ABC is congruent to triangle ADCConstruction: Join A to C. A BProof: In triangles ABC and ADC, AB is parallel to CD and AC is the transversal Angle BAC = Angle DCA (alternate angles) Angle BCA = Angle DAC (alternate angles) AC = AC (common side) Therefore, triangle ABC is congruent to triangle ADC byASA rule.
  • ThEoREm: In a parallelogram , opposite sides areequal. D CGiven: A parallelogram ABCD.To Prove: AB = DC and AD = BCConstruction: Join A to CProof: In triangles ABC and ADC, A B AB is parallel to CD and AC is the transversal. Angle BAC = Angle DCA (alternate angles) Angle BCA = Angle DAC (alternate angles) AC = AC (common side) Therefore, triangle ABC is congruent to triangle ADC by ASA rule. Now AB = DC and AD = BC (C.P.C.T)
  • ThEoREm: If the opposite sides of a quadrilateral are equal, then it is a parallelogram.Given: A quadrilateral ABCD in which AB=CD & AD=BC D CTo Prove: ABCD is a parallelogram.Construction: Join A to C.Proof: In triangle ABC and triangle ADC , AB = CD (given) A B AD = BC (given) AC = AC (common side)Therefore triangle ABC is congruent to triangle ADC bySSS ruleSince the triangles of a quadrilateral are equal,therefore it is a parallelogram.
  • ThEoREm: In a parallelogram opposite angles are equal.Given: A parallelogram ABCD. D CTo prove: Angle A = Angle C & angle B=angle DProof: In the parallelogram ABCD, Since AB is parallel to CD & AD is transversal angles A+D=180 degrees (co-interior angles)-1A B In the parallelogram ABCD, Since BC is parallel to AD & AB is transversal angles A+B=180 degrees (co-interior angles)-2 From 1 and 2, angles A+D=angles A+B. angle D= angle B.Similarly we can prove angle A= angle C.
  • ThEoREm: If in a quadrilateral, each pair of oppositeangles is equal, then it is a parallelogram.Given: In a quadrilateral ABCD angle A=angle C & angle B=angle D. D CTo prove: It is a parallelogram.Proof: By angle sum property of a quadrilateral, angles A+B+C+D=360 degrees B angles A+B+A+B=360 degrees (since, angle AA=C and angle B=D) 2angle A+ 2angle B=360 degrees 2(A+B)=360 degrees angles A+B= 180 degrees. (co-interior angles.) Therefore, AD is parallel to BCSimilarly’ we can prove AB is parallel to CD.This shows that ABCD is a parallelogram.
  • ThEoREm: The diagonals of a parallelogram bisect each other.Given: A parallelogram ABCDTo prove: AO= OC & BO= OD. D CProof: AD is parallel to BC & BD is transversal. angles CBD= ADB (alternate angles) O AB is parallel to CD & AC is transversal. angles DAC= ACB (alternate angles) A B Now, in triangles BOC and AOD, CBD=ADB DAC=ACB BC=AD (opposite sides of a parallelogram)Therefore, triangle BOC is congruent to triangle AOD by ASA rule.Therefore, AO=OC & BO=OD [C.P.C.T]This implies that diagonals of a parallelogram bisect each other.
  • ThEoREm: If the diagonals of a quadrilateral bisect each other then it is a parallelogram.Given: In a quadrilateral ABCD, D C AO = OC & BO = OD OTo Prove: ABCD is a parallelogram.Proof: In triangles AOD & BOC AO = OC (given) A B BO = OD (given) angles AOD = BOC (vertically opposite angles)Therefore, triangle BOC is congruent to triangle AOD by SAS ruleTherefore angle ADB = CBD & angle 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 proves that ABCD is a parallelogram .
  • ThEoREm: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.Given: In a quadrilateral ABCD, D C AB is parallel to CD AB = CDTo prove: ABCD is a parallelogram.Construction: Join A to C.Proof: In triangles ABC & ADC, A B AB = CD ( given) angle BAC = angle DCA (alternate angles.) AC= AC ( common)Therefore, triangle ABC is congruent to triangle ADC by SAS rule.Therefore, angle ACB=DAC and AD=BC [C.P.C.T]Since, AD is parallel to BC and AD=BC,ABCD is a parallelogram.
  • ThEoREm: The line segment joining the mid-points of twosides of a triangle is parallel to the third side.Given: A triangle ABC in which D and E are Athe mid- points of AB and Ac respectively.To prove: DE is parallel to BC & DE=1/2BC E DProof: In triangles AED and CEF F AE = CE (given) B C ED = EF (construction) angle AED = angle CEF (verticallyopposite angles) Therefore, triangle AED is congruent totriangle CEF by SAS rule.Thus, AD=CF [ C.P.C.T] angle ADE = angle CFE [C.P.C.T]
  • Now, AD= CFAlso, AD = BDTherefore, CF = BDAgain angle ADE = angle CFE (alternate angles)This implies that AD is parallel to FCSince, BD is parallel to CF (since, AD is parallel to CF andBD=AD).And, BD=CFTherefore, BCFD is a parallelogram.Hence, DF is parallel to BC and DF=BC (opposite sides of aparallelogram).Since, DF=BC;DE=1/2 BCSince, DE=DF (given)Therefore, DE is parallel to DF.
  • ThEoREm: The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.Given: E is the mid- point of AB, line ‘l’ ispassing through E and is parallel to BC and CM isparallel to BA. A M 1To prove: AF=CF 3 FProof: Since, Cm is parallel to BA and EFD is E l 4 Dparallel to BC, therefore BEDC is a 2parallelogram. B CBE= CD( opposite sides of a parallelogram)But, BE = AE, therefore AE=CD.In triangles AEF & CDF: angle 1=2 (alt.angles) angle 3=4 (alt.angles) AE=CD (proved)Therefore,triangle AEF is congruent to CDF(ASA)AF=CF [C.P.C.T]. Hence, proved.