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REMAINDER THEOREM STATEMENT : REMAINDER THEOREM STATEMENT Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number . If p(x) is divided by the linear polynomial x-a, then the remainder is p(a) . REMAINDER THEOREM PROOF : Suppose that when p(x) is divided by x-a the quotient is q(x) and the remainder is r(x) , i.e., p(x) = (x-a) q(x) + r(x) Since x-a is a 1 & the degree of r(x) is less than the degree of x-a, degree of r(x) =0 Therefore p(x) = (x-a) q(x) + r If x = a p(a) = (a-a) q(a) + r REMAINDER THEOREM PROOF PYTHAGORAS THEOREM STATEMENT : In a right angled triangle the square of the long side (the "hypotenuse") is equal to the sum of the squares of the other two sides.It is stated in this formula:a2 + b2 = c2 PYTHAGORAS THEOREM STATEMENT PYTHAGORAS THEOREM PROOF : PYTHAGORAS THEOREM PROOF a b c The square has a square hole with the side (a - b). Summing up its area (a - b)² and 2ab, the area of the four triangles (4·ab/2), we get FACTOR THEOREM : FACTOR THEOREM If p(x) is a polynomial of degree n > 1 & a is any real number then x-a is a factor of p(x) , if p(a) = 0 & p(a) = 0 , if x-a is a factor of p(x) By remainder theorem , p(x) = (x-a) p(x) + p(a) If p(a) =0 then p(x) = 0 (x-a) p(x) , which shows that x-a is a factor of p(x) . STATEMENT PROOF CHAPTER - 6 LINES & ANGLES : CHAPTER - 6 LINES & ANGLES Total Theorem - 8 THEOREM – 6.1 : THEOREM – 6.1 If two lines intersect each othjer , then the vertically opposite angles are equal . In the statements above , it is given that ‘two lines intersect each other’ . So, let AB and CD be two lines intersecting at O . They lead to two pairs of vertically Opposite angles , namely, PROOF STATEMENT THEOREM – 6.2STATEMENT : THEOREM – 6.2STATEMENT If a transversal intersect two parallel lines , then each pair of alternate interior angles are equal . THEOREM –6.2PROOF : THEOREM –6.2PROOF The transversal PS intersects lines AB & CD at points Q & R respectively such that BQR = QRC BQR = PQA BQR = QRC So we may conclude that PQA = QRC SO AB CD R P C S Q B D A THEOREM – 6.3 : THEOREM – 6.3 IF a transversal intersects two lines such that a pair of alternate interior angles is equal , then the two lines are parallel . THEOREM – 6.4 : THEOREM – 6.4 IF a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary . THEOREM –6.5 : THEOREM –6.5 IF a transversal intersects two lines, such that a pair of interior angles on the same side of the transversal is supplementary , then the two lines are parallel THEOREM – 6.6 : THEOREM – 6.6 Lines which are parallel to the same line are parallel to each other . THEOREM – 6.7 STATEMENT : THEOREM – 6.7 STATEMENT The sum of the angles of a triangle is 180 0 . P R Q R Q P Y X 2 1 3 4 3 2 1 5 THEOREM – 6.7 PROOF : THEOREM – 6.7 PROOF The triangle PQR and angle 1,2,3 are angles of triangle PQR . A line XPY parallel to QR through the opposite vertex P . So that we can use the properties related to the parallel lines . Therefore , |_ 4 + |_ 1 + |_5 = 180 0 (1) But XPY parallel to QR & PQ , PR are transversal. |_2 + |_1 + |_3 = 180 0 so, |_1 + |_2 + |_3 = 180 0 THEOREM –6.8 : THEOREM –6.8 If a side of a triangle is produced , then the exterior angle so formed is equal to the sum of the two interior opposite angles . An exterior angles of a triangle is greater than either of its interior opposite angles . CHAPTER – 7 TRIANGLES : CHAPTER – 7 TRIANGLES Total Theorem - 8 THEOREM – 7.1STATEMENT : THEOREM – 7.1STATEMENT Two triangles are congruent if two angles & the included side of one triangle are equal to two angles and the included side of other triangle . ASA CONGRUENCE RULE THEOREM – 7.1PROOF : THEOREM – 7.1PROOF Let AB = DE AB = DE |_B = |_E BC = EF SO, ABC = DEF A B E C F D THEOREM – 7.2STATEMENT : THEOREM – 7.2STATEMENT Angle opposite to equal sides of an isosceles triangle are equal . THEOREM – 7.2PROOF : THEOREM – 7.2PROOF The bisector of |_A & let D be the point of intersection of this bisector of |_A & BC . IN BAD & CAD , AB = AC |_BAD = |_CAD AD = AD So , BAD = CAD A B D C THEOREM – 7.3STATEMENT : THEOREM – 7.3STATEMENT The sides opposite to equal angles of a triangle are equal . THEOREM – 7.3PROOF : THEOREM – 7.3PROOF This theorem is Proved by ASA congruence rule . For example :- when In any |_BAD = |_CAD AD = AD |_ADB = |_ADC = 90 0 So, ABD = ACD ( ASA Rule ) So, AB = AC ( CPCT ) A D C B THEOREM – 7.4STATEMENT : THEOREM – 7.4STATEMENT SSS CONGRUENCE RULE If three sides of one triangle are equal to the three sides of another triangle , then the two triangles are congruent. THEOREM – 7.4PROOF : THEOREM – 7.4PROOF SSS CONGRUENCE RULE This theorem can be proved by suitable construction A Q B R C P 4.5 cm 3.5 cm 4 cm 4.5 cm 4 cm 3.5 cm THEOREM – 7.5 STATEMENT : THEOREM – 7.5 STATEMENT RHS CONGRUENCE RULE If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle , then the two triangle are congruent . THEOREM – 7.5 PROOF : THEOREM – 7.5 PROOF RHS CONGRUENCE RULE The two triangles cover each other completely & so they are congruent . Two right triangles are congruent if one pair of sides & the hypotenuse are equal . 4 cm 4 cm 5 cm 5 cm THEOREM – 7.6STATEMENT : THEOREM – 7.6STATEMENT If two sides of a triangle are unequal , the angle opposite to the longer side is larger . THEOREM – 7.6PROOF : THEOREM – 7.6PROOF A Q R S T U B P This theorem can be proved by this diagram. THEOREM – 7.7 : THEOREM – 7.7 In any triangle , the side opposite to the larger angle is longer . STATEMENT Take a triangle ABC & find AB+BC, BC + AC, AC+AB Observation : AB+BC >AC , BC+AC >AB , And AB+AC >BC . A B C PROOF THEOREM – 7.8 : THEOREM – 7.8 The sum of any two sides of a triangle is greater than the third side . PROOF D A B C This theorem is proved by this diagram STATEMENT CHAPTER – 8 QUADRILATERAL : CHAPTER – 8 QUADRILATERAL Total Theorem - 10 THEOREM – 8.1STATEMENT : THEOREM – 8.1STATEMENT A diagonal of a parallelogram divides it into two congruent triangles . THEOREM – 8.1PROOF : THEOREM – 8.1PROOF Let ABCD be a parallelogram and AC be a diagonal . We will finddiagonal AC divides parallelogram ABCD into two triangles ABC & CDA . In triangle ABC & CDA BC//AD and AC is transversal SO, |_BCA = |_DAC AB = DC & AC is a transversal . So, |_BCA = |_ DCA and AC = CA Therefore , Diagonal AC divides parallelogram ABCD into two congruent triangles ABC & CDA. D B A C THEOREM – 8.2 : THEOREM – 8.2 STATEMENT In a parallelogram , opposite sides are equal . PROOF In the previous theorem We have already learn that If a quadrilateral is a parallelogram then each pair of its opposite sides is equal . So opposite sides of parallelogram are equal . D B A C Therefore In this diagram AB = DC & AD = BC THEOREM – 8.3 : THEOREM – 8.3 If each pair of opposite sides of a quadrilateral is equal , then it is a parallelogram . STATEMENT PROOF Take quadrilateral ABCD & draw diagonal AC .We find that : ABC = CDA So, |_ DCA = |_ BAC |_BCA = |_ DAC and AB = CD , AD = BC D B A C THEOREM – 8.4 : THEOREM – 8.4 In a parallelogram , opposite angles are equal . STATEMENT PROOF This diagram is a proof of this theorem . D B A C THEOREM – 8.5STATEMENT : THEOREM – 8.5STATEMENT If in a quadrilateral , each pair of opposite angles is equal , then it is a parallelogram . THEOREM – 8.6STATEMENT : THEOREM – 8.6STATEMENT The diagonals of a parallelogram bisect each other . C A D B O THEOREM – 8.7 : THEOREM – 8.7 If the diagonal of a quadrilateral bisect each other then it is a parallelogram . PROOF OA=OC & OB=OD So, AOB = COD Therefore, |_ABO = |_CDO So , AB || CD , BC || AD Therefore ABCD is a parallelogram . C A D B O STATEMENT THEOREM – 8.8 : THEOREM – 8.8 A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel . PROOF In this fig. AB = CD and AB || CD And ABC = CDA So , BC || AD C A D B STATEMENT THEOREM – 8.9 : THEOREM – 8.9 The line segment joining the mid-points of two sides of a triangle is parallel to the third side . PROOF In this fig. E & F are mid points of AB & AC respectively and CD || BA And AEF = CDF So , EF || DF & BE = AE = DC This gives EF || BC A F E D C B STATEMENT THEOREM – 8.10 STATEMENT : THEOREM – 8.10 STATEMENT The line drawn through the mid point of one side of a triangle parallel to another side bisects the third side . PROOF A F E D C B This diagram is a proof of this theorem . CHAPTER – 9 AREAS OF PARALLELOGRAM & TRIANGLES : CHAPTER – 9 AREAS OF PARALLELOGRAM & TRIANGLES Total Theorem - 3 THEOREM – 9.1 STATEMENT : THEOREM – 9.1 STATEMENT Parallelogram on the same base and between the same parallels are equal in area . THEOREM – 9.1 PROOF : THEOREM – 9.1 PROOF Two parallelograms ABCD and EFCD , on the same base DC and between the same parallels AF and DC are given . |_DAE = |_CBF ( corresponding |_ from AD || BC & trans. AF |_AED = |_BFC ( corresponding |_ from AD || BC & trans.AF Therefore, |_ADE = |_BCF ( Angle sum property of a triangle ) Also, AD = BC ( Oppst. Sides of the parallelogram ABCD ) So ADE = BCF ( ASA rule ) So, Parallelograms ABCD and EFCD are equal in area . A F B D E A C THEOREM – 9.2 STATEMENT : THEOREM – 9.2 STATEMENT Two triangles on the same base ( or equal base ) and between the same parallels are equal in area . THEOREM – 9.2 PROOF : THEOREM – 9.2 PROOF A F B D E A C Suppose ABCD is a parallelogram whose one of the diagonal is AC . Let AN _|_ DC . ADC = CBA So ar(ADC) = ar (CBA) Therefore, ar(ADC) = ½ ar (ABCD) So, area of ADC = ½ base DC corresponding altitude AN THEOREM – 9.3 : THEOREM – 9.3 Two triangles having the same base ( or equal bases ) and equal areas lie between the same parallels . STATEMENT CHAPTER – 10 CIRCLES : CHAPTER – 10 CIRCLES Total Theorem - 12 THEOREM – 10.1 STATEMENT : THEOREM – 10.1 STATEMENT Equal chords of a circle subtend equal angles at the centre . THEOREM – 10.1 PROOF : THEOREM – 10.1 PROOF We have two equal chords AB and CD of a circle with centre O . In triangle AOB and COD OA = OC ( Radii of a circle ) OB = OD (Radii of a circle ) AB = CD Therefore , AOB = COD ( SSS rule ) This gives |_AOB = |_COD ( Corresponding parts of congruent triangles ) O D B C A THEOREM – 10.2 STATEMENT : THEOREM – 10.2 STATEMENT If the angles subtended by the chords of a circle at the centre are equal , then the chords are equal . PROOF |_AOB = |_COD ( By above theorem ) AOB = COD So, the chords AB & CD are equal . O B C A D THEOREM – 10.3 STATEMENT : THEOREM – 10.3 STATEMENT The perpendicular from the centre of a circle to a chord bisects the chord . O A M B THEOREM – 10.4 STATEMENT : THEOREM – 10.4 STATEMENT The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord . THEOREM – 10.4 PROOF : THEOREM – 10.4 PROOF O A M B Let AB be a chord of a circle with centre O and O is joined to the mid point M of AB . We have to prove that OM _|_ AB . Join OA and OB . In triangles OAM and OBM , OA = OB AM = BM OM = OM Therefore OAM = OBM This gives, |_OMA = |_OMB = 900 THEOREM – 10.5 : THEOREM – 10.5 O Q There is one and only one circle passing through three given non-collinear points . R P THEOREM – 10.6 : THEOREM – 10.6 O M Equal chords of a circle ( or of congruent circles ) are equidistant from the centre ( or centres ) . L O M L S Q P R THEOREM – 10.7 : THEOREM – 10.7 Chords equidistant from the centre of a circle are equal in length . PROOF If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection so, the chords are equal . O D L M E Q A B P C STATEMENT THEOREM – 10.8 STATEMENT : THEOREM – 10.8 STATEMENT The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle . THEOREM – 10.8 PROOF : THEOREM – 10.8 PROOF Arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle . In all cases, |_BOQ = |_QAO + |_AQO OA = OQ ( Radii of a circle ) Therefore, |_OAQ = |_OQA So, |_BOQ = 2 |_OAQ , |_BOP = 2 |_OAP By above |_BOP + |_BOQ = 2( |_OAP + |_OAQ ) |_POQ = 2 |_PAQ B P P Q A O O O B A B A Q Q P THEOREM – 10.9 : THEOREM – 10.9 Angles in the same segment of a circle are equal . A O Q C P We know that , |_PAQ = ½ |_POQ |_PAQ = ½ 180 0 = 90 0 Same as |_PCQ which is equal to 900 So |_PAQ = |_PCQ PROOF THEOREM – 10.10 STATEMENT : THEOREM – 10.10 STATEMENT If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment , the four points lie on a circle ( i.e. they are concyclic ) THEOREM – 10.10 PROOF : THEOREM – 10.10 PROOF C B E’ A AB is a line segment , which subtends equal angles at two point C & B . That is |_ACB = |_ADB If points A, C, E & B lie on circle . |_ACB = |_AEB |_ACB = |_ADB Therefore |_AEB = |_ADB So, this is not possible E coincides with D Similarly E’ should also coincides with D E D THEOREM – 10.11 : THEOREM – 10.11 The sum of either pair of opposite angles of a cyclic quadrilateral is 1800 A C B D |_A + |_C = 1800 |_B + |_D = 1800 THEOREM – 10.12 : THEOREM – 10.12 If the sum of a pair of opposite angles of a quadrilateral is 1800 , the quadrilateral is cyclic A C B D |_A + |_C = 1800 |_B + |_D = 1800 For that reason this quadrilateral is cylic Slide 69: PRESENTED BY :- HOPE YOU HAVE UNDERSTAND WELL NILESH DASHORE CLASS - X You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.