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It is a student’s helpline: Questions/Statements of any mathematical problem can be sent to:(a) Our Comments Sections. (b) [email protected] © [email protected] (d) [email protected] Articles/Reports that can help students in learning of general mathematics are cordially invited. Student’s helpline (Online) OKRAM LEARNER’ S web-zine. An autonomous service dedicated to students’ community. www.okramwebzine.org Absolutely Free If you are a student of 10th grade or below and need somebody to assist you in learning mathematics/solving any mathematical problem, please log on to www.okramwebzine.org Also Free tutorial classes for mathematics with general counseling have been being provided casually as well as annually to the selected local students of 10th grade at the premises of the students/our offices since 1990, the time even before the introduction of our educational helpline to you -Sd- ( RAMU THAPA ) Co-ordinator LEARNER’S CHOICE ( In this section, you can get the solutions of our mathematical questions ) Question by:- Alok Gupta Rasra (ballai), Uttar Pradesh. Q1. Four times the first of three consecutive odd integers is six more than the product of two and the third integer. Find the integers. Solution by:- Shyam Thapa, Kohima. Soln:- Let the three consecutive odd integers be x, x + 2 and x + 4. As per the question: 4Xx =[2X(x+4)]+6 4x = 2x + 8 + 6 4x = 2x + 14 4x – 2x = 14 X ‗ 14 2 7 x=7 The required three consecutive odd integers are:7, 9 and 11. Originality in Algebra (I) The original form of the present word “algebra” was “al-jebrw” almugabala: that was introduced by Mohammad ibu Musa Al Khowarizmi, an Arab astronomer in ninth century. Some of the original exponential forms:Year Mathematician Nationalism Exponential form. 1550 Rafael Bombelli Italian 1 (for 2x) 2 2 (for 2x2) 2 3 (for 2x3) 2 1600 Nicolas Chuquet France 20 (for 2), 21(for 2x) 22 (for 2x2) Ø (for x0), 2 (x2), 3 (for x3) 1610 Pietro Cataldi 1593 Adrianus Romanus Dutch 1 ( 15 ) (for x15) 1619 Jobst Burgi Swiss 7iii – 2v + 9iv (for 7x3 – 2x5 + 9x4) Question by:- RuopfÜvinuo Khezhie Class:- 10 Little Flowe Higher Secondary School, Kohima. Q:- Prove that. (I) Sin A – Cos A + 1 ‗ Cos A Sin A + Cos A – 1 1 – Sin A (II) ( 1 – 2Sin2A)2 ‗ 2Cos A – 1 Cos4A – Sin4A 2 Solution by:- H. Rameswar Khuman Christ Joti School Mantri Pukhri, Imphal East. Soln:(I) Sin A – Cos A + 1 ‗ Cos A Sin A + Cos A – 1 1 – Sin A LHS ‗ Sin A – Cos A + 1 Sin A + Cos A – 1 = Sin A – Cos A + 1 Cos A Sin A + Cos A - 1 Cos A Sin A – Cos A + 1 = Cos A Cos A Cos A Sin A + Cos A 1 Cos A Cos A Cos A ‗ Tan A – 1 + Sec A Tan A + 1 – Sec A ‗ (Sec A + Tan A) – 1 Tan A + 1 – Sec A Dividing num. & deno. By Cos A ‗ (Sec A + Tan A) – (Sec2A – Tan2A) Tan A + 1 – Sec A ‗ (Sec A + Tan A) – (Sec A – Tan A) (Sec A + Tan A) Tan A + 1 – Sec A ‗ (Sec A + Tan A) (1 – Sec A + Tan A) (1 – Sec A + Tan ) = Sec A + Tan A 1 + Sin A Cos A Cos A = = 1 + Sin A Cos A = (1 + Sin A) (1 – Sin A) Cos A (1 – Sin A) = 1 – Sin2A Cos A (1 – Sin A) = Cos2A Cos A (1 – Sin A) = Cos A = RHS 1 – Sin A Multiplying num & done by (1 – Sin A) (a + b) (a – b) = a2 – b2 Proved. ( 1 – 2Sin2A)2 ‗ 2Cos2A – 1 Cos4A – Sin4A (II) 2 2 LHS ‗ ( 1 – 2Sin A) (Cos4A – Sin4A) ‗ (1 – Sin2A – Sin2A)2 (Cos2A)2 – (Sin2A)2 = (Cos2A – Sin2A)2 (Cos A – Sin2A) (Cos2A + Sin2A) 2 2 2 2 A – Sin2A) = (Cos A – 2Sin A) (Cos 2 (Cos A – Sin A) X 1 = Cos2A – Sin2A = Cos2A – (1 – Cos2A) = Cos2A – 1 + Cos2A = 2cos2A – 1 = RHS Proved Sin2Ɵ + Cos2Ɵ = 1 a2 – b2 = (a – b) (a + b) Sin2Ɵ + Cos2Ɵ = 1 Continuation from the previous issue:Question by:- Q:- Thoubal District How to do a geometrical proof easily? Solution:- (a) M. Imtiaz Ahmad Thoubal Moijng, Okram Sunder Singh. St. Joseph’s School, Kohima. Theorem:- It is a statement of a geometrical truth that has to be proved deductively (by deductive method) Example:- The sum of the three angles of a triangle is 180o. The statement need a deductive proof:X A Y Now, let us consider a ABC. XY BC is drawn through A. Then, it is found that B C < XAB + <BAC + <YAC = 180o [ Forming a straight angle ] <B + <BAC + <C = 180o [ <XAB = <B, <YAC = <C; all < , XY BC ] s in ABC, <A + <B + <C = 180o So, the statement is a theorem, because it needs a proof and the fact is supported by a deductive proof. (b) Corollary/Criterion:It is a truth statement derived from a theorem (s) EX:“A triangle cannot have more than one right angle”. From the above theorem it can be extended that <A + <B = 180o - <C i.e, <A + <B 180o [ ( <A + <B ) is less than 180o ] [ <C ≠ 0 ] It shows that angles can neither be both right angles, an obtuse angle and a right angle nor both obtuse angles. So the statement is a corollary. Parts of a theorem:(a) General enunciation:- It is the general statement without having a reference to the parts (angles/points/segment) of a particular figure. (b) Particular enunciation:- It is a statement referring to a particular figure. It consists of the hypothesis / what is given and the conclusion / what is to be proved. EX:(i) The sum of all the angles of a triangle is 180o. This statement is an example of the general enunciation. (ii) In ABC, <A + <B + <C = 180o is an example of particular enunciation. Steps in performing a geometrical proof:I. II. III. Figure:- Transformation of the statement by a diagram using letters. Given:- It should be a particular enunciation referring to the figure drawn above already. To prove:- [ Originally a phrase was used as “It is required to prove that ….” ] What is required to be proved for the particular figure has to mention. IV. V. Construction:- If needed, the figure can be transformed into an equivalent figure or can be extra parts of the figure. Proper statements should be mentioned for every part of construction. Proof:- It is an analytical part leading to the required conclusion. Proof:It is an analytical part done by using suitable axiom (s) Theorem (s)/any other universal facts (s) of deductive reasoning. This section consists of two parts:(a) Argument:It is a stepwise arrangement of logical facts that connects between the given and the conclusion. (b) Reasons:Each reason is a short statement expressed by an acceptable abbreviation (s) for every application of an assumption to be written just to the right side of the particular step in the Argument. This section is the end part of a geometrical proof. Originally, an abbreviation, Q.E.D (Quod erat demonstrandum) was written at the end showing that the proof had been completed. But, in modern geometry it is replaced by hence proved/proved. A Format for Geometrical Proof. Figure Given:- ________ ________ ________ ________ To prove:- ________ ________ ________ ________ Construction:- ________ ________ ________ ________ Proof:- Argument Reasons Hence proved To be contd.. This section contains messages from resource persons Articles/Reports that can help students in learning of general mathematics are cordially invited. Continuation from the previous issue:- Vedic mathematics Part – II Okram Sunder Singh Now, let us have a generalized algorithm by observing the examples of the multiplication of whole number with base 100 from previous issue. Algorithm ( For base 100 ) (I) Base:- 100 (II) Subtraction of the base (100) from the whole numbers W1 and W2. W1 – 100 = S1 W2 – 100 = S2 (III) Addition of ( W1 and S2 ) / ( W2 and S1 ) ( W1 + S2 ) = ( W2 + S1 ) = A (IV) Multiplication of A and 100. A X 100 = M (V) To find the product of S1 and S2. S1 X S2 = P (VI) Sum of M and P. ( M + P ) = Required product. Multiplication of whole numbers with base 1000/10000/……… It has similar algorithm to that of base 100. Multiplication of A by 100 in the algorithm of base -100 should be replaced by 1000/10000/……. Now let us see an example:Calculation of 957 X 1024 975 , 1024 (+) 24 X -43 = -1032 981000 - 1032 979968 So, 957 X 1024 = 979968 Base = 1000 (a) Subtraction of base 957 – 1000 = - 43 1024 – 1000 = 24 (b) cross addition 957 + 24 = 981 © Multiplication by 1000 ( as base is 1000 ) 981 X 1000 = 981000 (d) Addition of ( -1032 ) 981000 + ( -1032 ) = 979968 Multiplication of whole numbers with base 50:Algorithm for base 50 has some modifications from that of base 100 as given below. Algorithm ( for base 50 ):(I) Original base:- 50 (II) Subtraction of the bases from the whole numbers W1 and W2. W1 – 50 = S1 W2 – 50 = S2 (III) Addition of ( W1 and S2 ) / ( W2 and S1 ):( W1 + S2 ) = ( W2 + S1 ) = A (IV) To adopt a reference base ( RB ):RB is a multiple of 10, i.e. 10/100/100….. According to our convenience, we can choose any of these as RB. (V) To find adjusting factor ( AF ):AF = Original base ( OB ) Reference base ( RB ) For example:- (VI) RB AF ‗ OB RB AF 10 50 10 5 100 50 100 1 2 Multiplication of A and AF:A X AF = M1 (VII) Multiplication of M1 and 10/100/100….. Multiplication is done according to RB. RB 10 100 M1 X ( 10/100/1000 ) = M2 Multiplier 10 100 (VIII) To find the product of S1 and S2:S1 X S2 = P (IX) Sum of M2 and P (M2 + P ) = Required Product. Illustrative explanation Calculation of 42 X 47 (i) Choosing 10 as RB:- (a) Subtract the base 50. 42 – 50 = -8 47 – 50 = -3 (b) Add the difference ( -8 ) to 47 or ( -3 ) to 42. (cross addition) 42 , 47 (+) - 3 , - 8 39 © Find AF as, AF ‗ OB ‗ 50 ‗ 5 RB 10 (d) Multiply 39 from step (b) by 5 (AF) 42 , 47 -3 , -8 39 X 5 (AF) 195 (e) Multiply 195 (from above) by 10 (as RB is 10) 42 , 47 -3 , -8 39 X 5 195 X 10 1950 (f) Find the product of ( -3 ) and ( -8 ). 42 , 47 - 3 X - 8 = 24 39 X 5 195 X 10 1950 (g) Add the product 24 [ from (d) ] to 1950. 42 , 47 - 3 X - 8 = 24 39 X 5 195 X 10 1950 (+) 24 1974 So, 42 X 47 = 1974. (ii) Choosing 100 as RB. a) Same as in (i) b) Same as in (i) c) Find AF ‗ 50 ‗ 1 100 2 (d) Multiply 39 from (b) by ½ (AF) 42 , 47 -3 , -8 39 (X) ½ (AF) 19.5 (e) Multiple 19.5 by 100 (as RB is 100) 42 , 47 -3 , -8 39 (X) ½ 19.5 (X) 100 1950 (f) Find the product of ( -3 ) and ( -8 ) 42 , 47 - 3 X - 8 = 24 39 (X) ½ 19.5 (X) 100 1950 (g) Add 24 from (f) to 1950 42 , 47 - 3 X - 8 = 24 39 (X) ½ 19.5 (X) 100 1950 (+) 24 1974 So, 42 X 47 = 1974 Multiplication of whole numbers greater than 50. Calculation of 53 X 57, using RB = 10 53 , 57 (+) 7 X 3 = 21 60 X 5 300 X 10 3000 + 21 3021 Subtraction of base (50) 53 – 50 = 3 57 – 50 = 7 RB = 10 , AF = 50 = 5 10 Multiplication by RB (10) So, 53 X 57 = 3021 Multiplication of whole numbers, one is less than 50 and the other is greater than 50.(RB = 10) Calculation of 44 X 56 Subtraction of base (50) 44 , 56 6 X - 6 = -36 50 X 5 (AF) 250 X 10 2500 (+) - 36 2464 44 – 50 = -6 56 – 50 = 6 So, 44 X 56 = 2464 AF = 50 = 5 10 multiplication by 10 (as RB is 10) To be contd.. Learning from our mistakes Parts of the contents in this section have errors either in questions or answers or in both. Your work is to find out the errors and then to do corrections. Polygon A closed figure formed by joining line segments is known as a polygon. A polygon has at least three sides. Depending upon the number of sides/amount of the angle measures, specific names are given as:- No. of sides 3 4 5 6 Name Triangle Quadrilateral Pentagon Hexagon No. of sides 7 8 9 10 Name Heptagon Octagon Nonagon Decagon A segment joining between two non consecutive vertices is known as a diagonal. In fig (a), in quadrilateral ABCD AC and BD are diagonals. D C A B Fig (a) Every polygon has at least a diagonal. Every two diagonals of a polygon intersect each other. Number of diagonals of a polygon cannot be greater than the number of sides. In a heptagon, every vertices has four diagonals, so, the total number of diagonals of a heptagon is 4 X 7 or 28 The polygon that has 35 diagonals is a decagon. A polygon is a convex if all the diagonals are inside it D C but it is a concave if any of its diagonals is outside it In the fig (b), ABCDA, BOCB are convex polygons O but ABCDOA, AODCOBA are concave polygons. A B Fig (b) A polygon is a regular polygon if all sides are equal in length. Each angle inside a polygon is known as interior angle but any angle outside polygon formed by any two sides of the polygon is known as an exterior angle. A 1 In fig ©, <1, <2, <3 are interior angles But a and b are exterior angles. a 2 3 B b C Fig © Exterior angle theorem:In a triangle an exterior angle is equal to the sum of interior opposite angles. So, B In fig (d), ABC <a = <b + <c. b a c D C B Fig (d) Similarly, in fig (e), In quadrilateral PQRS <5 = <1 + <2 + <3. P Q 1 5 4 S 2 3 Fig (e) R Angle sum property:The sum of all the interior angles of a polygon of n sides is (n – 2) 180o. In case of a triangle, n = 3, so (3 – 2) X 180o or 180o is the sum of all the three angle of a triangle. The sum of interior angles of a polygon has a minimum value of 180o. 100o cannot be the sum of all the interior angles of a polygon as it is less than 180 o but 1275o can be the sum of all the interior angles of a polygon. If each interior angle of a regular polygon is 144o, then the number of sides is 8. Each exterior angle of a regular polygon of sides is given by 360o N Please do correction as it may misguide. Research and Developments. This section consists of logical and creative works done by young learners. Submitted by:Name:- Reshikala Jamir Class:- 7 ‘A’ Roll. No:- 24 (Year – 2014) G. Rio Higher Secondary School, Kohima. Dear Sir, I have presented one of the activities from mathematics that I have done as my school work. Aim:- To determine the area of a circle. Requirements:- Woolen thread, glue, glass sheet, pencil, scale, protector, scissors etc. Procedures:1st step:Making a circular plate of coil with a piece of woolen thread that is moistened with glue by circulating it closely on a glass plate. The woolen circular plate is allowed to dry completely. 2nd step:- p P1 P2 O o A P 1 A P 2 The woolen circular plate is cut along PO from outer to the center (o) of the circular plate. Then the cut circular plate is slowly and carefully rollled out (without separating the threads) so that the pieces of the threads become straight. Observation:The whole circular woolen plate is transformed in a triangle, i,e Here, OA P1P2 OP 1P2 . Height of the triangle = OA = r (radius of the woolen circular plate) Base = P1P2 = 2π r (circumference) Calculation:Area of the circular plate = Area of P1OP2 = ½ X P1P2 X OA [ Area of a triangle = ½ X base X height ] =½X2πrxr = π r2 Observation:Area of a circular plate = π r2 Thank you Reshikala Jamir for your sharing. You have done it with clear and proper statements. We are very proud of you and your teachers efforts. We hope that you will present some more in future I order to share mathematical intelligence. COPYRIGHT @ 2013/2014 OKRAM LEARNERS WEBZINE