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**LESSON PLAN CLASS 10th SUBJECT MATHS TIME: 35min.**

Name of the topic : REAL NUMBERS Sub Topic : EUCLID’SDIVISION ALOGRITHIM

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Time Management. 1. P K Testing minutes. 2. Motivation minutes. 3. Presentation minutes. 4.Student Activity minutes. 5. Evaluation & Conclusion 8 minutes.

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PK TESTING Various questions about natural numbers , integers , whole numbers, and rational numbers should be asked from the students .If they are unable to answer the above said questions then teacher must explain these concepts to the students.

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MOTIVATION We shall motivate the students about division by putting an example which is practical to them s.t. if a basket contains 20 oranges and we are dividing these oranges to 4 students then how many oranges will each student get? If a basket contains 22 oranges and there are five students then how many oranges each student will get and how many oranges are left in the basket.

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PRESENTATION As we came to know from the students by putting two questions as above that we can distribute all the oranges to the students or there may be left some undistributed oranges. Dear students these undistributed oranges are called as REMAINDER and it is less than number of students. Where as number of students is called divisor ,

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Number of oranges is called dividend and number of oranges each student got is called quotient . From this discussion we are able to define Euclid's Division Lemma and Euclid's Division algorithm . We shall define them only for positive integers . Definition :Let a and b be any two positive integers s.t. a>b,then

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**There exists unique integers q and r s**

There exists unique integers q and r s.t a=bq+r, 0<r<b or r=0 As in example 20=5*4+0 or 22=5*4+2 Now we will use Euclid's Lemma to define Euclid's Algorithm . It is a technique to evaluate HCF of two positive integers . We shall use repeated steps

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**Euclid's division Lemma until we get remainder zero**

Euclid's division Lemma until we get remainder zero.In the step we get remainder zero,the divisor of last step is called HCF of two given positive integers .Now we will explain it by using an example as follows let two integers be 455 and =42* =35*1+7 35=7*5+0 therefore HCF of 455 and 42 is 7

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EVALUATION Now we shall formulate four groups for the assessment of the students in the concerned topic by putting different questions . Thereafter we shall compete the groups with each other on the same topic to enhance their self confidence and self assessment .

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**Group1: task for group 1 is to take any two numbers at random**

Group1: task for group 1 is to take any two numbers at random. Say 96,18 Group2:task for group 2 is to divide 96 by 18 repeatedly unless we get remainder zero.

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**Group3: Task for group 3 is to write the division steps in the form a = bq+r ,r=0 or 0<r<b**

Group 4: Task for group 4 is to compute HCF of two numbers .

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PREPARED BY SH.LALIT KUMAR LECTURER MATHS GSSS NAGWAIN DISTT. MANDI SH. YOG RAJ LECTURER MATHS GSSS BARYARA DISTT. MANDI SH. SATYA PRAKASH SHARMA LECTURER MATHS GSSS DEHAR DISTT. MANDI SH. BHIM SINGH THAKUR LECTURER MATHS GSSS DRANG DISTT. MANDI SH KAMAL KISHORE LECTURER MATHS GSSS MAHADEV DISTT. MANDI SH ROOP SINGH LECTURER MATHS GSSS RANDHARA DISTT. MANDI SH BHAGAT SINGH LECTURER MATHS GSSS TANGLING DISTT. KINNAUR SH PAWAN KUMAR LECTURER MATHS GSSS SIHUNTA DISTT. CHAMBA SH OM PRAKASH TGT GSSS CHANDI DISTT. SOLAN. SH. SURENDER CHAUHAN LECTURER MATHS GSSS PORTMORE DISTT. SHIMLA

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THANKS

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