Nerul, Mumbai, India - 400706.
2
Details verified of Samar S.✕
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English
Hindi
Mumbai University
1989
Bachelor of Science (B.Sc.)
Mumbai University
1997
Bachelor of Education (B.Ed.)
Nerul, Mumbai, India - 400706
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
20
Board
ICSE, State, CBSE
CBSE Subjects taught
Mathematics
ICSE Subjects taught
Physics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
20
Board
ICSE, State, CBSE
CBSE Subjects taught
Mathematics
ICSE Subjects taught
Physics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
20
Board
State, CBSE
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
20
Experience in School or College
.......................................
1 Student
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in CET Coaching classes
20
Subjects
Mathematics
Class type catered to
Integrated coaching classes, Regular Classes
1. Which school boards of Class 10 do you teach for?
ICSE, State and CBSE
2. Do you have any prior teaching experience?
No
3. Which classes do you teach?
I teach CET Coaching, Class 10 Tuition, Class 11 Tuition, Class 12 Tuition and Class 9 Tuition Classes.
4. Do you provide a demo class?
Yes, I provide a free demo class.
5. How many years of experience do you have?
I have been teaching for 20 years.
Answered on 28/02/2018 Learn CBSE/Class 10/Mathematics
Answered on 28/02/2018 Learn CBSE/Class 10/Mathematics
Answered on 27/02/2018 Learn CBSE/Class 9/Mathematics
Let?ABC be required triangle
AB=AC=5cm
BC=8 cm
Let AD be the altitude
We know that perpendicular drawn from vertex bisect the base.
Therefore BD=4 cm
Using Pythagoras theorem we get AD= 3 cm
Area of ?ABC=1/2 × base×height
=1/2×8×3=12 cm2
Answered on 23/02/2018 Learn CBSE/Class 9/Mathematics
Answered on 23/02/2018 Learn CBSE/Class 9/Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
20
Board
ICSE, State, CBSE
CBSE Subjects taught
Mathematics
ICSE Subjects taught
Physics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
20
Board
ICSE, State, CBSE
CBSE Subjects taught
Mathematics
ICSE Subjects taught
Physics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
20
Board
State, CBSE
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
20
Experience in School or College
.......................................
1 Student
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in CET Coaching classes
20
Subjects
Mathematics
Class type catered to
Integrated coaching classes, Regular Classes
Answered on 28/02/2018 Learn CBSE/Class 10/Mathematics
Answered on 28/02/2018 Learn CBSE/Class 10/Mathematics
Answered on 27/02/2018 Learn CBSE/Class 9/Mathematics
Let?ABC be required triangle
AB=AC=5cm
BC=8 cm
Let AD be the altitude
We know that perpendicular drawn from vertex bisect the base.
Therefore BD=4 cm
Using Pythagoras theorem we get AD= 3 cm
Area of ?ABC=1/2 × base×height
=1/2×8×3=12 cm2
Answered on 23/02/2018 Learn CBSE/Class 9/Mathematics
Answered on 23/02/2018 Learn CBSE/Class 9/Mathematics
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Sakinaka East,
Mumbai
