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Hemant Kashyap Class 11 Tuition trainer in Bhopal/>

Hemant Kashyap

Maths tutor with 4 years experience

Maharana Pratap Nagar, Bhopal, India - 462001.

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Referral Discount: Get ₹ 500 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

I am an engineer by qualification and currently freelance online tutor. I have been taking online classes for last 4 years. And mentoring students up to grade 12. I teach mathematics and also help in assignment. My key skills are mathematics (all level), Assignment help (school assignments), reference study materials etc.

Languages Spoken

Hindi Mother Tongue (Native)

English Proficient

Education

RGPV 2018

Bachelor of Engineering (B.E.)

Address

Maharana Pratap Nagar, Bhopal, India - 462001

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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

4

Board

State, IGCSE, ISC/ICSE, CBSE, International Baccalaureate

IB Subjects taught

Mathematics

ISC/ICSE Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

IGCSE Subjects taught

Mathematics

Experience in School or College

Have mentored students of different level, along with tutoring I straightened their weakness by providing them guidance and reference study materials.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

IGCSE, CBSE, International Baccalaureate, State, ICSE

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

IGCSE Subjects taught

Mathematics

Experience in School or College

Can help students up to 12th in their school assignment, can provide reference study material if required, can guide them to their best.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

4

Board

IGCSE, CBSE, International Baccalaureate, State, ICSE

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

IGCSE Subjects taught

Mathematics

Experience in School or College

Have experience in teaching to students of all grades. Can help them in their school assignment and can guide them.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 8 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

4

Board

State, International Baccalaureate, IGCSE, CBSE, ICSE

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

IGCSE Subjects taught

Mathematics

Experience in School or College

Can help students up to 12th in their school assignment, can provide reference study material if required, can guide them to their best

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 12 do you teach for?

State, IGCSE, ISC/ICSE and others

2. Have you ever taught in any School or College?

Yes

3. Which classes do you teach?

I teach Class 10 Tuition, Class 11 Tuition, Class 8 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 4 years.

Contact

Answers by Hemant Kashyap (16)

Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.6

The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres... ...more

The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000  = 1l)

Given that 2r = 132 cm, therefore r = 21 cmWater holding capacity = Volume of cylindrical vessel = r2hNow V = 34.65 litres Thus, the vessel can hold 34.65 litres of water. ...more

Given that 2r = 132 cm, therefore r = 21 cm
Water holding capacity = Volume of cylindrical vessel = r2h
Now V = 34.65 litres

Thus, the vessel can hold 34.65 litres of water.
 
Answers 5 Comments
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Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Unit 4-Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

The steps of construction for the required triangles are as follows: Step I: Draw a line segment AB of 11 cm.(As XY + YZ + ZX = 11 cm) Step II: Construct an angle PAB of 30o at point A and an angle QBA of 90o at point B. Step III: Bisect PAB and QBA. Let these bisectors intersect each other at point... ...more
The steps of construction for the required triangles are as follows:
 
Step I: Draw a line segment AB of 11 cm.(As XY + YZ + ZX = 11 cm)
Step II: Construct an angle PAB of 30o at point A and an angle QBA of 90o at point B.
Step III: Bisect PAB and QBA. Let these bisectors intersect each other at point X.
Step IV: Draw perpendicular bisector ST of AX and UV of BX. Step V: Let ST intersects AB at Y and UV intersects AB at Z.Join XY, XZ.XYZ is the required triangle.
 
Answers 3 Comments
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Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Unit 4-Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ =2cm.

The steps of construction for the required triangles are as follows:Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say XQR.Step II: Cut a line segment QS of 2 cm from the line segment QT extended an opposite side of line segment XQ. (As PR> PQ and PR - PQ = 2cm). Join SR.Step... ...more
The steps of construction for the required triangles are as follows:
Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say XQR.
Step II: Cut a line segment QS of 2 cm from the line segment QT extended an opposite side of line segment XQ. (As PR> PQ and PR - PQ = 2cm). Join SR.
Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR. PQR is the required triangle.
 
 
Answers 3 Comments
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Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Unit 4-Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.

The steps of construction for the required triangles are as follows:Step I: Draw the line segment BC = 8 cm and at point B make an angle of 45o say XBC. Step II: Cut the line segment BD = 3.5 cm (equal to AB - AC) on ray BX. Step III: Join DC and draw the perpendicular bisector PQ of DC. Step IV:... ...more
The steps of construction for the required triangles are as follows:
Step I: Draw the line segment BC = 8 cm and at point B make an angle of 45o say XBC.     
Step II: Cut the line segment BD = 3.5 cm (equal to AB - AC) on ray BX.

Step III: Join DC and draw the perpendicular bisector PQ of DC.

Step IV: Let it intersect BX at point A. Join AC. ABC is the required triangle.
 
 
Answers 3 Comments
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Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Unit 4-Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

The steps of construction for the required triangles are as follows: Step I: Draw a line segment BC of 7 cm. At point B draw an angle of 75o sayXBC. Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX. Step III: Join DC and make an angle DCY equal to BDC Step IV: Let... ...more
The steps of construction for the required triangles are as follows:
Step I: Draw a line segment BC of 7 cm. At point B draw an angle of 75o sayXBC.  
Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX.  
Step III: Join DC and make an angle DCY equal to BDC  
Step IV: Let CY intersects BX at A. ABC is the required triangle.  
Answers 3 Comments
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Contact

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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

4

Board

State, IGCSE, ISC/ICSE, CBSE, International Baccalaureate

IB Subjects taught

Mathematics

ISC/ICSE Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

IGCSE Subjects taught

Mathematics

Experience in School or College

Have mentored students of different level, along with tutoring I straightened their weakness by providing them guidance and reference study materials.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

IGCSE, CBSE, International Baccalaureate, State, ICSE

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

IGCSE Subjects taught

Mathematics

Experience in School or College

Can help students up to 12th in their school assignment, can provide reference study material if required, can guide them to their best.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

4

Board

IGCSE, CBSE, International Baccalaureate, State, ICSE

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

IGCSE Subjects taught

Mathematics

Experience in School or College

Have experience in teaching to students of all grades. Can help them in their school assignment and can guide them.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 8 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

4

Board

State, International Baccalaureate, IGCSE, CBSE, ICSE

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

IGCSE Subjects taught

Mathematics

Experience in School or College

Can help students up to 12th in their school assignment, can provide reference study material if required, can guide them to their best

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

No Reviews yet!

Answers by Hemant Kashyap (16)

Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.6

The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres... ...more

The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000  = 1l)

Given that 2r = 132 cm, therefore r = 21 cmWater holding capacity = Volume of cylindrical vessel = r2hNow V = 34.65 litres Thus, the vessel can hold 34.65 litres of water. ...more

Given that 2r = 132 cm, therefore r = 21 cm
Water holding capacity = Volume of cylindrical vessel = r2h
Now V = 34.65 litres

Thus, the vessel can hold 34.65 litres of water.
 
Answers 5 Comments
Dislike Bookmark

Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Unit 4-Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

The steps of construction for the required triangles are as follows: Step I: Draw a line segment AB of 11 cm.(As XY + YZ + ZX = 11 cm) Step II: Construct an angle PAB of 30o at point A and an angle QBA of 90o at point B. Step III: Bisect PAB and QBA. Let these bisectors intersect each other at point... ...more
The steps of construction for the required triangles are as follows:
 
Step I: Draw a line segment AB of 11 cm.(As XY + YZ + ZX = 11 cm)
Step II: Construct an angle PAB of 30o at point A and an angle QBA of 90o at point B.
Step III: Bisect PAB and QBA. Let these bisectors intersect each other at point X.
Step IV: Draw perpendicular bisector ST of AX and UV of BX. Step V: Let ST intersects AB at Y and UV intersects AB at Z.Join XY, XZ.XYZ is the required triangle.
 
Answers 3 Comments
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Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Unit 4-Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ =2cm.

The steps of construction for the required triangles are as follows:Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say XQR.Step II: Cut a line segment QS of 2 cm from the line segment QT extended an opposite side of line segment XQ. (As PR> PQ and PR - PQ = 2cm). Join SR.Step... ...more
The steps of construction for the required triangles are as follows:
Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say XQR.
Step II: Cut a line segment QS of 2 cm from the line segment QT extended an opposite side of line segment XQ. (As PR> PQ and PR - PQ = 2cm). Join SR.
Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR. PQR is the required triangle.
 
 
Answers 3 Comments
Dislike Bookmark

Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Unit 4-Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.

The steps of construction for the required triangles are as follows:Step I: Draw the line segment BC = 8 cm and at point B make an angle of 45o say XBC. Step II: Cut the line segment BD = 3.5 cm (equal to AB - AC) on ray BX. Step III: Join DC and draw the perpendicular bisector PQ of DC. Step IV:... ...more
The steps of construction for the required triangles are as follows:
Step I: Draw the line segment BC = 8 cm and at point B make an angle of 45o say XBC.     
Step II: Cut the line segment BD = 3.5 cm (equal to AB - AC) on ray BX.

Step III: Join DC and draw the perpendicular bisector PQ of DC.

Step IV: Let it intersect BX at point A. Join AC. ABC is the required triangle.
 
 
Answers 3 Comments
Dislike Bookmark

Answered on 13/09/2019 Learn CBSE/Class 9/Mathematics/Unit 4-Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

The steps of construction for the required triangles are as follows: Step I: Draw a line segment BC of 7 cm. At point B draw an angle of 75o sayXBC. Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX. Step III: Join DC and make an angle DCY equal to BDC Step IV: Let... ...more
The steps of construction for the required triangles are as follows:
Step I: Draw a line segment BC of 7 cm. At point B draw an angle of 75o sayXBC.  
Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX.  
Step III: Join DC and make an angle DCY equal to BDC  
Step IV: Let CY intersects BX at A. ABC is the required triangle.  
Answers 3 Comments
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Hemant Kashyap describes himself as Maths tutor with 4 years experience. He conducts classes in Class 10 Tuition, Class 11 Tuition and Class 8 Tuition. Hemant is located in Maharana Pratap Nagar, Bhopal. Hemant takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 4 years of teaching experience . Hemant has completed Bachelor of Engineering (B.E.) from RGPV in 2018. He is well versed in Hindi and English.

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