UrbanPro
Login Ask & Answer Success Stories Paid Courses Free Classes Tuition & Classes Fees Write a Review All Categories Help Center
Post your Learning Need Signup as a Tutor
✕

Popular Cities

Bangalore

Hyderabad

Delhi

Mumbai

Chennai

Pune

Kolkata

Gurgaon

Ahmedabad

Jaipur

Noida

✕
true

Class 11 Tuition > Class 11 Tuition in Bhopal > Hemant Kashyap

Hemant Kashyap Class 11 Tuition trainer in Bhopal/>

Hemant Kashyap

Maths tutor with 4 years experience

Maharana Pratap Nagar, Bhopal, India - 462001.

Contact
Referral Discount: Get ₹ 250 off when you make a payment to start classes. Get started by Booking a Demo.

Details verified of Hemant Kashyap✕

Identity

Education

Know how UrbanPro verifies Tutor details

Identity is verified based on matching the details uploaded by the Tutor with government databases.

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

Verified Info

Phone Verified

Email Verified

Report this Profile

✕
x

Is this listing inaccurate or duplicate? Any other problem?

Please tell us about the problem and we will fix it.

Please describe the problem that you see in this page.

Type the letters as shown below *

Please enter the letters as show below

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, ISC/ICSE, International Baccalaureate, IGCSE, CBSE

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.

Subjects taught

Mathematics

Taught in School or College

Yes

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

ICSE, IGCSE, State, CBSE, International Baccalaureate

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.

Subjects taught

Mathematics

Taught in School or College

Yes

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

ICSE, IGCSE, State, CBSE, International Baccalaureate

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.

Subjects taught

Mathematics

Taught in School or College

Yes

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

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

Subjects taught

Mathematics

Taught in School or College

Yes

Reviews

No Reviews yet!

FAQs

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

State, ISC/ICSE, International Baccalaureate 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...

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/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
Dislike Bookmark

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

<p>Construct a triangle ABC in which BC = 7cm, &ang;B = 75&deg; 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
Dislike Bookmark

Contact

View All

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, ISC/ICSE, International Baccalaureate, IGCSE, CBSE

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.

Subjects taught

Mathematics

Taught in School or College

Yes

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

ICSE, IGCSE, State, CBSE, International Baccalaureate

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.

Subjects taught

Mathematics

Taught in School or College

Yes

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

ICSE, IGCSE, State, CBSE, International Baccalaureate

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.

Subjects taught

Mathematics

Taught in School or College

Yes

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

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

Subjects taught

Mathematics

Taught in School or College

Yes

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...

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/Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle XYZ in which &ang;Y = 30&deg;, &ang;Z = 90&deg; 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
Dislike Bookmark

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

Construct a triangle PQR in which QR = 6cm, &ang;Q = 60&deg; and PR &ndash; 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/Geometry/Construction/NCERT Solutions/Exercise 11.2

Construct a triangle ABC in which BC = 8cm, &ang;B = 45&deg; and AB &ndash; 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/Geometry/Construction/NCERT Solutions/Exercise 11.2

<p>Construct a triangle ABC in which BC = 7cm, &ang;B = 75&deg; 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
Dislike Bookmark

Contact

Load More

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.

X

Share this Profile

Recommended Profiles

Abhishek Agrawal photo

Abhishek Agrawal

Abhishek Agrawal photo Maharana Pratap Nagar, Bhopal

Yashpal Mahdele photo

Yashpal Mahdele

Yashpal Mahdele photo Maharana Pratap Nagar, Bhopal

Diksha C.

Diksha C. photo Maharana Pratap Nagar, Bhopal

Ravi Bhatnagar photo

Ravi Bhatnagar

Ravi Bhatnagar photo Maharana Pratap Nagar, Bhopal

Sakshi P.

Sakshi P. photo Maharana Pratap Nagar, Bhopal

Shubham Shrivastava

Shubham Shrivastava photo Maharana Pratap Nagar, Bhopal

Also have a look at

Learn More

NCERT Solutions for Class 9 CBSE Class 10 Syllabus NCERT Class 10 Solutions NCERT Solutions For Class 11 NCERT Solutions for Class 8 CBSE Syllabus for Class 8 NCERT Solutions for Class 9 Maths NCERT Solutions for Class 9 Science Chapter 1 - Matter In Our Surroundings Chapter 2 - Is Matter Around Us Pure?

Reply to 's review

Enter your reply*

1500/1500

Please enter your reply

Your reply should contain a minimum of 10 characters

Your reply has been successfully submitted.

✕

Certified

The Certified badge indicates that the Tutor has received good amount of positive feedback from Students.

✕

Different batches available for this Course

Book a Free Demo

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more

ThinkVidya Learning Pvt Ltd © 2010-2025 All Rights Reserved