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Practical Geometry

Practical Geometry relates to CBSE/Class 6/Maths

Top Tutors who teach Practical Geometry

1
Deepashree Class 6 Tuition trainer in Bangalore Featured
Basaveshwara Nagar, Bangalore
Super Tutor
12 yrs of Exp
500per hour
Classes: Class 6 Tuition, Class 9 Tuition and more.

I have 3 years of experience teaching for 6th standard students in a reputed school am very much fond of solving problems and explaining experiments...

2
Richa B. Class 6 Tuition trainer in Bangalore Featured
Sarjapura, Bangalore
Super Tutor
6 yrs of Exp
700per hour
Classes: Class 6 Tuition, Yoga and more.

I am an experienced qualified teacher with over 7 years of teaching. I have been Certified by Urban Pro too. Students show good improvement from CBSE,...

3
Purnima Joshi Class 6 Tuition trainer in Nainital Featured
Haldwani, Nainital
Super Tutor
20 yrs of Exp
500per hour
Classes: Class 6 Tuition, Class 7 Tuition

I have more than 30 years experience of teaching in a convent school and almost two years experience of online teaching.Teaching is my passion.I am...

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4
Ottapalam, Ottapalam
Super Tutor
15 yrs of Exp
380per hour
Classes: Class 6 Tuition, Class 9 Tuition and more.

I am a resident of Kerala working in a CBSE school.I got the opportunity of working with Kendreya Vidyalaya (Ottapalam) , Cambridge Public School...

5
Aditya Pratap Singh . Class 6 Tuition trainer in Lucknow Featured
Lalkuan, Lucknow
Super Tutor
3 yrs of Exp
300per hour
Classes: Class 6 Tuition, Class 8 Tuition and more.

I have a three years of experience teaching Class 6 to 10 students, I specialize in Mathematics, Science and Social Studies. My approach includes...

6
Sanjukta I. Class 6 Tuition trainer in Kolkata Featured
VIP RD, Kolkata
Super Tutor
7 yrs of Exp
300per hour
Classes: Class 6 Tuition, Class 8 Tuition and more.

It was fabulous

7
Amsumathi K Class 6 Tuition trainer in Taliparamba Featured
Kunhimangalam, Taliparamba
Super Tutor
9 yrs of Exp
300per hour
Classes: Class 6 Tuition, Malayalam Speaking and more.

I'm a primary teacher.I am giving online tuition since 2020. I have a degree in M.A.malayalam with TTC .I have 9 years of teaching experience in CBSE...

8
Akshay Aggarwal Class 6 Tuition trainer in Delhi Featured
Badarpur, Delhi
Super Tutor
8 yrs of Exp
1000per hour
Classes: Class 6 Tuition, Class 7 Tuition and more.

I am an experienced tutor with nine years of teaching under my belt. I hold a BSc degree in geography. My passion for teaching has always been my...

9
Deepak Joshi Class 6 Tuition trainer in Gurgaon Featured
Sec-66, Gurgaon
Super Tutor
10 yrs of Exp
800per hour
Classes: Class 6 Tuition, BBA Tuition and more.

Ph.D candidate with over 10 years of teaching exp. (7 years at BYJU's & 3 years at a IB/IGCSE school). Specialized in IB, IGCSE, CBSE & ICSE cu...

10
Thanuja G. Class 6 Tuition trainer in Bangalore Featured
Rustam Bagh Layout, Bangalore
Super Tutor
1 yrs of Exp
350per hour
Classes: Class 6 Tuition, German Language and more.

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Practical Geometry Questions

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Answered on 24 Feb Learn CBSE/Class 6/Maths/Practical Geometry

Sadika

To construct a line segment equal to AB+CDAB+CD, you would first draw a line segment ABAB with a length of 3.6 cm and then draw a line segment CDCD with a length of 1.6 cm. Then, you would place the endpoint of CDCD at the endpoint of ABAB, so they connect end to end. The resulting line segment would... read more

To construct a line segment equal to AB+CDAB+CD, you would first draw a line segment ABAB with a length of 3.6 cm and then draw a line segment CDCD with a length of 1.6 cm.

Then, you would place the endpoint of CDCD at the endpoint of ABAB, so they connect end to end. The resulting line segment would be the sum of the lengths of ABAB and CDCD.

Now, to measure the total length of the constructed line segment:

Total length=AB+CD=3.6 cm+1.6 cm=5.2 cmTotal length=AB+CD=3.6 cm+1.6 cm=5.2 cm

So, the total length of the constructed line segment is 5.2 cm.

 
 
 
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Answered on 24 Feb Learn CBSE/Class 6/Maths/Practical Geometry

Sadika

To construct a perpendicular to a given line segment at a point on it, you can follow these steps: Draw the given line segment. Choose a point on the line segment where you want the perpendicular to pass through. Let's call this point PP. With point PP as the center, draw a circle with any radius... read more

To construct a perpendicular to a given line segment at a point on it, you can follow these steps:

  1. Draw the given line segment.
  2. Choose a point on the line segment where you want the perpendicular to pass through. Let's call this point PP.
  3. With point PP as the center, draw a circle with any radius that intersects the line segment at two points, let's call them AA and BB.
  4. Draw a straight line through points AA and BB. This line will be perpendicular to the given line segment at point PP.

Now, you have constructed a perpendicular to the given line segment at the chosen point PP.

 
 
 
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Answered on 24 Feb Learn CBSE/Class 6/Maths/Practical Geometry

Sadika

To construct an angle of 60 degrees and bisect it, you can follow these steps using a compass and straightedge: Start by drawing a straight line segment ABAB using the straightedge. With AA as the center, draw an arc using the compass to create an angle of any size, let's say 60 degrees. Mark the... read more

To construct an angle of 60 degrees and bisect it, you can follow these steps using a compass and straightedge:

  1. Start by drawing a straight line segment ABAB using the straightedge.
  2. With AA as the center, draw an arc using the compass to create an angle of any size, let's say 60 degrees. Mark the intersection point with the line segment as CC.
  3. Without changing the compass width, place the compass at point CC and draw an arc intersecting the first arc at point DD.
  4. Draw a straight line segment connecting point CC to point DD. This line segment CDCD bisects the angle CABCAB, creating two 30-degree angles.

Now, you have constructed an angle of 60 degrees and bisected it into two 30-degree angles.

 
 
 
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Answered on 24 Feb Learn CBSE/Class 6/Maths/Practical Geometry

Sadika

To draw an angle of 120° and then construct an angle of 105°, follow these steps: Draw an Angle of 120°: Use a protractor to draw a straight line (ray) and mark a point on it to represent the vertex of the angle. Place the center of the protractor at the vertex and align the baseline... read more

To draw an angle of 120° and then construct an angle of 105°, follow these steps:

  1. Draw an Angle of 120°:

    • Use a protractor to draw a straight line (ray) and mark a point on it to represent the vertex of the angle.
    • Place the center of the protractor at the vertex and align the baseline of the protractor with the ray.
    • Mark a point at the 120° mark on the protractor.
    • Draw a line from the vertex to the marked point to complete the angle.
  2. Construct an Angle of 105°:

    • Place the center of the protractor at the vertex of the 120° angle.
    • Align the baseline of the protractor with one side of the 120° angle.
    • Mark a point at the 105° mark on the protractor.
    • Draw a line from the vertex to the marked point to construct the 105° angle.

Now, you have both the 120° angle and the constructed 105° angle.

 
 
 
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Answered on 24 Feb Learn CBSE/Class 6/Maths/Practical Geometry

Sadika

To draw triangle ABC and its perpendiculars from each vertex onto the opposite sides, follow these steps: Draw triangle ABC: Draw three non-parallel lines intersecting at points A, B, and C to form triangle ABC. Draw perpendiculars from each vertex onto the opposite sides: From vertex A, draw... read more

To draw triangle ABC and its perpendiculars from each vertex onto the opposite sides, follow these steps:

  1. Draw triangle ABC:

    • Draw three non-parallel lines intersecting at points A, B, and C to form triangle ABC.
  2. Draw perpendiculars from each vertex onto the opposite sides:

    • From vertex A, draw a line perpendicular to side BC. Label the point of intersection with side BC as D.
    • From vertex B, draw a line perpendicular to side AC. Label the point of intersection with side AC as E.
    • From vertex C, draw a line perpendicular to side AB. Label the point of intersection with side AB as F.

Now, you have triangle ABC with perpendiculars AD, BE, and CF drawn from vertices A, B, and C respectively onto the opposite sides BC, AC, and AB.

To determine if these perpendiculars are concurrent (passing through the same point), you can construct the altitudes of the triangle, which are known to be concurrent at the orthocenter.

If the perpendiculars AD, BE, and CF intersect at a single point, then they are concurrent, and that point is the orthocenter of triangle ABC. If they do not intersect at a single point, then they are not concurrent.

Please note that the concurrency of perpendiculars in a triangle depends on the specific properties of the triangle. In some cases, the perpendiculars may be concurrent, while in others, they may not be. You would need to verify this for the specific triangle you've drawn.

 
 
 
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