Saturday, December 21, 2024

Practical Geometry

- Advertisement -spot_imgspot_img
- Advertisement -spot_img

Practical Geometry, a chapter in your 6th-grade math textbook, focuses on using tools to construct and measure geometric shapes. Here’s a breakdown of the key concepts:

Tools of the Trade:

  • Ruler: Measures line segments and helps draw straight lines.
  • Compasses: Used for constructing circles and arcs.
  • Set Squares: Have two right angles (90°) and help draw perpendicular lines and angles.
  • Protractor: Measures angles in degrees.

Constructions:

  • Circles: You’ll learn to draw circles of a specific radius using the compass.
  • Line Segments: Techniques to construct line segments of a desired length with the help of the ruler and compass are covered.
  • Perpendicular Lines: You’ll practice drawing perpendicular lines from a point on a line segment and from a point outside the line segment using ruler and compass.

By the end of the chapter, you’ll be able to apply these techniques to solve problems involving geometric shapes

Exercise 14.1

1. Draw a circle of radius 3.2 cm.

Ans : 

2. With the same centre O, Draw two circles of radius 4 cm and 2.5 cm.

Ans : 

3. Draw a circle and any two its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?

Ans : 

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry

(i) Draw a circle with centre 0 with suitable radius.

(ii) AB and CD are any two diameters.

(iii) On joining the end points of the diameters, we get a quadrilateral ACBD.

(iv) We note that OA = OB = OC = OD [Same radius]

and AC = DB, AD = BC

∠A = ∠C = ∠B = ∠D = 90°

Thus ACBD is a rectangle.

Again if the diameters are perpendicular to each other then on measuring, we get

AC = DB = AD = BC

Thus, ACBD is a square.

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry

4. Draw any circle and mark points A, B and C such that

(a) A is on the circle

(b) B is in the interior of the circle

(c) C is in the exterior of the circle.

Ans : 

5. Let A, B be the centres of the two circles of equal radii. Draw them so that each one of them passes through the centre of the other. Let them intersect at C and D.

Examine whether  AB and CD are at right angles.

Ans : 

In the given figure, two circles of equal radii intersect each other at points C and D. Upon measurement, we see that AB‾ and CD‾ intersect each other at right angles.

Exercise 14.2

1. Draw a line segment of length 7.3 cm using ruler.

Ans : 

2. Construct a line segment of length 5.6 cm using ruler and compass.

Ans : 

3. Construct ABof length 7.8 cm. From this, cut off ACof length 4.7 cm. Measure BC

Ans : 

Given that AB = 7.8 and AC=4.7 cm.

Step I: Place the zero mark of the ruler at point A.

Step II: Mark point B at a distance of 7.8 cm from A.

Step III: Mark another point C at a distance of 4.7 cm from A, ensuring AC=4.7 cm.

Step IV: On measuring the length of BC, we find that BC=3.1 cm.

4. Given AB of length 3.9 cm. Construct PQ Such that the length of PQ is twice that of AB. Verify by measurement. 

Ans : 

5. Given AB of length 7.3 cm and CDof length 3.4 cm, construct  a line segment XYsuch that the length of XY is equal to the differences between the length of ABand AC . Verify the measurement. 

Ans : 

Thus , PQ = AB – CD

Exercise 14.3

1.Draw any line segment PQ . Without measuring PQ , construct a copy of PQ .

Ans :

2. Give some line segment AB whose length you do not know, construct PQ such that the length of PQ is twice that of AB

Ans : 

Exercise 14.4

1. Draw the line segment AB . Make any point M on it. Through M, draw a perpendicular to AB

Ans : 

2.Draw any line segment PQ  Take any point R not on it. Through R, draw a perpendicular to PQ . (Use ruler and set square).

Ans : 

Draw the Line Segment PQ:

  • Use a ruler to draw a line segment and label its endpoints as P and Q.

Mark the Point R:

  • Choose any point R that is not on the line segment PQ.

Place the Set Square:

  • Position the set square so that one of its right-angled sides is aligned with the line segment PQ.

Align the Ruler:

  • Place the ruler along the other right-angled side of the set square.

Slide the Set Square:

  • Keeping the ruler fixed, slide the set square along the ruler until the edge of the set square intersects point R.

Draw the Perpendicular Line:

  • Using the edge of the set square that intersects point R, draw a line from R to the line segment PQ. This line will be perpendicular to PQ.

Label the Intersection:

  • Label the point where the perpendicular line intersects PQ as S.

3. Draw a line l and a point X on it. Through X, draw a line segment \overline { XY } XY perpendicular to l. Now draw a perpendicular to at XY  y. (Use ruler and compasses)

Ans :

Exercise 14.5 

1. Draw AB of length 7.3 cm and find its axis of symmetry.

Ans : 

2. Draw a line segment of length 9.5 cm and construct its perpendicular bisector.

Ans : 

3. Draw the perpendicular bisector of XY whose length is 10.3 cm.

(a) Take any point P on the bisector drawn. Examine whether PX = PY.

(b) If M is the midpoint of XY  What can you say about the length of MX and MY?

Ans : 

(a) Take Any Point P on the Bisector:

  • Place a point P anywhere on the perpendicular bisector AB.
  • Measure the distance PX and PY. Since P is on the perpendicular bisector, PX should be equal to PY.
  • Thus, PX= PY.

(b) Midpoint M:

  • M is the midpoint of XY, so by definition, MX = MY.
  • Since M bisects XY, each segment MX and MY is half the length of  XY.
  • Therefore, MX = MY = 10.3/2 = 5.15cm

4. Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.

Ans : 

Step I: Draw a line segment AB = 12.8 cm

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry

Step II : With centre A and B and radius more than half of AB, draw two arcs which meet each other at D and E.

Step III : Join D and E which meets AB at C which is the midpoint of AB

Step IV : With centre A and C and radius more than half of AC, draw two arcs which meet each other at F and G.

Step V: Join F and G which meets AC at H which is the midpoint of  AC 

Step VI : With centre C and B and radius more than half of CB, draw two arcs which meet each other at J and K.

Step VII : Join J and K which meets CB at L which is the midpoint of CB 

Thus, on measuring, we find

AH = HC =  CL = LB = 3.2 cm.

5. With PQ of length 6.1 cm as diameter, draw a circle.

Ans : 

6. Draw a circle with centre C and radius 3.4 cm. Draw any chord AB . Construct the perpendicular bisector of AB and examine if it passes through C.

Ans : 

7. Repeat Question number 6, if AB happens to be a diameter.

Ans :

8. Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?

Ans : 

On producing the two perpendicular bisectors meet each other at the centre O of the circle.

9. Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of OA and OB. Let them meet at P. Is PA = PB?

Ans : 

Step I: Draw an angle XOY with O as its vertex.

Step II : Take any point A on OY and B on OX, such that OA + OB.

Step III : Draw the perpendicular bisectors of OA and OB which meet each other at a point P.

Step IV : Measure the lengths of PA and PB Yes. PA = PB 

Exercise 14.6

1. Draw ∠POQ of measure 75° and find its line of symmetry.

Ans : 

Draw a ray OP: Using a ruler, draw a straight line and mark a point O on it. This line will be one side of the angle.

Construct a 90° angle: Using a protractor, measure and draw a ray OQ making a 90° angle with OP.

Bisect the 90° angle:

  • With O as the center, draw an arc cutting both rays OP and OQ.
  • With the same radius, draw arcs from the points where the first arc intersected the rays, such that they intersect each other.
  • Draw a ray from O passing through the intersection point of the two arcs. This bisects the 90° angle, creating a 45° angle.

Bisect the 45° angle: Repeat the process of bisecting the angle to obtain a 22.5° angle.

Construct the 75° angle: Add the 45° angle and the 30° angle (half of 60°) to get a 75° angle. This is ∠POQ.

The line of symmetry of an angle is the line that divides the angle into two equal parts. For ∠POQ, the line of symmetry is the angle bisector.

2. Draw an angle of measure 147° and construct its bisector.

Ans :

  1. Draw a ray OA.
  2. Place the center of the protractor at point O and align the base line of the protractor with ray OA.
  3. Mark a point B at 147° on the protractor.
  4. Draw a ray OB passing through point B.

Angle AOB is the required angle of 147°.

3. Draw a right angle and construct its bisector.

Ans :

4. Draw an angle of 153° and divide it into four equal parts.

Ans : 

5. Construct with ruler and compasses, angles of the following measures:

(a) 60°

(b) 30°

(c) 90°

(d) 120°

(e) 45°

(f) 135°

Ans : 

(a) Angle of 60°

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry

(b) Angle of 60°

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry

Thus ∠ABF = 60/2 = 30°.

(c) Angle of 90°

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry

d) Angle of 120°

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry

(e) Angle of 45s, i.e., 90/2 = 45°

(f) An angle of 135°

Since 135° = 90° + 45°

= 90° + (90/2 )°

In this figure ∠ABC = 135°

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry

6. Draw an angle of measure 45° and bisect it.

Ans : 

7. Draw an angle of measure 135° and bisect it.

Ans : 

8. Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.

Ans : 

 Draw ∠COB = 70° using protractor.

Draw a ray PQ

9. Draw an angle of 40°. Copy its supplementary angle.

Ans : 

Construct ∠AOB = 40° using protractor.

∠COF is the supplementary angle of ∠AOB.

Draw a ray PR and take any point Q on it.

Thus, ∠PQS is the copy of the supplementary angle COB.

- Advertisement -spot_imgspot_img
Dr. Upendra Kant Chaubey
Dr. Upendra Kant Chaubeyhttps://education85.com
Dr. Upendra Kant Chaubey, An exceptionally qualified educator, holds both a Master's and Ph.D. With a rich academic background, he brings extensive knowledge and expertise to the classroom, ensuring a rewarding and impactful learning experience for students.
Latest news
- Advertisement -spot_img
Related news
- Advertisement -spot_imgspot_img