NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 are the part of NCERT Solutions for Class 6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6.



Ex 14.6 Class 6 Maths Question 1.

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

Solution:
Step 1: Draw a line segment PO.
Step 2: With centre O and any suitable radius, draw an arc to cut PO at R.

Step 3: With centre R and radius of the same length, mark two arcs S and T on the previous arc.
Step 4: With centres S and T and with the same radius, draw two arcs which intersect each other at U.
Step 5: Join OU such that
POU = 90°.
Step 6: With centres S and W, draw two arcs of the same radius which intersect each other at Q.
Step 7: Join O and Q such that
POQ = 75°.
Step 8: Bisect
POQ with OV.
Thus, OV is the line of symmetry of
POQ.

 

Ex 14.6 Class 6 Maths Question 2.

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

Solution:
Step 1: Draw ABC = 147° with the help of a protractor.

Step 2: With centre B and radius of any suitable length, draw an arc which intersects AB and BC at E and F, respectively.
Step 3: With centres E and F and the radius more than half of the length of arc EF, draw two arcs which intersect each other at D.
Step 4: Join B and D.
Thus, BD is the bisector of
ABC.

 

Ex 14.6 Class 6 Maths Question 3.

Draw a right angle and construct its bisector.

Solution:
Step 1: Draw a line segment AB.
Step 2: With centre B and any suitable radius, draw an arc to intersect AB at C.

Step 3: With centre C and same radius, mark two arcs D and E on the previous arc.
Step 4: With centres D and E and the same radius, draw two arcs which intersect each other at G.
Step 5: Join B and G such that
ABG = 90°.
Step 6: Draw BH as the bisector of
ABG such that ABH = 45°.
Thus,
ABG is the right angle and BH is the bisector of ABG.

 

Ex 14.6 Class 6 Maths Question 4.

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

Solution:
Step 1: Draw ABP of measure 153° with the help of a protractor.

Step 2: Draw BC as the bisector of ABP which divides ABP into two equal parts.
Step 3: Draw BD and BE as the bisectors of
ABC and CBP, respectively.
Thus, the bisectors BD, BC and BE divide the
ABP into four equal parts.

 

Ex 14.6 Class 6 Maths Question 5.

Construct with ruler and compasses, angles of the following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 120°
(e) 45°
(f) 135°

Solution:
(a) Construction of 60°:

Step 1: Draw a line segment AB.
Step 2: With centre B and any suitable radius, draw an arc which intersect AB at D.
Step 3: With centre D and radius of the same length, draw an arc E on the previous arc.
Step 4: Join B to E and produce it up to C. Thus, 
ABC is the required angle of measure 60°.


(b) Construction of 30°:

Step 1: Draw ABC = 60° as we have drawn in part (a).
Step 2: Draw BF as the bisector of
ABC.

Thus, ABF = 60/2 = 30°.


(c) Construction of 90°:
Step 1: Draw a line segment AB.
Step 2: With centre B and any suitable radius, draw an arc to intersect AB at C.
Step 3: With centre C and same radius, mark two arcs D and E on the previous arc.
Step 4: With centres D and E and the same radius, draw two arcs which intersect each other at G.
Step 5: Join B and G such that
ABG = 90°.

(d) Construction of 120°:
Step 1: Draw a line segment AB.
Step 2: With centre A and radius of any suitable length, draw an arc which intersect AB at D.

Step 3: With centre D and the same radius, draw two arcs E and F on previous arc.
Step 4: Join A to F and produce it up to C. Thus, 
CAB = 120°.


(e) Construction of 45°:

To make an angle of 45°, we bisect the angle of 90°.
Step 1: Construct an angle of 90° as in part (c).

Step 2: Bisect the angle of 90° to get an angle of 45°.

In the figure, ABD = 45°


(f) Construction of 135°:
Since 135° = 90° + 45° = 90° + (90/2
In this figure,
ABC = 135°.

 

Ex 14.6 Class 6 Maths Question 6.

Draw an angle of measure 45° and bisect it.

Solution:
Step 1: Draw a line AB and take any point O on it.
Step 2: Construct an angle of 90° on point O. Construct
AOE = 45° by bisecting AOC.

Step 3: With centre O and any suitable radius, draw an arc GF.
Step 4: With centres G and F and a suitable radius, draw two arcs which intersect each other at D.
Step 5: Join O to D.
Thus, 
AOE = 45° and OD is its bisector.


Ex 14.6 Class 6 Maths Question 7.

Draw an angle of measure 135° and bisect it.

Solution:
Steps 1: Draw a line OA and take any point P on it.

Step 2: Construct APQ = 135°.
Step 3: Draw PD as the bisector of angle APQ.
Thus,
APD = 135/2 = 67½°.

 

Ex 14.6 Class 6 Maths Question 8.

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

Solution:
Step 1: Draw a line AB and take any point O on it.
Step 2: Draw
COB = 70° using a protractor.

Step 3: Draw a ray PQ.
Step 4: With centre O and a suitable radius, draw an arc which meets OB and OC at E and F, respectively.

Step 5: With the same radius and centre at P, draw an arc meeting PQ at R.
Step 6: With centre R and radius equal to EF, draw an arc intersecting the former arc at S.
Step 7: Join P and S and produce it. Thus, QPS is the copy of
COB = 70°.

 

Ex 14.6 Class 6 Maths Question 9.

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

Solution:
Step 1: Construct AOB = 40° using protractor.
COB is the supplementary angle of AOB.

Step 2: Draw a ray PR and take a point Q on it.
Step 3: With centre O and a suitable radius, draw an arc which intersects OC and OB at E and F, respectively.

Step 4: With centre Q and same radius, draw an arc which intersects PQ at L.
Step 5: With centre L and radius equal to EF, draw an arc which intersects the previous arc at S.
Step 6: Join Q and S and produce it.
Thus, 
PQS is the copy of the supplementary angle of 40°.



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NCERT Solutions for Maths Class 9

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