NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exz 14.6 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6.
- Practical Geometry Class 6 Ex 14.1
- Practical Geometry Class 6 Ex 14.2
- Practical Geometry Class 6 Ex 14.3
- Practical Geometry Class 6 Ex 14.4
- Practical Geometry Class 6 Ex 14.5
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 6 |
| Subject | Maths |
| Chapter | Chapter 14 |
| Chapter Name | Practical Geometry |
| Exercise | Ex 14.6 |
| Number of Questions Solved | 9 |
| Category | NCERT Solutions |
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6
Ex 14.6 Class 6 Maths Question 1.
Draw ZPOQ of measure 75° and find its line of symmetry.
Solution:
Steps of Construction:
- Draw a ray OP.
- Draw ∠POR = 60° and ∠POS = 90°.
- Draw OQ, the bisector of ∠ROS.
Then,
Ex 14.6 Class 6 Maths Question 2.
Draw an angle of measure 147° and construct its bisector.
Solution:
Steps of Construction:
- Draw a ray OA.
- Place the protractor on OA such that its centre falls on the initial point O and 0 -180 line lies along OA.
- Mark a point B on the paper against the mark of 147° on the protractor.
- Remove the protractor and draw OB. Then, the ∠AOB so obtained is the required angle such that ZAOB = 147°.
To construct its bisector:
- With centre O and a convenient radius draw an arc cutting sides OA and OB at P and Q respectively.
- With centre P and radius > \(\frac { 1 }{ 2 } \) PQ, draw an arc.
- With centre Q and the same radius, as in the previous step, draw another arc intersecting the arc drawn in the previous step at R.
- Join OR and produce it to form ray OC.
Then, the ∠AOC so obtained is the bisector of ∠AOB.
Ex 14.6 Class 6 Maths Question 3.
Draw a right angle and construct its bisector.
Solution:
Steps of Construction:
To draw an angle of 90°:
- Draw a ray OA.
- With O as centre and any convenient radius, draw an arc cutting at P.
- With P as centre and same radius, draw an arc cutting the arc of step 2 at Q.
- With Q as centre and the same radius as in steps 2 and 3, draw an arc cutting the arc drawn in step 3 at R.
- With Q as centre and the same radius, draw an arc.
- With R as centre and the same radius, draw an arc cutting the arc drawn in step 5 at B.
To draw its bisector:
- With P as centre and radius > \(\frac { 1 }{ 2 } \) PT, draw an arc in the interior of ∠AOC.
- With T as centre and the same radius, an in step 1, draw another arc intersecting the arc in step 1 at D.
- Join OD and produce it to any point E.
Then, ∠AOE so obtained is the bisector of ∠AOC.
Ex 14.6 Class 6 Maths Question 4.
Draw an angle of measure 153° and divide it into four equal parts.
Solution:
Steps of Construction:
- With the help of the protractor, draw ∠AOB = 153°.
- With centre O and any convenient radius, draw an arc cutting OA and OB at P and Q respectively.
- With centre P and radius > \(\frac { 1 }{ 2 } \) PQ, draw an arc in the interior of ∠AOB.
- With centre Q and the same radius, as in step 3, draw another arc intersecting the arc in step 3 at B1.
- Join OB1 and produce it to any point C.
Then, ∠AOC = \(\frac { 1 }{ 2 } \) x ∠AOB i.e., bisector of ∠AOB. - Draw OD, the bisector of ∠AOC. Then ∠AOD = ∠DOC.
- Draw OE, the bisector of ∠COB. Then, ∠COE = ∠COB.
Combining these results, we have
∠AOD = ∠DOC = ∠COE = ∠EOB.
Thus, ∠AOB is divided into four equal parts by the rays OD,OC and OE.
Ex 14.6 Class 6 Maths Question 5.
Construct with ruler and compasses, angles of following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 45°
(f) 135°
Solution:
(a) Steps of Construction:
- Draw a ray OA.
- With centre O and any radius, draw an arc PQ with the help of compasses, cutting the ray at P.
- With centre P and the same radius draw an arc cutting the arc PQ at R.
- Join OR and produce it to obtain ray OB.
Then, ∠AOB so obtained is of 60°.
(b) Steps of Construction:
- Draw a ray OA.
- With centre O any radius draw an arc PT with the help of compasses, cutting ray OA at P.
- With centre P and the same radius draw an arc cutting the arc PT at Q.
- Join OQ and produce it to obtain ray OB.
Then, ∠AOB = 60°. - With centre P and radius > \(\frac { 1 }{ 2 } \) PQ, draw an arc in the interior of ∠AOB.
- With centre Q and the same radius, as in step 5, draw another arc intersecting the arc in step 5 at R.
- Join OR and produce it on any point C.
Then, ∠AOC = 30°
(c) Step of Construction:
- Draw a ray OA.
- With O as centre and any convenient radius, draw an arc cutting at P.
- With P as centre and same radius, draw an arc cutting the arc of step 2 at Q.
- With Q as centre and the same radius as in steps 2 and 3, draw an arc cutting the arc drawn in step 3 at R.
- With Q as centre and the same radius, draw an arc.
- With R as centre and the same radius, draw an arc cutting the arc drawn in step 5 at B.
- Join O to B and produce it to any point C.
Then, ∠AOC =90°
(d) Steps of Construction:
- Draw a ray OA.
- With O as centre and any convenient radius, draw an arc cutting OA at P.
- With P as centre and the same radius draw an arc cutting the first arc at Q.
- With Q as centre and the same radius, draw an cutting the arc drawn in step 2 at R.
- Join AR and produce it to any point C.
Then, ∠AOC so obtained is of 120°.
(e) Steps of Construction:
- Draw ∠AOC = 90° by following the steps given in part (iii) above,
- Draw OE, the bisector of ∠AOC.
Then, ∠AOD so obtained is the required angle of 45°.
(f) Steps of Construction:
- Draw ∠AOC = 120°
∠AOB =150°. - Draw OD, bisector of ∠COB.
Then,
Or
Steps of Construction:
- Draw a line AB and mark a point O on it.
- With centre O and any convenient radius draw a semi-circle cutting OA and OB at P and Q respectively.
- With Q as centre and same radius, draw an arc cutting the semi-circle as R.
- With R as centre and same radius, draw an arc cutting the semi-circle of step 2 at S.
- With R as centre and same radius draw an arc.
- With S as centre and same radius draw an arc cutting the arc drawn in step 5 at T. Join OT and produce it to D such that ∠BOD – ∠AOD = 90°.
- Draw OE, the bisector of ∠AOD.
Ex 14.6 Class 6 Maths Question 6.
Draw an angle of measure 45° and bisect it.
Solution:
Steps of Construction:
- Draw ∠AOB = 90° by the steps given in question 5 (c).
- Draw OC, the bisector of ∠AOB. Then, ∠AOC = 45°.
- Draw OD, the bisector of ∠AOC. Then, ∠AOD = ∠DOC.
Ex 14.6 Class 6 Maths Question 7.
Draw an angle of measure 135° and bisect it.
Solution:
Steps of Construction:
- Draw ∠EOB = 135° by the steps given in question 5 (f).
- Draw OF, the bisector of ∠EOB.
Then, ∠BOF = ∠FOE.
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:
Steps of Construction:
- Draw an angle 70° with protractor, i. e. ∠POQ = 70°
- Draw a ray \(\overline { AB } \)
- Place the compasses at O and draw an arc to cut the ray of ∠POQ at I and M.
- Use the same compasses, setting to draw an arc with A as centre, cutting AB at X.
- Set your compasses setting to the length LM with the same radius.
- Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.
- Join A7.
Thus, ∠YAX =70°
Ex 14.6 Class 6 Maths Question 9.
Draw an angle of 40° copy its supplementary angle.
Solution:
Steps of construction:
- Draw an angle of 40° with the help of protractor, naming ∠ AOB.
- Draw a line PQ.
- Take any point M on PQ.
- Place the compasses at O and draw an arc to cut the rays of ∠ AOB at L and N.
- Use the same compasses setting to draw an arc O as centre, cutting MQ at X.
- Place the compasses at X and draw the arc to cut the arc drawn earlier Y.
- Join MY.
Thus, ∠QMY = 40° and ∠PMY is supplementary of it.
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