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

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,

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 > 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.

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 > 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.

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 > 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 B
_{1}. - Join OB
_{1}and produce it to any point C.

Then, ∠AOC = 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.

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 > 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.

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.

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.

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
- 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°

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.

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6, drop a comment below and we will get back to you at the earliest.

## Leave a Reply