NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 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 ∠POQ of measure 75° and find its line of symmetry.

Solution :

**Step 1.** Draw a ray \(\overline { OQ }\).

**Step 2.** Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).

**Step 3.** Start with 0 near Q. Mark point P at 75°

**Step 4.** Join \(\overline { OP }\). Then, ∠POQ = 75°.

**Step 5.** With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.

**Step 6.** With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.

**Step 7.** With the same radius and with F as center, draw another arc in the interior of ∠POQ. Let

the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ which is also the line of symmetry of ∠POQ as ∠POR = ∠ROQ.

Question 2.

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

Solution :

**Step 1.** Draw \(\overline { OQ }\) of any length.

**Step 2.** Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).

**Step 3.** Start with 0 near Q. Mark a point P at 147°.

**Step 4.** Join OP. Then, ∠POQ = 147°.

**Step 5.** With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.

**Step 6.** With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.

**Step 7.** With the same radius and with F as center, draw another arc in the interior of ∠POQ. Let the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.

Question 3.

Draw a right angle and construct its bisector.

Solution :

**Step 1.** Draw a ray OQ.

**Step 2.** Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).

**Step 3.** Start with 0 near Q. Mark point P at 90°.

**Step 4.** Join \(\overline { OP }\). Then, ∠POQ = 90°.

**Step 5.** With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.

**Step 6.** With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half th length Q’F.

**Step 7.** With the same radius and with P center, draw another arc in the interior of ∠POQ the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.

Question 4.

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

Solution :

**Step 1.** Draw a ray \(\overline { OQ }\).

**Step 2.** Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).

**Step 3.** Start with 0 near Q. Mark a point P at 153°.

**Step 4.** Join OP. Then, ∠POQ = 153°.

**Step 5.** With O as the center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q.

**Step 6.** With Q’ as the center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.

**Step 7.** With the same radius and with F as a center, draw another arc in the interior of ∠POQ. Let the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.

**Step 8.** With O as a center and using compasses, draw an arc that cuts both rays of ∠ROQ. Label the points of intersection as B and A.

**Step 9.** With A as a center, draw (in the interior of ∠ROQ) an arc whose radius is more than half the length AB.

**Step 10.** With the same radius and with B as a center, draw another arc in the interior of ∠ROQ. Let the two arcs intersect at S. Then, \(\overline { OS }\) is the bisector of ∠ROQ.

**Step 11.** With O as a center and using compasses, draw an arc that cuts both rays of ∠POR. Label the points of intersection as D and C.

**Step 12.** With C as a center, draw (in the interior of ∠POR) an arc whose radius is more than half the length CD.

**Step 13.** With the same radius and with D as centre, draw another arc in the interior of ∠POR. Let the two arcs intersect at T. Then, \(\overline { OT }\) is the bisector of ∠POR. Thus, \(\overline { OS }\), \(\overline { OR }\) and \(\overline { OT }\) divide ∠POQ = 153° into four equal parts.

Question 5.

Construct with ruler and compasses, angles of following measures:

**(a)** 60°

**(b)** 30°

**(c)** 90°

**(d)** 120°

**(e)** 45°

**(f)** 135°.

Solution :

**(a)** **Construction of an angle of measure 60°**

**Step 1.** Draw a line PQ and mark a point O on it.

**Step 2.** Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line PQ at a point say A.

**Step 3.** With the pointer at A (as center), now draw an arc that passes through O.

**Step 4.** Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.

**(b)** **Construction of an angle of measure 30°**

**Step 1.** Draw a line PQ and mark a point O on it.

**Step 2.** Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line PQ at a point say A.

**Step 3.** With the pointer at A (as center), now draw an arc that passes through O.

**Step 4.** Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.

**Step 5.** With O as a center and using compasses, draw an arc that cuts both rays of ∠BOA. Label the points of intersection as D and C.

**Step 6.** With C as a center, draw (in the interior of ∠BOA) an arc whose radius is more than half the length CD.

**Step 7.** With the same radius and with D as a center, draw another arc in the interior of ∠BOA. Let the two arcs intersect at E. Then, \(\overline { OE }\) is the bisector of ∠BOA, i.e., ∠BOE = ∠EOA = 30°.

**(c) Construction of an angle of measure 90°**

**Step 1.** Draw any line PQ and take a point O on it.

**Step 2.** Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.

**Step 3.** Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.

**Step 4.** Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.

**Step 5.** Join OB and OC.

**Step 6.** With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.

**Step 7.** With E as a center, draw (in the interior of ∠COB) an arc where the radius is more than half the length ED.

**Step 8.** With the same radius and with D as a center, draw another arc in the interior of ∠COL. Let

the two arcs intersect at F. Join \(\overline { OF }\). Then, \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠ FOQ = 90°.

**(d) Construction of an angle of measure 120°**

**Step 1.** Draw any line PQ and take a point O on it.

**Step 2.** Place the pointer of the compass es at O and draw an arc of convenient radius which puts the line at A. • ‘

**Step 3.** Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.

**Step 4.** Again without disturbing the i alius > on the compasses and with B as a center, draw an arc which cuts the first arc at C. .

**Step 5.** Join OC. Then, ∠COA is the required angle whose measure is 120°.

**(e) Construction of an angle of measure 45°**

**Step 1.** Draw any line PQ and take a point O on it.

**Step 2.** Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.

**Step 3.** Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.

**Step 4.** Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.

**Step 5.** Join OB and OC.

**Step 6.** With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.

**Step 7.** With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.

**Step 8.** With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let the two arcs intersect at F. Join OF. Then, ∠FOQ = 90°.

**Step 9.** With O as a center and using compasses, draw an arc that cuts both rays to ∠FOQ. Label the points of the intersection as G and H.

**Step 10.** With H as a center, draw (in the interior of. ∠FOQ) an arc whose radius is more than half the length HG.

**Step 11.** With the same radius and with G as a center, draw another arc in the interior of ∠FOQ. Let the two arcs intersect at I. Join OI. Then, OI is the bisector of ∠FOH, i.e., ∠FOI = ∠IOH. Now,

∠FOI = ∠IOH = 45°.

**(f) Construction of an angle of measure 135° it.**

**Step 1. **Draw any line PQ and take a point O on it.**
Step 2.** Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.

**Step 3.**Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.

**Step 4.**Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.

**Step 5.**Join OB and OC.

**Step 6.**With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.

**Step 7.**With E as the center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.

**Step 8.**With the same radius and with D as the center, draw another arc in the interior of ∠COB. Let the two arcs intersect at F. Join \(\overline { OF }\). Then, \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠FOQ = 90°.

**Step 9.**With O as a center and using compasses, draw an arc that cuts both rays of ∠POF. Label the points of intersection as G and H.

**Step 10.**With H as center draw (in the interior of ∠POF) an arc whose radius is more than half the length HG.

**Step 11.**With the same radius and with G as a center, draw another arc in the interior of ∠POF. Let

the two arcs intersect at I. Join 01. Then, \(\overline { OI }\) is the bisector of ∠POF, i.e., ∠GOI = ∠IOF. Now, ∠IOQ = 135°.

Question 6.

Draw an angle of measure 45° and bisect it.

Solution :

**Step 1.** Draw any line PQ and take a point O on it.

**Step 2.** Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A. ’

**Step 3.** Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.

**Step 4.** Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.

**Step 5.** Join OB and OC.

**Step 6.** With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.

**Step 7.** With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.

**Step 8.** With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let

the two arcs intersect at F. Join OF. Then ∠FOQ = 90°.

**Step 9.** With O as a center and using compasses, draw an arc that cuts both rays of ∠FOQ. Label the points of intersection on G and H.

**Step 10.** With G as a center, draw in the interior of ∠FOQ) an arc whose radius is more than half the length HG.

**Step 11.** With the same radius and with G as a center, draw another arc in the interior of ∠FOQ. Let the two arcs intersect at I. Join OI. Then OI is the bisector of ∠FOQ, i.e., ∠FOI = ∠IOH. Now, ∠FOI = ∠IOH = 45°.

**Step 12.** With O as a center and using compasses, draw an arc that cuts both rays of ∠IOH. Label the points of intersection as J and K.

**Step 13.** With K as a center, draw (in the interior of ∠IOH) an arc whose radius is more than half the length KJ.

**Step 14.** With the same radius and with J as a center, draw another arc in the interior of ∠IOH. Let the two arcs intersect at L. Join OL. Then OL is the bisector of ∠IOH, i.e., ∠IOL = ∠LOK \(22\frac { 1^{ \circ } }{ 2 }\) .

Question 7.

Draw an angle of measure 135° and bisect it.

Solution :

**Step 1.** Draw any line PQ and take a point O on.

**Step 2.** Place the pointer of the compasses at’ and draw an arc of convenient radius which cul line at A.

**Step 3.** Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.

**Step 4.** Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.

**Step 5.** Join OB and OC.

**Step 6.** With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.

**Step 7.** With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.

**Step 8.** With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let

the two arcs intersect at F. Join \(\overline { OF }\). Then \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠FOQ = 90°.

**Step 9.** With O as a center and using compasses, draw an arc that cuts both rays of ∠POF. Label the points of intersection as G and H.

**Step 10.** With G as a center, draw in the interior of ∠POF an arc whose radius is more than half the length HG.

**Step 11.** With the same radius and with H as a centre, draw another arc in the interior of ∠POF. Let

the two arcs intersect at I. Join \(\overline { OI }\). Then \(\overline { OI }\) is the bisector of ∠POF, i.e., ∠POI = ∠IOF. Now, ∠IOQ = 135°. .

**Step 12.** With O as centre and using compasses, draw an arc that cuts both rays of ∠IOQ. Label the points of intersection as J and K.

**Step 13.** With K as centre, draw (in the interior of ∠IOQ) an arc whose radius is more than half the length KJ.

**Step 14.** With the same radius and with J as centre, draw another arc in the interior of ∠IOQ. Let the two arcs intersect at L. Join \(\overline { OL }\). Then \(\overline { OL }\) is the bisector of ∠IOQ, i.e., ∠IOL = ∠LOQ.

Question 8.

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

Solution :

**Steps of construction**

1. Construct an angle ABC = 70°.

2. Take a line z and mark a point D on it.

3. Fix the compasses pointer on B and draw an arc which cuts the sides of ∠ABC at D and E.

4. Without changing the compasses setting, place the pointer on P and draw an arc which cuts ∠ at Q.

5. Open the compasses equal to length DE.

6. Without disturbing the radius on compasses, place its pointer at Q and draw an arc which cuts the previous arc at R.

7. Join PR and draw ray PR. It gives ∠RPQ which is the required angle whose measure is equal to the measure of ∠ABC.

Question 9.

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

Solution :

**Steps of construction**

1. Draw ∠CAB = 40°.

2. Draw a line I and mark a point P on it.

3. Place the pointer of the compasses on A and draw an arc which cuts extended BA at E and AC at F.

4. Without changing the radius on compasses, place its pointer at P and draw an arc which cuts l at Q.

5. Open the length of compasses equal to EF.

6. Without disturbing the radius on compasses, place its pointer at Q and draw an arc which cuts the previous arc at R.

7. Join QR and draw ray QR. It gives ∠RQS which is the required angle whose measure

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