**According to the latest update on the CBSE Syllabus 2023-24, this chapter has been removed.*

**NCERT Solutions or Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6** covers the answers to the questions present in this exercise. Angles and their construction, the bisector of an angle and angles of special measures, along with illustrative examples, are covered under Exercise 14.6 of Chapter 14. Students are provided with exercise-wise NCERT Solutions to get better acquainted with the concepts, which are important from the exam point of view.

## NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6

### Access NCERT Solutions for Class 6 Chapter 14: Practical Geometry Exercise 14.6

**1. Draw âˆ POQ of measure 75Â° and find its line of symmetry.**

**Solutions:**

The following steps are used to construct an angle of 75^{0} and its line of symmetry:

(i) Draw a line l and mark two points, O and Q, on it. Draw an arc of convenient radius while taking centre as O. Let this intersect line l at R.

(ii) Taking R as the centre and with the same radius as before, draw an arc such that it intersects the previously drawn arc at S.

(iii) By taking the same radius as before and S as the centre, draw an arc intersecting the arc at point T, as shown in the figure.

(iv) Take S and T as the centre, and draw an arc of the same radius such that they intersect each other at U.

(v) Join OU. Let it intersect the arc at V. Now, take S and V as centres draw arcs with a radius of more than 1/2 SV. Let these intersect each other at P, then Join OP. Now, OP is the ray making 75^{0} with the line l.

(vi) Let this ray intersect the major arc at point W. By taking R and W as centres, draw arcs with a radius of more than 1/2 RW in the interior angle of 75^{0}. Let these intersect each other at point X. Join OX.

OX is the line of symmetry for the âˆ POQ = 75^{0}.

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

**Solutions:**

The following steps are used to construct an angle of measure 147^{0} and its bisector:

(i) Draw a line l and mark point O on it. Place the centre of the protractor at point O and the zero edge along line l.

(ii) Mark point A at an angle of measure 147^{0} and join OA. Now, OA is the required ray making 147^{0} with line l.

(iii) By taking point O as the centre, draw an arc of the convenient radius. Let this intersect both rays of angle 147^{0} at points A and B.

(iv) By taking A and B as centres, draw arcs of radius more than 1/2 AB in the interior angle of 147^{0}. Let these intersect each other at point C. Join OC.

OC is the required bisector of 147^{0} angle.

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**3. Draw a right angle and construct its bisector.**

**Solutions:**

The following steps are used to construct a right angle and its bisector:

(i) Draw a line l and mark a point P on it. Draw an arc of the convenient radius by taking point P as the centre. Let this intersect line l at R.

(ii) Draw an arc by taking R as the centre and with the same radius as before such that it intersects the previously drawn arc at S.

(iii) Take S as the centre and with the same radius as before, draw an arc intersecting the arc at T, as shown in the figure.

(iv) By taking S and T as centres, draw arcs of the same radius such that they intersect each other at U.

(v) Join PU. PU is the required ray making a right angle with the line l. Let this intersect the major arc at point V.

(vi) Now, take R and V as centres, and draw arcs with a radius of more than 1/2 RV to intersect each other at point W. Join PW.

PW is the required bisector of this right angle.

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

**Solutions:**

The following steps are used to construct an angle of measure 153^{0} and its bisector:

(i) Draw a line l and mark a point O on it. Place the centre of the protractor at point O and the zero edge along line l.

(ii) Mark a point A at the measure of angle 153^{0}, and join OA. Now, OA is the required ray making 153^{0} with line l.

(iii) Draw an arc of the convenient radius by taking point O as the centre. Let this intersect both rays of angle 153^{0} at points A and B.

(iv) Take A and B as centres and draw arcs of radius more than 1/2 AB in the interior of an angle of 153^{0}. Let these intersect each other at C. Join OC.

(v) Let OC intersect the major arc at point D. Draw arcs of radius more than 1/2 AD with A and D as centres and also D and B as centres. Let these are intersecting each other at points E and F, respectively. Now join OE and OF.

OF, OC and OE are the rays dividing the 153^{0} angle into four equal parts.

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

**(a) 60Â° **

**(b) 30Â° **

**(c) 90Â° **

**(d) 120Â° **

**(e) 45Â° **

**(f) 135Â°**

**Solutions:**

(a) 60^{0}

The following steps are followed to construct an angle of 60^{0}:

(i) Draw a line l and mark a point P on it. Take P as the centre, and with a convenient radius, draw an arc of a circle such that it intersects the line l at Q.

(ii) Take Q as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.

(iii) Join PR. PR is the required ray making 60^{0} with the line l.

(b) 30^{0}

The following steps are followed to construct an angle of 30^{0}:

(i) Draw a line l and mark a point P on it. By taking P as the centre and with a convenient radius, draw an arc of a circle such that it intersects the line l at Q.

(ii) Take Q as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.

(iii) By taking Q and R as centres and with a radius of more than 1/2 RQ draw arcs such that they intersect each other at S. Join PS, which is the required ray making 30^{0} with the line l.

(c) 90^{0}

The following steps are used to construct an angle of measure 90^{0}:

(i) Draw a line l and mark a point P on it. Take P as the centre, and with a convenient radius, draw an arc of a circle such that it intersects the line l at Q.

(ii) Take Q as the centre, and with the same radius as before, draw an arc intersecting the previously drawn arc at R.

(iii) By taking R as the centre and with the same radius as before, draw an arc intersecting the arc at S, as shown in the figure.

(iv) Now, take R and S as the centre, and draw an arc of the same radius to intersect each other at T.

(v) Join PT, which is the required ray, making 90^{0} with line l.

(d) 120^{0}

The following steps are used to construct an angle of measure 120^{0}:

(i) Draw a line l and mark a point P on it. Taking P as the centre and with a convenient radius, draw an arc of the circle such that it intersects the line l at Q.

(ii) By taking Q as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.

(iii) Take R as the centre and with the same radius as before, draw an arc such that it intersects the arc at S, as shown in the figure.

(iv) Join PS, which is the required ray making 120^{0} with the line l.

(e) 45^{0}

The following steps are used to construct an angle of measure 45^{0}:

(i) Draw a line l and mark a point P on it. Take P as the centre, and with a convenient radius, draw an arc of a circle such that it intersects the line l at Q.

(ii) Take Q as the centre, and with the same radius as before, draw an arc intersecting the previously drawn arc at R.

(iii) By taking R as the centre and with the same radius as before, draw an arc such that it intersects the arc at S, as shown in the figure.

(iv) Take R and S as centres and draw arcs of the same radius such that they intersect each other at T.

(v) Join PT. Let this intersect the major arc at point U.

(vi) Now, take Q and U as centres and draw arcs with a radius of more than 1/2 QU to intersect each other at point V. Join PV.

PV is the required ray making 45^{0} with the line l.

(f) 135^{0}

The following steps are used to construct an angle of measure 135^{0}:

(i) Draw a line l and mark a point P on it. Taking P as the centre and with a convenient radius, draw a semicircle which intersects the line l at Q and R, respectively.

(ii) By taking R as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(iii) Taking S as the centre and with the same radius as before, draw an arc such that it intersects the arc at T, as shown in the figure.

(iv) Take S and T as centres, and draw arcs of the same radius to intersect each other at U.

(v) Join PU. Let this intersect the arc at V. Now, take Q and V as centres and with a radius of more than 1/2 QV, draw arcs to intersect each other at W.

(vi) Join PW, which is the required ray making 135^{0} with the line l.

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

**Solutions:**

The following steps are used to construct an angle of measure 45^{0} and its bisector:

(i) Using the protractor, a âˆ POQ of 45^{0} measure may be formed on a line l.

(ii) Draw an arc of the convenient radius with the centre as O. Let this intersect both rays of angle 45^{0} at points A and B.

(iii) Take A and B as centres and draw arcs of radius more than 1/2 AB in the interior of an angle of 45^{0}. Let these intersect each other at C. Join OC.

OC is the required bisector of 45^{0} angle.

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

**Solutions:**

The following steps are used to construct an angle of measure 135^{0} and its bisector:

(i) By using a protractor, a âˆ POQ of 135^{0} measure may be formed on a line l.

(ii) Draw an arc of the convenient radius by taking O as the centre. Let this intersect both rays of angle 135^{0} at points A and B, respectively.

(iii) Take A and B as centres and draw arcs of a radius of more than 1/2 AB in the interior of an angle of 135^{0}. Let these intersect each other at C. Join OC.

OC is the required bisector of 135^{0} angle.

**8. Draw an angle of 70 ^{0}. Make a copy of it using only a straight edge and compasses.**

**Solutions:**

The following steps are used to construct an angle of measure 70^{0} and its copy:

(i) Draw a line l and mark a point O on it. Now, place the centre of the protractor at point O and the zero edge along line l.

(ii) Mark a point A at an angle of measure 70^{0}. Join OA. Now, OA is the ray making 70^{0} with line l. With point O as the centre, draw an arc of a convenient radius in the interior of 70^{0} angle. Let this intersect both rays of angle 70^{0} at points B and C, respectively.

(iii) Draw a line m and mark a point P on it. Again, draw an arc with the same radius as before and P as the centre. Let it cut the line m at point D.

(iv) Adjust the compasses up to the length of BC. With this radius, draw an arc taking D as the centre, which intersects the previously drawn arc at point E.

(v) Join PE. Here, PE is the required ray which makes the same angle of measure 70^{0} with the line m.

**9. Draw an angle of 40 ^{0}. Copy its supplementary angle.**

**Solutions:**

The following steps are used to construct an angle of measure 45^{0} and a copy of its supplementary angle:

(i) Draw a line segment

and mark a point O on it. Place the centre of the protractor at point O and the zero edge along the line segment.

.

(ii) Mark a point A at an angle of measure 40^{0}. Join OA. Here. OA is the required ray, making 40^{0} with

. âˆ POA is the supplementary angle of 40^{0}.

(iii) With point O as the centre, draw an arc of convenient radius in the interior of âˆ POA. Let this intersect both rays of âˆ POA at points B and C, respectively**.**

(iv) Draw a line m and mark a point S on it. Again draw an arc by taking S as the centre with the same radius as used before. Let it cut the line m at point T.

(v) Now, adjust the compasses up to the length of BC. Taking T as the centre, draw an arc with this radius which will intersect the previously drawn arc at point R**.**

(vi) Join RS. Here, RS is the required ray which makes the same angle with the line m as the supplementary of 40^{0} [i.e., 140^{0}].

**Also, explore â€“**

NCERT Solutions for Class 6 Maths Chapter 14

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