# NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

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**Chapter 14 Practical Geometry Class 6 Maths NCERT Solutions**through which you can score good marks in test and excel in the examinations. NCERT Solutions for Class 6 Maths is one of the best way which will get you deeper understanding of the concepts embedded in the problems and hence helps in obtaining more marks everytime. It will make you understand the topics in most simple manner and grasp it easily to perform better.These

**Chapter 14 NCERT Solutions**are prepared by Studyrankers experts that will help in building a great foundation of concepts and make easy for the students to understand basics. You can study in an organized manner and outperform your classmates.**Exercise 14.1**

1. Draw a circle of radius 3.2 cm.

**Answer**

Steps of construction:

(a) Open the compass for the required radius of 3.2 cm.

(b) Make a point with a sharp pencil where we want the centre of circle to be.

(c) Name it O.

(d) Place the pointer of compasses on O.

(e) Turn the compasses slowly to draw the required circle.

2.Â With the same centre O, draw two circles of radii 4 cm and 2.5 cm.

**Answer**

Steps of construction:

(a) Marks a point â€˜Oâ€™ with a sharp pencil where we want the centre of the circle.

(b) Open the compasses 4 cm.

(c) Place the pointer of the compasses on O.

(d) Turn the compasses slowly to draw the circle.

(e) Again open the compasses 2.5 cm and place the pointer of the compasses on D.

(f) Turn the compasses slowly to draw the second circle.

Hence, it is the required figure.

3.Â Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?

**Answer**

(i) By joining the ends of two diameters, we get a rectangle. By measuring, we find AB = CD = 3 cm, BC = AD = 2 cm i.e., pairs of opposite sides are equal and also âˆ A = âˆ B = âˆ C = âˆ D = 90Â°Â i.e., each angle is of 90Â°.

Hence, it is a rectangle.

(ii) If the diameters are perpendicular to each other, then by joining the ends of two diameters, we get a square. By measuring, we find that AB = BC = CD = DA = 2.5 cm, i.e., all four sides are equal. Also âˆ A = âˆ B = âˆ C = âˆ D = 90Â°, i.e., each angle is of 90Â°.

Hence, it is a square.

4. Draw any circle and mark points A, B and C such that:

(a) A is on the circle.

(b) B is in the interior of the circle.

(c) C is in the exterior of the circle.

**Answer**

(i) Mark a point â€˜Oâ€™ with sharp pencil where we want centre of the circle.

(ii) Place the pointer of the compasses at â€˜Oâ€™. Then move the compasses slowly to draw a circle.

(a) Point A is on the circle.

(b) Point B is in interior of the circle.

(c) Point C is in the exterior of the circle.

5. Let A, B be the centres of two circles of equal radii; draw them so that each one of them passes through the centre of the other. Let them intersect at C and D. Examine whetherÂ Â are at right angles.

**Answer**

Draw two circles of equal radii taking A and B as their centre such that one of them passes through the centre of the other. They intersect at C and D. Join AB and CD.

Yes, AB and CD intersect at right angle as âˆ COB is 90Â°.

**Exercise 14.2**

1.Â Draw a line segment of length 7.3 cm, using a ruler.

**Answer**

Steps of construction:

(i) Place the zero mark of the ruler at a point A.

(ii) Mark a point B at a distance of 7.3 cm from A.

(iii) Join AB.

Hence,Â is the required line segment of length 7.3 cm.

2. Construct a line segment of length 5.6 cm using ruler and compasses.

**Answer**

(i) Draw a line 'l'. l Mark a point A on this line.

(ii) Place the compasses pointer on zero mark of the ruler. Open it to place the pencil point up to 5.6 cm mark.

(iii) Without changing the opening of the compasses. Place the pointer on A and cut an arc 'l'Â at B.Â Â is the required line segment of length 5.6 cm.

3. ConstructÂ Â of length 7.8 cm. From this, cut offÂ of length 4.7 cm. MeasureÂ .

**Answer**

(i) Place the zero mark of the ruler at A.

(ii) Mark a point B at a distance 7.8 cm from A.

(iii) Again, mark a point C at a distance 4.7 from A.

Hence, by measuringÂ Â , we find that BC = 3.1 cm.

4. GivenÂ of length 3.9 cm, constructÂ Â such that the length ofÂ Â is twice that ofÂ Â . Verify by measurement.

(Hint: constructÂ Â such that length ofÂ Â = length ofÂ Â then cut offÂ Â such thatÂ Â also has the length ofÂ .

**Answer**

(i) Draw a line 'l'.

(ii) ConstructÂ Â such that length ofÂ Â = length ofÂ

(iii) Then cut ofÂ Â such thatÂ Â also has the length ofÂ Â .

(iv) Thus the length ofÂ Â and the length ofÂ Â added together make twice the length ofÂ .

Verification: Hence, by measurement we find that PQ = 7.8 cm

= 3.9 cm + 3.9 cm

=Â Â +Â = 2 Ã—Â .

5. GivenÂ Â of length 7.3 cm andÂ Â of length 3.4 cm, construct a line segmentÂ Â such that the length ofÂ Â is equal to the difference between the lengths ofÂ Â andÂ .Â Verify by measurement.

**Answer**

(i) Draw a line 'l' and take a point X on it.

(ii) ConstructÂ Â such that lengthÂ Â = length ofÂ Â = 7.3 cm.

(iii) Then cut offÂ Â = length ofÂ Â = 3.4 cm.

(iv) Thus the length ofÂ Â = length ofÂ Â â€“ length ofÂ .

Verification:

Hence, by measurement we find that length ofÂ Â

= 3.9 cm

= 73. cm â€“ 3.4 cm

=Â Â -Â .

**Exercise 14.3**

1.Â Draw any line segmentÂ .Â Without measuringÂ Â , construct a copy ofÂ Â .

**Answer**

(i) GivenÂ Â whose length is not known.

(ii) Fix the compasses pointer on P and the pencil end on Q. The opening of the instrument now gives the length ofÂ .

(iii) Draw any line 'l'. Choose a point A on 'l'. Without changing the compasses setting, place the pointer on A.

(iv) Draw an arc that cuts 'l' at a point, say B.

Hence,Â is the copy ofÂ .

2.Â Given some line segmentÂ Â Â whose length you do not know, constructÂ Â such that the length ofÂ Â is twice that ofÂ .

**Answer**

(i) GivenÂ Â whose length is not known.

(ii) Fix the compasses pointer on A and the pencil end on B. The opening of the instrument now gives the length ofÂ .

(iii) Draw any line 'l'. Choose a point P on 'l'. Without changing the compasses setting, place the pointer on Q.

(iv) Draw an arc that cuts 'l' at a point R.

(v) Now place the pointer on R and without changing the compasses setting, draw another arc that cuts 'l' at a point Q.

Hence,Â Â is the required line segment whose length is twice that of AB.

**Exercise 14.4**

1.Â Draw any line segmentÂ . Mark any point M on it. Through M, draw a perpendicular toÂ Â . (use ruler and compasses)

**Answer**

(i) Draw a line segmentÂ Â and mark a point M on it.

(ii) Taking M as centre and a convenient radius, construct an arc intersecting the line segmentÂ Â at points X and Y respectively.

(iii) By taking centres as X and Y and radius greater than XM, construct two arcs such that they intersect each other at point D.

(iv) Join DM. NowÂ Â is perpendicular toÂ .

2.Â Draw any line segmentÂ Â .Â Take any point R not on it. Through R, draw a perpendicular toÂ .Â (Use ruler and set-square)

**Answer**

(i) Draw a given line segmentÂ Â and mark a point R outside the line segmentÂ .

(ii)Â Place a set square onÂ Â such that one of its right angles arm aligns alongÂ .

(iii) Now, place the ruler along the edge opposite to right angle of set square.

(v) Draw a line along this edge of set square which passes through point R. Now, it is the required line perpendicular toÂ .

3. Draw a line l and a point X on it. Through X, draw a line segmentÂ Â perpendicular to l. Now draw a perpendicular toÂ Â to Y. (use ruler and compasses)

**Answer**

(i) Draw a line l and mark a point X on it.

(ii) By taking X as centre and with a convenient radius, draw an arc intersecting the line l at points A and B respectively.

(iii) With A and B as centres and a radius more than AX, construct two arcs such that they intersect each other at point Y.

(iv) Join XY. HereÂ Â is perpendicular to l.

Similarly, by taking C and D as centres and radius more than CY, construct two arcs intersecting at point Z. Join ZY. The lineÂ Â is perpendicular toÂ Â at Y.

**Exercise 14.5**

1.Â DrawÂ Â of length 7.3 cm and find its axis of symmetry.

**Answer**

(i) Draw a line AB of length 7.3 cm.Â

(ii) Take a point O as the mid-point of AB.

(iii) Draw OP perpendicular to ABÂ

Thus, OP is a line of symmetry.

2. Draw a line segment of length 9.5 cm and construct its perpendicular bisector.

**Answer**

(i) Draw a line segment AB = 9.5 cm.Â

(ii) With A as centre and radius more than half of AB, draw an arc on both sides of AB.

(iii) With B as centre and the same radius as in step 2, draw an arc intersecting the arc drawn in step 2, at C and D.Â

(iv) Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

3.Â Draw the perpendicular bisector ofÂ Â whose length is 10.3 cm.

(a) Take any point P on the bisector drawn. Examine whether PX = PY.

(b) If M is the mid-point ofÂ Â , what can you say about the lengths MX and XY?

**Answer**

(i) Draw a line segmentÂ Â = 10.3 cm

(ii) Taking X and Y as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.

(iii) Join CD. Then CD is the required perpendicular bisector ofÂ .

Now:

(a) Take any point P on the bisector drawn. With the help of divider we can check thatÂ Â =Â .

(b) If M is the mid-point ofÂ Â thenÂ = 1/2Â .

4. Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.

**Answer**

(i) Draw a line segment AB = 12.8 cm

(ii) Draw the perpendicular bisector ofÂ Â which cuts it at C. Thus, C is the midpoint ofÂ .

(iii) Draw the perpendicular bisector ofÂ Â which cuts it at D. Thus D is the midpoint of.

(iv) Again, draw the perpendicular bisector ofÂ Â which cuts it at E. Thus, E is the mid-point ofÂ

(v) Now, point C, D and E divide the line segmentÂ Â in the four equal parts.

(vi) By actual measurement, we find thatÂ Â =Â Â =Â Â =Â Â = 3.2 cm.

5. WithÂ Â of length 6.1 cm as diameter, draw a circle.

**Answer**

(i) Draw a line segmentÂ Â = 6.1 cm.

(ii) Draw the perpendicular bisector of PQ which cuts, it at O. Thus O is the mid-point ofÂ .

(iii) Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segmentÂ .

6.Â Draw a circle with centre C and radius 3.4 cm. Draw any chordÂ Â . Construct the perpendicular bisectorÂ Â and examine if it passes through C.

**Answer**

(i) Mark any point C on the sheet

(ii) Adjust the compasses up to 3.4 cm and by putting the pointer of compasses at point C, turn compasses slowly to draw the circle. This is the required circle of 3.4 cm radius.

(iii) Mark any chordÂ Â in the circle.

(iv) Now, taking A and B as centres, draw arcs on both sides ofÂ Â . Let these intersect each other at points D and E.

(v) Join DE. Now DE is the perpendicular bisector of AB.

IfÂ Â is extended, it will pass through point C.

7.Â Repeat Question 6, ifÂ Â happens to be a diameter.

**Answer**

(i) Mark any point C on the sheet.

(ii) Adjust the compasses up to 3.4 cm and by putting the pointer of compasses at point C, Turn the compasses slowly to draw the circle. This is the required circle of 3.4 cm.

(iii) Now mark any diameterÂ Â in the circle.

(iv) Now taking A and B as centres, draw arcs on both sides ofÂ Â with radius more thanÂ Â .Â Let these intersect each other at points D and E.

(v) Join DE, which is perpendicular bisector of AB.

Now, we may observe thatÂ Â is passing through the centre C of the circle.

8. Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?

**Answer**

(i) Mark any point C on the sheet. Now adjust the compasses up to 4 cm and by placing the pointer of compasses at point C, turn the compasses slowly to draw the circle. This is the required circle of 4 cm radius

(ii) Take any two chordsÂ Â andÂ Â in the circle.

(iii) By taking A and B as centres and radius more than half ofÂ Â draw arcs on both sides of AB. The arcs are intersecting each other at point E and F. Join EF which is perpendicular bisector of AB.

(iv) Again take C and D as centres and radius more than half ofÂ Â draw arcs on both sides of CD such that they are intersecting each other at points G, H. Join GH which is perpendicular bisector of CD.

We may observe that when EF and GH are extended they meet at the point O, which is the centre of circle.

9.Â Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors ofÂ Â andÂ . Let them meet at P. Is PA = PB?

**Answer**

(i) Draw any angle with vertex as O.

(ii) By taking O as centre and with convenient radius, draw arcs on both rays of this angle. Let these points are A and B

(iii) Now take O and A as centres and with radius more than half of OA, draw arcs on both sides of OA. Let these intersects at points C and D respectively. Join CD

(iv) Similarly we may findÂ Â Â which is perpendicular bisector ofÂ Â . These perpendicular bisectorsÂ Â andÂ Â intersects each other at point P. Now measure PA and PB. They are equal in length.

**Exercise 14.6**

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

**Answer**

(i) Draw a line l and mark a point O on it.

(ii) Place the pointer of the compasses at O and draw an arc of any radius which intersects the line l at A.

(iii) Taking same radius, with centre A, cut the previous arc at B.

(iv) Join OB, then âˆ BOA = 60 .

(v) Taking same radius, with centre B, cut the previous arc at C.

(vi) Draw bisector of âˆ BOC. The angle is of 90 . Mark it at D. Thus, âˆ DOA = 90Â°Â

(vii) DrawÂ Â as bisector of âˆ DOB. Thus, âˆ POA = 75Â°.

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

**Answer**

The steps are followed to construct an angle of measure 147Â° and its bisector

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

(ii) Mark a point A at an angle of measure 147Â°. Join OA. Now OA is the required ray making 147Â° with line l

(iii) By taking point O as centre, draw an arc of convenient radius. Let this intersect both rays of angle 1470 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 147Â°. Let these intersect each other at point C. Join OC. OC is the required bisector of 147Â° angle.

3. Draw a right angle and construct its bisector.

**Answer**

(i) Draw a line PQ and take a point O on it.

(ii) Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.

(iii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.

(iv) Join OC. Thus, âˆ COQ is the required right angle.

(v) Taking B and E as centre and radius more than half of BE, draw two arcs which intersect each other at the point D.

(vi) Join OD. Thus,Â Â Â is the required bisector of âˆ COQ.

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

**Answer**

The steps are followed to construct an angle of measure 153Â°Â and its bisector

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

(ii) Mark a point A at the measure of angle 153Â°. Join OA. Now OA is the required ray making 153Â°Â with line l

(iii) Draw an arc of convenient radius by taking point O as centre. Let this intersects both rays of angle 153Â°Â 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 angle of 153Â°. Let these intersect each other at C. Join OC

(v) Let OC intersect 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, OE are the rays dividing 153Â°Â angle into four equal parts.

5. Construct with ruler and compasses, angles of following measures:Â

(a) 60Â°Â

(b) 30Â°Â

(c) 90Â°Â

(d) 120Â°

(e) 45Â°Â

(f) 135Â°

**Answer**

(a) 60Â°

(i) Draw a line segment OA.

(ii) With O as centre and any suitable radius draw an arc cutting OA at point D.

(iii) With D as centre and the same radius draw another arc cutting the previous arc at E.

(iv) Join OE and produce it to B. ZAOB = 60Â°.

(b) 30Â°Â

(i) Draw a line segment OA.

(ii) With O as centre and any suitable radius draw an arc cutting OA at point C.

(iii) With C as centre and same radius as before draw another arc cutting the previous arc at E.

(iv) Join OE and produce it to B. ZAOB = 60Â°.

(v) Bisect ZAOB. Thus, ZAOM = ZMOB = 30Â°.

(c) 90Â°Â

(i) Draw a line segment OA.

(ii) With O as centre and any suitable radius draw an arc cutting OA at point C.Â

(iii) With Cas centre and same radius cut off the arc P and then with P as centre and the same radius cut off the arc again at Q.

(iv) With Pand Q as centres and any suitable radius or even the same radius draw arcs cutting each other at R.Â

(v) Join OR and produce it to B. Then ZAOB = 90Â°.

(d) 120Â°

(i) Draw a line segment OA.

(ii) With O as centre and any suitable radius draw an arc cutting OA at point M.

(iii) With M as centre and same radius draw an arc which cuts the arc at Nand then with Nas centre and the same radius cut off the arc again at Q.

(iv) Join OQ and produce it to B. Thus, âˆ AOB = 120Â°.

(e) 45Â°Â

(i) Draw a line segment OA.Â

(ii) With O as centre and any suitable radius draw an arc cutting OA at point C.Â

(iii) With C as centre and same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q.Â

(iv) With P and Q as centres and any suitable radius (more than half of PQ) or even the same radius draw arc cutting each other at R.Â

(v) Join OR and produce it to B. Then âˆ AOB = 90Â°.Â

(vi) Bisect âˆ AOB. 7. OD is the bisector of âˆ AOB.

âˆ´ âˆ BOD = âˆ DOA = 45Â°.

(f) 135Â°

(i) Draw a line segment OA.Â

(ii) With O as centre and any suitable radius draw an arc cutting OA at point C.Â

(iii) With C as centre and the same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q and then with Q as centre and the same radius cut off the arc again at R.

(iv) With Q and R as centres and radius more than half of RQ draw arcs cutting each other at L.

(v) Join OL and produce it to B. Then âˆ AOB = 150Â°.Â

(vi) Take a point M on the arc where OL intersects the arc.Â

(vii) With M and Q as centres and radius more than half of MQ draw arcs cutting each other at N.Â

(viii) Join ON and produce it to E. Thus, âˆ AOE = 135Â°.

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

**Answer**

(i) Draw a line segment OA.Â

(ii) With O as centre and any suitable radius draw an arc cutting OA at C.Â

(iii) With C as centre and same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q.

(iv) With P and Q as centres and radius more than half of PQ draw arcs cutting each other at R.Â

(v) Join OR and produce it to B. Then âˆ AOB = 90Â°.Â

(vi) Bisect âˆ AOB.

(vii) OD is the bisector of âˆ AOB.

âˆ BOD = ZDOA = 45Â°.

(viii) Again draw OE bisector of ZDOA.

Thus, angle âˆ DOE = âˆ EOA = 22.5Â°.

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

**Answer**

The steps are followed to construct an angle of measure 135Â°Â and its bisector.

(i) By using a protractor âˆ POQ of 135Â°Â measure may be formed on a line l

(ii) Draw an arc of convenient radius by taking O as centre. Let this intersect both rays of angle 135Â°Â at points A and B respectively.

(iii) Take A and B as centres, draw arcs of radius more than 1/2 AB in the interior of angle of 135Â°.

Let these intersect each other at C. Join OC.

OC is the required bisector of 135Â°Â angle.

**Go Back To NCERT Solutions for Class 6 Maths**

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

NCERT Solutions will help you in improving the marks in the examinations and have edge over your mates. In this chapter, we will learn about various geometric tools and circles.

â€¢ The tools in our geometry box are:

(i) Ruler

(ii) Compass

(iii) Divider

(iv) Set squares

(v) Protractor

You can easily find exercisewise NCERT Solutions that can be useful in getting command over the concepts below via the links.

- Exercise 14.1 Chapter 14 Class 6 Maths NCERT Solutions
- Exercise 14.2 Chapter 14 Class 6 Maths NCERT Solutions
- Exercise 14.3 Chapter 14 Class 6 Maths NCERT Solutions
- Exercise 14.4 Chapter 14 Class 6 Maths NCERT Solutions
- Exercise 14.5 Chapter 14 Class 6 Maths NCERT Solutions
- Exercise 14.6 Chapter 14 Class 6 Maths NCERT Solutions

Class 6 Maths NCERT Solutions will help an individual to increase concentration and you can solve questions of supplementary books easily. It help students cope with the pressure of the large examination syllabus.

### NCERT Solutions for Class 6 Maths Chapters:

**FAQ onÂ ChapterÂ**

**14 Practical Geometry**

#### How many exercises in Chapter 14 Practical Geometry?

Chapter 14 contains total 6 exercises which help the students in learning complex topics and problems in an easy way. These NCERT Solutions are prepared as per the accordance of latest CBSE guidelines so you can score maximum marks.

#### What is a Ruler?

A ruler is a flat and straight-edged strip, whose one side is graduated into centimetres and the other into inches. A ruler is commonly called a scale. It is the most essential tool in geometry. It is used in all constructions.

#### What are Set squares?

The two triangular tools in the geometry box are called set squares. One of the set square is an isosceles triangle with two angles measuring 45Â° each. The other set square is a scalene triangle with two angles measuring 30 Â°and 60Â° each. The two perpendicular sides of either set square are graduated into centimetres.

#### What is a Protractor?

A semi-circular tool with degrees marked is called a protractor. The centre of the semicircle is called the midpoint of the protractor. This point helps as a reference point for the protractor. The horizontal line is called the base line or the straight edge of the protractor.