# NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

NCERT solutions for class 6 maths chapter 14 Practical Geometry: Constructions of figures or sketches are mandatory for the design of different types of equipment, instruments, buildings, etc. You may have seen in films or in your real-life, architect drawing sketches of the building using a different kind of geometrical instruments. In CBSE NCERT solutions for class 6 maths chapter 14 practical geometry, you will get questions related to constructions of figures using rulers and compass. In this digital world now sketches are designed in computers using different software available but even to draw sketches using the software, you should have an idea in mind to draw those sketches. Important topics are like the construction of the circle of a given radius using a compass, the construction of line segments and angles covered in this chapter. In the last exercise of solutions of NCERT for class 6 maths chapter 14 practical geometry, you will get questions related to measuring of angle using compass and ruler. Also, you will use the set square to construct perpendicular lines. In some questions, you will be drawing lines using the compass. There are many other constructions explained in 6 exercises of CBSE  NCERT solutions for class 6 maths chapter 14 practical geometry. You will get NCERT solutions from class 6 to 12 for science and maths by clicking on the above link.

Exercise 14.1

Exercise 14.2

Exercise 14.3

Exercise 14.4

Exercise 14.5

Exercise 14.6

The main topics of NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry along with the introduction are given below.

14.1 Introduction

14.2 The Circle

14.2.1 Construction of a circle when its radius is known

14.3 A Line Segment

14.3.1 Construction of a line segment of a given length

14.3.2 Constructing a copy of a given line segment

14.4 Perpendiculars

14.4.1 Perpendicular to a line through a point on it

14.4.2 Perpendicular to a line through a point not on it

14.4.3 The perpendicular bisector of a line segment

14.5 Angles

14.5.1 Constructing an angle of a given measure

14.5.2 Constructing a copy of an angle of unknown measure

14.5.3 Bisector of an angle

14.5.4 Angles of special measures

## NCERT solutions for class 6 maths chapter 14 practical geometry-Exercise: 14.1 Question:1 Draw a circle of radius 3.2 cm.

To draw a circle of radius 3.2 cm follow the steps given below:

(i) Take compasses for the required radius of 3.2 cm.

(ii) Mark point O where you want the centre of the circle.

(iii) After placing the compass on O, rotate the compasses slowly to make the circle.

To draw the required circle follow the steps as follows:

(i) First, Set the compass for radius 4 cm

(ii) Mark point O which will be the centre for the circles.

(iiI) Place the pointer of compasses on O.

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

Repeat above 4 steps with the compass set at 2.5 cm

By joining the ends of the two diameters, we get a rectangle.

Therefore, by measuring, we find that $AB=CD$ and $BC=AD$, i.e, pairs of opposite sides are equal and also $\angle A = \angle B = \angle C = \angle D = 90^{\circ}$.

and each angle is fo $90^{\circ}.$

Hence, it is a rectangle:

If the diameters are perpendicular to each other, then by joining the ends of two diameters, we get a square.

Then, by measuring the length of the side we find that $AB=BC=CD=DA$. that is all sides are of same length also, $\angle A = \angle B = \angle C = \angle D = 90^{\circ}$, that is each angle is $90^{\circ}.$

Hence, it is a square:

(a) A is on the circle.            (b) B is in the interior of the circle.
(c) C is in the exterior of the circle.

To draw the circle with the given conditions:

First, mark point 'O' as the centre and place the pointer of the compasses at 'O' then turn the compasses slowly to complete the circle.

Now, mark point A on the circle, mark the point 'B' in the interior of the circle, also point 'C' exterior to the circle.

First, draw two circles of equal radii taking A and B as their centre and they intersect at C and D. Then, join AB and CD.

From the figure, AB and CD intersect at right angle with

Solutions of NCERT for class 6 maths chapter 14 practical geometry-Exercise: 14.2

Let the line segment be 'AB':

Then, follows these steps:

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

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

(iii) Now, join AB.

Then, AB is the required line segment of the length of 7.3 cm.

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

(ii) Place the compasses pointer on the 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 at B on line l.

$\bar{AB}$ is the required line segment of length 5.6 cm.

The steps of constructions:

(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.

By measuring $\bar{BC}$ , we find that BC = 3.1 cm

(Hint: Construct $\overline{PX}$ such that length of $\overline{PX}$ = length of $\overline{AB}$ ; then cut off $\overline{XQ}$ such that $\overline{XQ}$ also has the length of $\overline{AB}$ .)

The steps of constructions are following:

(i) Draw a line ′l′.

(ii) Construct $\bar{PX}$ such that length of $\bar{PX}$ = length of $\bar{AB}$

(iii) Then cut of $\bar{XQ}$ such that $\bar{XQ}$ also has the length of $\bar{AB}$ .

(iv) Thus the length of $\bar{PX}$  and the length of $\bar{XQ}$ added together make twice the length of $\bar{AB}$.

Verification:

By measurement we find that PQ = 7.8 cm

= 3.9 cm + 3.9 cm = $\bar{AB} + \bar{AB}$   = $2\times\bar{AB} .$

The steps of construction are as follows:

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

(ii) Construct $\bar{XZ}$ such that length $\bar{XZ}$  = length of $\bar{AB}$ = 7.3 cm

(iii) Then cut off $\bar{ZY}$ = length of $\bar{CD}$ = 3.4 cm

(iv) Thus the length of $\bar{XY}$ = length of $\bar{AB}$ – length of $\bar{CD}$

Verification:

By measurement we find that length of $\bar{XY}$ = 3.9 cm

$= 7.3 cm - 3.4cm = \bar{AB} - \bar{CD}$

CBSE NCERT solutions for class 6 maths chapter 14 practical geometry-Exercise: 14.3

The steps of constructions are the following:

(i) Given $\bar{PQ}$ 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 of $\bar{PQ}$.

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

Draw an arc that cuts  at $'l'$ a point, say B. Now $\bar{AB}$ is a copy of $\bar{PQ}$ .

The steps of construction are followings:

(i) Given $\bar{AB}$ 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 of$\bar{AB}$.

(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.

Thus $\bar{PQ}$ is the required line segment whose length is twice that of AB.

NCERT solutions for class 6 maths chapter 14 practical geometry-Exercise: 14.4

The steps of constructions are the following:

(i) With M as the centre and a convenient radius, draw an arc intersecting the line AB at two points C and B.

(ii) With C and D as centres and a radius greater than MC, draw two arcs, which cut each other at P.

(iii) Join PM. Then PM is perpendicular to AB through point M.

The steps of constructions are:

(i) Place a set-square on $PQ$ such that one arm of its right angle aligns along $PQ$ .

(ii) Place a ruler along the edge opposite to the right angle of the set-square.

(iii) Hold the ruler fixed. Slide the set square along the ruler till the point R touches the other arm of the set square.

(iv) Join RM along the edge through R meeting $PQ$ at M. Then RM ⊥ PQ.

The steps of construction are as follows:

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

(ii) With X as the centre and a convenient radius, draw an arc intersecting the line $'l'$ at two points A and B.

(iii) With A and B as centres and a radius greater than XA, draw two arcs, which cut each other at C.

(iv) Join AC and produce it to Y. Then XY is perpendicular to $'l'$.

(v) With Y as the centre and a convenient radius, draw an arc intersecting XY at two points C and D.

(vi) With C and D as centres and radius greater than YD, draw two arcs which cut each other at F.

(vii) Join YF, then YF is perpendicular to XY at Y.

## Solutions of NCERT for class 6 maths chapter 14 practical geometry-Exercise: 14.5

The axis of symmetry of the line segment $\bar{AB }$ will be the perpendicular bisector of $\bar{AB }$. So, draw the perpendicular bisector of AB.

The steps of constructions are as follows:

(i) Draw a line segment $\bar{AB }$ = 7.3 cm

(ii) Taking A and B 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 axis of symmetry of the line segment AB.

The steps of constructions are:

(i) Draw a line segment $\bar{AB}$ = 9.5 cm

(ii) Taking A and B 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 perpendicular bisector of $\bar{AB}$.

Follow the steps to draw the perpendicular bisector of a line XY.

a) PX= PY

b)MX=MY

The steps of constructions are:

(i) Draw a line segment AB = 12.8 cm

(ii) Draw the perpendicular bisector of $\bar{AB}$ which cuts it at C. Thus, C is the mid-point of $\bar{AB}$ .

(iii) Draw the perpendicular bisector of $\bar{AC}$ which cuts it at D. Thus D is the mid-point of .

(iv) Again, draw the perpendicular bisector of $\bar{CB}$ which cuts it at E. Thus, E is the mid-point of $\bar{CB}$.

(v) Now, point C, D, and E divide the line segment $\bar{AB}$ in the four equal parts.

(vi) By actual measurement, we find that

$\bar{AD} = \bar{DC} = \bar{CE} = \bar{EB} = 3.2cm$

The steps of constructions are:

(i) Draw a line segment $\bar{PQ}$ = 6.1 cm.

(ii) Draw the perpendicular bisector of PQ which cuts, it at O. Thus O is the mid-point of $\bar{PQ}$ .

Taking O as the centre and OP or OQ as radius draw a circle where the diameter is the line segment $\bar{PQ}$.

The steps of constructions are:

(i) Draw a circle with centre C and radius 3.4 cm.

(ii) Draw any chord $\bar{AB}$.

(iii) Taking A and B as centres and radius more than half of $\bar{AB}$, draw two arcs which cut each other at P and Q.

(iv) Join PQ. Then PQ is the perpendicular bisector of $\bar{AB}$.

This perpendicular bisector of $\bar{AB}$ passes through the centre C of the circle.

The steps of constructions are:

(i) Draw a circle with centre C and radius 3.4 cm.

(ii) Draw its diameter $\bar{AB}$

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

(iv) Join PQ. Then PQ is the perpendicular bisector of $\bar{AB}$

We observe that this perpendicular bisector of $\bar{AB}$ passes through the centre C of the circle.

The steps of constructions are:

(i) Draw the circle with O and radius 4 cm.

(ii) Draw any two chords $\bar{AB}$ and $\bar{CD }$in this circle.

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

(iv) Join EF. Thus EF is the perpendicular bisector of chord $\bar{CD }$.

(v) Similarly draw GH the perpendicular bisector of chord $\bar{CD }$.

These two perpendicular bisectors meet at O, the centre of the circle.

The steps of constructions are:

(i) Draw any angle with vertex O.

(ii) Take a point A on one of its arms and B on another such that

(iii) Draw perpendicular bisector of $\bar{OA}$ and $\bar{OB}$.

(iv) Let them meet at P. Join PA and PB.

With the help of divider, we obtained  that $\bar{PA} = \bar{PB}.$

## NCERT solutions for class 6 maths chapter 14 practical geometry topic 14.5.4 angles of special measures

Question: How will you construct a 15° angle?

First, we make $30^{\circ},$ and then its bisector.

The steps of constructions are:

1. Draw a ray OA.

2. Taking O as the center and any radius of your own choice, draw an arc cutting OA at B.

3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.

4. Draw the ray OD passing through the C.

Thus, $\angle AOD = 60^{\circ}$

Now, we draw bisector of $\angle AOD$

5. Taking C and D as the center, with radius more than $\frac{1}{2}CD$, draw arcs intersecting at E.

6. Join OE.

Thus, $\angle AOE = 30^{\circ}$

Now, we draw the bisector of $\angle AOD$

7. Mark point P where the ray OE intersects the arc.

8. Taking P and B as center, with having the radius more than $\frac{1}{2}PB$, draw arcs intersecting at F.

9. Join OF

Thus, $\angle AOF = 15^{\circ}.$

Question: How will you construct a 150° angle?

Take a line segment, say, AB.

Choose a point C on it.

With center C, and radius BC, draw an arc.

With center B and radius BC, cut the previous arc at say DD.

$(\angle DCB=60^{\circ}\ because\ we\ just\ made\ CD=BC=BD )$

With center D, and radius BC draw an arc.

With center B and radius BC, cut this arc at, say, EE.

Then EC is the bisector of ∠BCD, and

Hence, $\angle BCE=30^{\circ}$

Then, $\angle ACE=180^{\circ} - \angle BCE=150^{\circ}$

Question: How will you construct a 45° angle?

The steps of constructions:

1. Draw a ray OA.

2. Taking O as center and any radius, draw an arc cutting OA at B.

3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.

4. With C as the center and the same radius, draw an arc cutting the arc at D.

\

5. With C and D as centers and radius more than $\frac{1}{2}CD$, draw two arcs intersecting at P.

6. Join OP.

Thus, $\angle AOP = 90^{\circ}$

Now, we draw the bisector of $\angle AOP$

7. Let OP intersects the original arc at point Q.

8. Now, taking B and Q as centers, and radius greater than $\frac{1}{2}BQ,$ draw two arcs intersecting at R.

9. Join OR.

Thus, $\angle AOR = 45^{\circ}$

CBSE NCERT solutions for class 6 maths chapter 14 practical geometry-Exercise: 14.6

Here, we will draw $75^{\circ}$ using a protractor.

1. Draw a ray OA.

2. Place the centre of the protractor on point O, and coincide line OA and Protractor line

3. Mark point B on 75 degrees.

4. Join OB

Therefore $\angle AOB = 75^{\circ}$

Now, we need to find its line of symmetry

that is, we need to find its bisector.

1. Mark points C and D where the arc intersects OA and OB

2. Now, taking C and D as centres and with the radius more than $\frac{1}{2}CD$, draw arcs to intersects each other.

3. Let them intersects at point E.

4. Join O and E.

Therefore OE is the line of symmetry of $\angle BOA = 75^{\circ}$

The steps of constructions are:

1. Draw a line  OA.
2. Using protractor and centre 'O draw an angle  AOB =147°.
3. Now taking 'O' as the centre and any radius draws an arc that intersects  'OA' and 'OB'  at p  and q.
4. Now take p and q as centres and radius more than half of PQ, draw arcs.
5. Both the arcs intersect at 'R'
6. Join 'OR' and produce it.
7. 'OR' is the required bisector of angle AOB.

The steps of construction:

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

(b) Taking O as the centre and convenient radius, draw an arc that intersects PQ at A and B.

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

(d) Join OC. Thus, ∠COQ is the required right angle.

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

(f) Join OD. Thus, the OD bar is the required bisector of ∠COQ.

The steps of constructions are:

(a) Draw a ray OA.

(b) At O, with the help of a protractor, construct ∠AOB = 153 degrees.

(c) Draw OC as the bisector of ∠AOB.

(d) Again, draw OD as bisector of ∠AOC.

(e) Again, draw OE as bisector of,∠BOC.

(f) Thus, OC, OD, and OE divide ∠AOB into four equal parts.

60o

The steps of constructions are:

1. Draw a ray OA

2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at P.

3. Taking P as the centre and the same radius, cut the previous arc at Q. Join OQ. Thus,∠BOA is the required angle of 60°

30o

The steps of constructions are:

1. Draw a ray OA.

2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at P.

3. Taking P as the centre and the same radius, cut the previous arc at Q. Join OQ. Thus, ∠BOA is the required angle of 60°.

4. Put the pointer on P and mark an arc.

5. Put the pointer on Q and with the same radius, cut the previous arc at C. Thus, ∠COA is required angle of 30°

90o

The steps of constructions are:

1. Draw a ray OA

2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at X.

3. Taking X as the centre and the same radius, cut the previous arc at Y. Taking Y as the centre and the same radius, draw another arc intersecting the same arc at Z.

4. Taking Y and Z as centres and the same radius, draw two arcs intersecting each other at S.

5. Join the OS. Thus, ∠SOA is a required angle of 90°

120o

The steps of constructions are:

1. Draw a ray OA

2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at P.

3. Taking P as the centre and same radius, cut previous arc at Q. Taking Q as the centre and the same radius cut the arc at S. Join OS. Thus, ∠AOS is the required angle of 120°.

45o

The steps of constructions are:

1. Draw a ray OA

2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at X.

3. Taking X as the centre and the same radius, cut the previous arc at Y. Taking Y as the centre and the same radius, draw another arc intersecting the same arc at Z.

4. Taking Y and Z as centres and the same radius, draw two arcs intersecting each other at S. Join OS. Thus, ∠SOA is a required angle of 90°.

5. Draw the bisector of SOA. Hence, ∠MOA = 45°

135o

The steps of constructions are:

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

2. Taking O as the centre and convenient radius, mark an arc, which intersects PQ at A and B.

3. Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R. Join OR. Thus, ∠QOR = ∠POR = 90°.

4. Draw OD the bisector of ∠POR. Thus, ∠QOD is required angle of 135°

The steps of constructions are:

1. Draw a ray OA

2. Taking O as the centre and convenient radius, mark an arc, which intersects OA at X.

3. Taking X as a centre and the same radius, cut the previous arc at Y. Taking Y as the centre and the same radius, draw another arc intersecting the same arc at Z.

4. Taking Y and Z as centres and the same radius, draw two arcs intersecting each other at S. Join OS. Thus, ∠SOA is a required angle of 90°.

5. Draw the bisector of ∠SOA. Hence, ∠MOA = 45°

6. Draw the bisector of ∠MOA. Hence, ∠NOA= 22.50

The steps of constructions are:

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

(b) Taking O as the centre and convenient radius, mark an arc, which intersects PQ at A and B.

(c) Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.

(d) Join OR. Thus, ∠QOR = ∠POQ = 90 .

(e) Draw OD the bisector of ∠POR. Thus, ∠QOD is the required angle of 135.

(f) Now, draw OE as the bisector of ∠QOD=1/2 of 135=67.5

The steps of constructions are:

(a) Draw an angle 70 degrees with a protractor, i.e., ∠POQ = 70 degrees

(b) Draw a ray AB.

(c) Place the compasses at O and draw an arc to cut the rays of ∠POQ at L and M.

(d) Use the same compasses, setting to draw an arc with A as the centre, cutting AB at X.

(e) Set your compasses setting to the length LM with the same radius.

(f) Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.

(g) Join AY. Thus, ∠YAX = 70 degree

The steps of constructions are:

(a) Draw an angle of 40 degrees with the help of protractor, naming ∠ AOB.

(b) Draw a line PQ.

(c) Take any point M on PQ.

(d) Place the compasses at O and draw an arc to cut the rays of ∠AOB at L and N.

(e) Use the same compasses setting to draw an arc O as the centre, cutting MQ at X.

(g) Place the compasses at X and draw the arc to cut the arc drawn earlier Y.

(h) Join MY.

(i) Thus, < QMY = 40 degree and < PMY is supplementary of it.

NCERT Solutions for Class 6 Mathematics - Chapter-wise

 Chapters No. Chapters Name Chapter - 1 Solutions of NCERT for class 6 maths chapter 1 Knowing Our Numbers Chapter - 2 CBSE NCERT solutions for class 6 maths chapter 2 Whole Numbers Chapter - 3 NCERT solutions for class 6 maths chapter 3 Playing with Numbers Chapter - 4 Solutions of NCERT for class 6 maths chapter 4 Basic Geometrical Ideas Chapter - 5 CBSE NCERT solutions for class 6 maths chapter 5 Understanding Elementary Shapes Chapter - 6 NCERT solutions for class 6 maths chapter 6 Integers Chapter - 7 Solutions of NCERT for class 6 maths chapter 7 Fractions Chapter - 8 NCERT solutions for class 6 maths chapter 8 Decimals Chapter - 9 CBSE NCERT solutions for class 6 maths chapter 9 Data Handling Chapter -10 NCERT solutions for class 6 maths chapter 10 Mensuration Chapter -11 Solutions of NCERT for class 6 maths chapter 11 Algebra Chapter -12 CBSE NCERT solutions for class 6 maths chapter 12 Ratio and Proportion Chapter -13 NCERT solutions for class 6 maths chapter 13 Symmetry Chapter -14 Solutions of NCERT for class 6 maths chapter 14 Practical Geometry

## Benefits of NCERT solutions for class 6 maths chapter 14 practical geometry-

• You will get NCERT solutions for this chapter very easily.
• You will get some different ways to construct geometrical in this solution article.
• There are some figures and images given in these solutions. It will help you in drawing geometrical figures easily.
• In CBSE NCERT solutions for class 6 maths chapter 14 practical geometry, you will find that questions are explained with step-by-step to construct lines and geometrical figures with the help of compass and rulers. So you can understand easily

Happy learning!!!