NCERT Solutions Class 6 Mathematics Chapter 14 Practical Geometry have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 6 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 6 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 6 Mathematics are an important part of exams for Class 6 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 6 Mathematics and also download more latest study material for all subjects. Chapter 14 Practical Geometry is an important topic in Class 6, please refer to answers provided below to help you score better in exams

## Chapter 14 Practical Geometry Class 6 Mathematics NCERT Solutions

Class 6 Mathematics students should refer to the following NCERT questions with answers for Chapter 14 Practical Geometry in Class 6. These NCERT Solutions with answers for Class 6 Mathematics will come in exams and help you to score good marks

### Chapter 14 Practical Geometry NCERT Solutions Class 6 Mathematics

**Practical Geometry**

**Exercise 14.1**

**Q.1) Draw a circle of radius 3.2 ππ.**

**Sol.1)** 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 circle.

It is required circle.

**Q.2) With the same centre O, draw two circles of radii 4 ππ and 2.5 ππ.**

**Sol.2)** (a) Marks a point βOβ 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.

It is the required figure.

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

**Sol.3)** (i) By joining the ends of two diameters, we get a rectangle.

By measuring, we find π΄π΅ = πΆπ· = 3 ππ, π΅πΆ = π΄π· = 2 ππ, i.e., pairs of opposite sides are equal and also β π΄ = β π΅ = β πΆ = β π· = 90Β° , i.e. each angle is of 90Β°.

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 π΄π΅ = π΅πΆ = πΆπ· = π·π΄ = 2.5 ππ, i.e., all four sides are equal.

Also β π΄ = β π΅ = β πΆ = β π· = 90Β° , i.e. each angle is of 90Β°.

Hence, it is a square.

**Q.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.**

**Sol.4)** (i) Mark a point βOβ with sharp pencil where we want centre of the circle.

(ii) Place the pointer of the compasses at βOβ. 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.

**Q.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 whether AB and CD are at right angles**

**Sol.5)** 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 β πΆππ΅ is 90Β°.

**Exercise 14.2**

**Q.1) Draw a line segment of length 7.3 ππ using a ruler.**

**Sol.1) **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 ππ from A.

(iii) Join AB.

AB is the required line segment of length 7.3 ππ.

**Q.2) Construct a line segment of length 5.6 ππ using ruler and compasses**

**Sol.2)** Steps of construction:

(i) Draw a line π 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 ππ mark.

(iii) Without changing the opening of the compasses. Place the pointer on A and cut an arc π at B.

AB is the required line segment of length 5.6 ππ

**Q.3) Construct AB of length 7.8 ππ. From this cut off AC of length 4.7 ππ. Measure BC.**

**Sol.3)** Steps of construction:

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

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

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

By measuring BC, we find that π΅πΆ = 3.1 ππ

**Q.4) Given AB of length 3.9 ππ, construct PQ such that the length PQ is twice that of AB.**

**Verify by measurement.**

**Sol.4)** Steps of construction:

(i) Draw a line π.

(ii) Construct PX such that length of PX = length of AB

(iii) Then cut of XQ such that XQ also has the length of AB

(iv) Thus the length of PX and the length of XQ added together make twice the length of AB.

Verification:

By measurement we find that ππ = 7.8 ππ

= 3.9 ππ + 3.9 ππ = AB + AB = 2 Γ AB

**Q.5) Given AB of length 7.3 cm and CD of length 3.4 ππ, construct a line segment XY such that the length of XY is equal to the difference between the lengths of AB and CD.**

**Verify by measurement.**

**Sol.5)** Steps of construction:

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

(ii) Construct XZ such that length XZ = length of AB = 7.3ππ

(iii) Then cut off ZY = length of CD = 3.4 ππ

(iv) Thus the length of XY = length of AB β length of CD

Verification:

By measurement we find that length of XY = 3.9 ππ

= 7.3 ππ β 3.4 ππ = AB β CD

**Exercise 14.3**

**Q.1) Draw any line segment PQ. Without measuring PQ, construct a copy of PQ.**

**Sol.1)** Steps of construction:

(i) Given PQ whose length is not known.

(ii) Fix the compasses pointer on π and the pencil end on π. The opening of the instrument now gives the length of PQ.

(iii) Draw any line π. Choose a point π΄ on π. Without changing the compasses setting, place the pointer on π΄.

Draw an arc that cuts π at a point, say π΅. Now AB is a copy of PQ

**Q.2) Given some line segment AB, whose length you do not know, construct PQ such that the ****length of PQ is twice that of AB.**

**Sol.2)** Steps of construction:

(i) Given AB whose length is not known.

(ii) Fix the compasses pointer on π΄ and the pencil end on π΅. The opening of the instrument now gives the length of AB.

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

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

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

Thus PQ is the required line segment whose length is twice that of π΄π΅.

**Exercise 14.4**

**Q.1) Draw any line segment AB. Mark any point π on it. Through π, draw a perpendicular to AB. (Use ruler and compasses)**

**Sol.1)** Steps of construction:

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

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

(iii) Join ππ. Then ππ is perpendicular to π΄π΅ through the point π.

**Q.2) Draw any line segment PQ. Take any point R not on it. Through R, draw a perpendicular to PQ. (Use ruler and set-square)**

**Sol.2)** Steps of construction:

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

(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 π
π along the edge through R meeting ππ at π. Then π
π β₯ ππ.

**Q.3) Draw a line π and a point π on it. Through π, draw a line segment Μ
πΜ
Μ
πΜ
perpendicular to π.**

**Now draw a perpendicular to Μ
πΜ
Μ
πΜ
to π. (use ruler and compasses)**

**Sol.3)** Steps of construction:

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

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

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

(iv) Join π΄πΆ and produce it to Y. Then XY is perpendicular to π.

(v) With D as centre and a convenient radius, draw an arc intersecting ππ 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 ππ at Y.

**Exercise 14.5**

**Q.1) Draw AB of length 7.3 ππ and find its axis of symmetry.**

**Sol.1)** The below given steps will be followed to construct a line segment π΄π΅ of length 7.3 ππ & to find its axis of symmetry

i) Draw a lone segment of 7.3 ππ

ii) Taking π΄ as centre, draw a circle by using compasses. The radius of circle should be more than half the length of π΄π΅

**Q.2) Draw a line segment of length 9.5 ππ & construct its perpendicular bisector.
Sol.2) **The below given steps will be followed to construct a line segment ππ of length 9.5 ππ & its perpendicular bisector.

**Q.3) Draw the perpendicular bisector of XY whose length is 10.3 ππ.
(a) Take any point π on the bisector drawn. Examine whether PX = PY.
(b) If π is the mid-point of XY, what you say about the lengths MX and XY
Sol.3)** i) Draw a line segment XY = 10.3 ππ

**Q.4) Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts.**

**Verify by actual measurement.**

**Sol.4)** Steps of construction:

(i) Draw a line segment π΄π΅ = 12.8 ππ

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

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

(iv) Again, draw the perpendicular bisector of CB which cuts it at E. Thus, E is the midpoint of CB .

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

(vi) By actual measurement, we find that

AD = DC = CE = EB = 3.2 ππ

**Q.5) With PQ of length 6.1 ππ as diameter, draw a circle.**

**Sol.5)** Steps of construction:

(i) Draw a line segment ππ = 6.1 ππ.

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

Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment ππ.

**Q.6) Draw a circle with centre C and radius 3.4 cm. Draw any chord AB. Construct the ****perpendicular bisector of AB and examine if it passes through C.**

**Sol.6)** Steps of construction:

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

(ii) Draw any chord AB.

(iii) Taking A and B as centres and radius more than half of AB, draw two arcs which cut

each other at P and Q.

(iv) Join PQ. Then PQ is the perpendicular bisector of AB.

This perpendicular bisector of AB passes through the centre C of the circle.

**Q.7) Repeat Question 6, if AB happens to be a diameter**

**Sol.7)** Steps of construction:

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

(ii) Draw its diameter π΄π΅.

(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 ππ. Then ππ is the perpendicular bisector of π΄π΅.

We observe that this perpendicular bisector of π΄π΅ passes through the centre C of the circle.

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

**Sol.8)** Steps of construction :-

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

(ii) Draw any two chords π΄π΅ and πΆπ· in this circle.

(iii) Taking A and B as centres and radius more than half π΄π΅, draw two arcs which

intersect each other at E and F.

(iv) Join πΈπΉ. Thus πΈπΉ is the perpendicular bisector of chord πΆπ·.

(v) Similarly draw πΊπ» the perpendicular bisector of chord πΆπ·.

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

**Q.9) Draw any angle with vertex O. Take a point A on one of its arms and B on another such ****that ππ΄ = ππ΅. Draw the perpendicular bisectors of ππ΄ and ππ΅. Let them meet at P. Is ****ππ΄ = ππ΅?**

**Sol.9)** Steps of construction:

(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 ππ΄ and ππ΅

(iv) Let them meet at P. Join ππ΄ and ππ΅.

With the help of divider, we check that ππ΄ = ππ΅

**Exercise 14.6**

**Q.1) Draw β POQ of measure 75Β° and find its line of symmetry.**

**Sol.1)** Steps of construction:

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

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

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

(d) Join ππ΅, then β π΅ππ΄ = 60Β°

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

(f) Draw bisector of β π΅ππΆ. The angle is of 90Β°. Mark it at D. Thus, β π·ππ΄ = 90Β°

(g) Draw ππ as bisector of β π·ππ΅.

Thus, β πππ΄ = 75Β°

**Q.2) Draw an angle of measure 147Β° and construct its bisector.**

**Sol.2)** Steps of construction:

(a) Draw a ray OA

(b) With the help of protractor, construct β π΄ππ΅ = 147Β°.

(c) Taking centre O and any convenient radius, draw an arc which intersects the arms ππ΄ and ππ΅Β― at P and Q respectively.

(d) Taking P as centre and radius more than half of ππ, draw an arc.

(e) Taking Q as centre and with the same radius, draw another arc which intersects the previous at R.

(f) Join ππ
and produce it.

Thus, ππ
is the required bisector of β π΄ππ΅.

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

**Sol.3)** Steps of construction:

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

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

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

(d) Join ππΆ. Thus, β πΆππ is the required right angle.

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

(f) Join ππ·.

Thus, ππ· is the required bisector of β πΆππ.

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

**Sol.4)** Steps of construction:

(a) Draw a ray OA .

(b) At O, with the help of a protractor, construct β π΄ππ΅ = 153Β°.

(c) Draw ππΆ as the bisector of β π΄ππ΅.

(d) Again, draw ππ· as bisector of β π΄ππΆ.

(e) Again, draw ππΈ as bisector of β π΅ππΆ.

(f) Thus, ππΆ, ππ· and ππΈ divide β π΄ππ΅ in four equal parts.

**Q.5) Construct with ruler and compasses, angles of following measures:**

**(a) 60Β° (b) 30Β° (c) 90Β° (d) 120Β° (e) 45Β° (f) 135Β°**

**Sol.5)** Steps of construction:

(a) 60Β°

(i) Draw a ray OA .

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at P.

(iii) Taking P as centre and same radius, cut previous arc at Q.

(iv) Join OQ.

Thus, β π΅ππ΄ is required angle of 60Β°.

(b) 30Β°

(i) Draw a ray OA .

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at P.

(iii) Taking P as centre and same radius, cut previous arc at Q.

(iv) Join OQ. Thus, β π΅ππ΄ is required angle of 60Β°.

(v) Put the pointer on P and mark an arc.

(vi) Put the pointer on Q and with same radius, cut the previous arc at C.

Thus, β πΆππ΄ is required angle of 60Β°.

(c) 90Β°

(i) Draw a ray OA .

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at X.

(iii) Taking X as centre and same radius, cut previous arc at Y.

(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.

(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.

(vi) Join OS and produce it to form a ray OB.

Thus, β π΅ππ΄ is required angle of 90Β°

(d) 120Β°

(i) Draw a ray OA

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA at P.

(iii) Taking P as centre and same radius, cut previous arc at Q.

(iv) Taking Q as centre and same radius cut the arc at S.

(v) Join OS.

Thus, β π΄ππ· is required angle of 120Β°

(e) 45Β°

(i) Draw a ray ππ΄

(ii) Taking O as centre and convenient radius, mark an arc, which intersects π π΄ at X.

(iii) Taking X as centre and same radius, cut previous arc at Y.

(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.

(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.

(vi) Join OS and produce it to form a ray ππ΅. Thus, β π΅ππ΄ is required angle of 90Β°

(vii) Draw the bisector of β π΅ππ΄.

Thus, β πππ΄ is required angle of 45Β°

(f) 135Β°

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

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

(iii) Taking A and B as centres and radius more than half of π΄π΅, draw two arcs intersecting

each other at R.

(iv) Join OR. Thus, β πππ
= β πππ = 90Β°

(v) Draw π π· the bisector of β πππ
.

thus, β πππ· is required angle of 135Β°

**Q.6) Draw an angle of measure 45Β° and bisect it.**

**Sol.6)** Steps of construction:

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

(b)Taking O as centre and a convenient radius, draw an arc which intersects PQ at two points 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 ππΆ. Then β πΆππ is an angle of 90Β°

(e) Draw π π as the bisector of β πΆππΈ. Thus, β πππΈ = 45Β°

(f) Again draw π πΊ as the bisector of β πππΈ.

Thus, β πππΊ = β πΈππΊ = 22(1Β°/2)Β°

**Q.7) Draw an angle of measure 135Β° and bisect it.**

**Sol.7)** Steps of construction:

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

(b) Taking O as 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 π΄π΅, draw two arcs intersecting each other at R.

(d) Join ππ
. Thus, β πππ
= β πππ = 90Β°

(e) Draw π π· the bisector of β πππ
. Thus, β πππ· is required angle of 135Β°

(f) Now, draw π πΈ as the bisector of β β QOD.

Thus, β πππΈ = β π·ππΈ = 67(1Β°/2)

**Q.8) Draw an angle of 70Β°. Make a copy of it using only a straight edge and compasses.**

**Sol.8)** Steps of construction:

(a) Draw an angle 70Β°. with protractor, i.e., β πππ = 70Β°.

(b) Draw a ray AB

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

(d) Use the same compasses, setting to draw an arc with A as 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, β ππ΄π = 70Β°.

**Q.9) Draw an angle of 40Β°. Copy its supplementary angle.**

**Sol.9)** Steps of construction:

(a) Draw an angle of 40Β° with the help of protractor, naming β π΄ππ΅.

(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 β π΄ππ΅ at L and N.

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

(f) Set your compasses to length LN with the same radius.

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

(h) Join MY.

Thus, β πππ = 40Β° and β πππ is supplementary of it.

NCERT Solutions Class 6 Mathematics Chapter 1 Knowing our Numbers |

NCERT Solutions Class 6 Mathematics Chapter 2 Whole Numbers |

NCERT Solutions Class 6 Mathematics Chapter 3 Playing with Numbers |

NCERT Solutions Class 6 Mathematics Chapter 4 Basic Geometrical Ideas |

NCERT Solutions Class 6 Mathematics Chapter 5 Understanding Elementary Shapes |

NCERT Solutions Class 6 Mathematics Chapter 6 Integers |

NCERT Solutions Class 6 Mathematics Chapter 7 Fractions |

NCERT Solutions Class 6 Mathematics Chapter 8 Decimals |

NCERT Solutions Class 6 Mathematics Chapter 9 Data Handling |

NCERT Solutions Class 6 Mathematics Chapter 10 Mensuration |

NCERT Solutions Class 6 Mathematics Chapter 11 Algebra |

NCERT Solutions Class 6 Mathematics Chapter 12 Ratio and Proportion |

NCERT Solutions Class 6 Mathematics Chapter 13 Symmetry |

NCERT Solutions Class 6 Mathematics Chapter 14 Practical Geometry |

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