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    NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

    Edited By Ramraj Saini | Updated on Nov 30, 2023 10:46 AM IST

    NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry are discussed in this article. These NCERT solutions are created by the expert team at careers360 considering the latest CBSE syllabus 2023. Constructions of figures or sketches are mandatory for the design of different types of equipment, instruments, buildings, etc. 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.

    As per the NCERT syllabus for Class 6 Maths, the important topics of this chapter are the construction of the circle of a given radius using a compass, the construction of line segments and angles, etc. In the last exercise of this chapter as mentioned in the NCERT Books for Class 6 Maths, you will get questions related to measuring of angle using a compass and ruler. In some questions in NCERT, you will be drawing lines using the compass. There are many other constructions explained in 6 exercises of NCERT solutions for practical geometry class 6. You will get NCERT Solutions from Class 6 to 12 for Science and Math by clicking on the above link.

    NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry - Important Points

    Geometrical shapes: This class 6th maths chapter 14 focuses on the methods used to draw various geometrical shapes. The shapes are constructed using mathematical instruments such as a graduated ruler, compasses, divider, set-squares, and a protractor.

    Constructions using ruler and compasses:

    (i) Circle: A circle can be drawn when the length of its radius is known. The compasses are used to measure and draw the circle.

    (ii) Line segment: With the ruler and compasses, a line segment can be constructed when its length is given.

    (iii) Copy of a line segment: The compasses are used to create an exact copy of a given line segment.

    For more, Download Ebook - NCERT Class 6 Maths: Chapterwise Important Formulas And Points

    (iv) Perpendicular to a line through a point:

    • (a) On the line: A perpendicular line can be drawn through a point that lies on the given line.
    • (b) Not on the line: A perpendicular line can be drawn through a point that does not lie on the given line.

    (v) Perpendicular bisector: The perpendicular bisector of a line segment, which divides it into two equal halves, can be constructed.

    (vi) Angle of a given measure: An angle with a specific measure can be constructed using the ruler and compasses.

    (vii) Copy of an angle: With the ruler and compasses, an exact copy of a given angle can be drawn.

    (viii) Bisector of an angle: The bisector of a given angle, which divides it into two equal angles, can be constructed.

    (ix) Special angles: The class 6 maths chapter 14 chapter also covers the construction of angles with special measures such as 90 degrees, 45 degrees, 60 degrees, 30 degrees, 120 degrees, and 135 degrees. These angles are constructed using the ruler and compasses

    Free Download NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry For CBSE Exam

    NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

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    NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry (Intext Questions and Exercise)

    NCERT Class 6 Maths Chapter 14 Practical Geometry Exercise 14.1

    Question: 1 Draw a circle of radius 3.2 cm.

    Answer: 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. 1643782376732

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

    Answer: 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 1643782419227

    Question:3 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: 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: 1643782458502

    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 the 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: 1643782484441

    Question: 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: 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. 1643782537160

    Question: 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 \overline{AB} and \overline{CD} are at right angles.

    Answer: 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.

    1643782571220 From the figure, AB and CD intersect at right angle with

    NCERT Class 6 Maths Chapter 14 Practical Geometry Exercise 14.2

    Question: 1 Draw a line segment of length 7.3 cm using a ruler.

    Answer: 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.

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

    Answer:

    1643782642506 (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.


    Question: 3 Construct AB of length 7.8 cm. From this, cut off \overline{AB} of length 4.7 cm. Measure \overline{BC} .

    Answer: The steps of constructions:

    1643782667509 (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

    Question: 4 Given \overline{AB} of length 3.9 cm, construct \overline{PQ} such that the length of \overline{PQ} is twice that of \overline{AB} . Verify by measurement.

    1657530982905

    ( 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} .)

    Answer: The steps of constructions are following:

    1643782692823

    (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} .

    Answer: 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}

    1643782715163

    Verification :

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

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

    NCERT Class 6 Maths Chapter 14 Practical Geometry Exercise 14.3

    Question: 1 Draw any line segment \overline{PQ} . Without measuring \overline{PQ} , construct a copy of \overline{PQ} .

    Answer: The steps of constructions are the following: 1643782797660

    (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} .

    Question:2 Given some line segment \overline{AB} , whose length you do not know, construct \overline{PQ} such that the length of \overline{PQ} is twice that of \overline{AB} .

    Answer: The steps of construction are followings: 1643782823669

    (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 Class 6 Maths Chapter 14 Practical Geometry Exercise 14.4

    Question: 1 Draw any line segment \overline{AB} . Mark any point M on it. Through M, draw a perpendicular to \overline{AB} . (use ruler and compasses)

    Answer: The steps of constructions are the following: 1643782859391

    (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.

    Question: 2 Draw any line segment \overline{PQ} . Take any point R not on it. Through R, draw a perpendicular to \overline{PQ} . (use ruler and set-square)

    Answer: The steps of constructions are:

    1643782899040

    (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.

    Question: 3 Draw a line l and a point X on it. Through X, draw a line segment \overline{XY} perpendicular to l . Now draw a perpendicular to \overline{XY} at Y. (use ruler and compasses)

    Answer: The steps of construction are as follows:

    1643782927968

    (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.

    NCERT Class 6 Maths Chapter 14 Practical Geometry Exercise 14.5

    Question: 1 Draw \overline{AB} of length 7.3 cm and find its axis of symmetry.

    Answer: 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: 1643782960670

    (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.

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

    Answer: The steps of constructions are: 1643782985978

    (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} .

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

    Answer: The steps of constructions are:

    1643783061230

    (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

    Question: 5 With PQ of length 6.1 cm as diameter, draw a circle.

    Answer: The steps of constructions are: 1643783103487

    (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} .

    Question: 6 Draw a circle with centre C and radius 3.4 cm. Draw any chord \overline{AB} . Construct the perpendicular bisector of \overline{AB} and examine if it passes through C.

    Answer: The steps of constructions are: 1643783143570

    (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.

    Question: 7 Repeat Question 6, if AB happens to be a diameter.

    Answer: The steps of constructions are:

    1643783196739

    (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.

    Question: 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: The steps of constructions are: 1643783225321

    (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.

    Question:9 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 of \overline{OA} and \overline{OB} . Let them meet at P. Is PA = PB ?

    Answer: The steps of constructions are:

    1643783250917

    (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 Class 6 Maths Chapter 14 Practical Geometry Topic 14.5.4 Angles of Special Measures

    Question: How will you construct a 15° angle?

    Answer: 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.

    1657531206681

    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

    1657531394717

    1657531423090

    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

    1657531454828

    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?

    Answer: 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?

    Answer:

    1. Draw a ray OA.

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

    1643783313354

    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.

    1643783313039\

    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}

    1643783313781

    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.

    1643783314111

    Thus, \angle AOR = 45^{\circ}

    NCERT Class 6 Maths Chapter 14 Practical Geometry Exercise: 14.6

    Question: 1 Draw \angle POQ of measure 75° and find its line of symmetry.

    Answer:

    1. Draw a ray OA.

    1643783364905

    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}

    1643783365649

    Now, we need to find its line of symmetry

    that is, we need to find its bisector.

    We follow these steps

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

    1643783365879

    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.

    1643783366336

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

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

    Answer: 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.
    1643783405828

    Question: 3 Draw a right angle and construct its bisector.

    Answer: 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.

    1643783431661

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

    Answer: 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.

    1643783457223

    Question: 5(a) Construct with ruler and compasses, angles of following measures:

    60 o

    Answer: 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°

    1643783491983

    Question:5(b) Construct with ruler and compasses, angles of following measures:

    30 o

    Answer: 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°.

    1643783531952

    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° 1643783546175

    Question: 5(c) Construct with ruler and compasses, angles of following measures:

    90 o

    Answer: 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.

    1643783585629

    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. 1643783607939 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°

    1643783627827

    Question: 5(d) Construct with ruler and compasses, angles of following measures:

    120 o

    Answer: 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.

    1643783661012

    3. Taking P as the centre and same radius, cut the 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°. 1643783687173

    Question:5(e) Construct with ruler and compasses, angles of following measures:

    45 o

    Answer: 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. 1643783863071

    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°. 1643783885127

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

    1643783929219

    Question: 5(f) Construct with ruler and compasses, angles of following measures:

    135 o

    Answer: 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.

    1643784000285

    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°. 1643784019617

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

    1643784039544

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

    Answer: 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.

    1643784073108

    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°. 1643784092564

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

    6. Draw the bisector of ∠MOA. Hence, ∠NOA= 22.5 0

    1643784112573

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

    Answer: The steps of constructions are:

    1643784150134

    (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

    Question: 8 Draw an angle of 70 o . Make a copy of it using only a straight edge and compasses.

    Answer: The steps of constructions are: 1643784187334

    (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

    Question: 9 Draw an angle of 40 o . Copy its supplementary angle.

    Answer: 1643784219376

    The steps of constructions are:

    (a) Draw an angle of 40 degrees with the help of a 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.

    (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.

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

    Main Topics of NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

    • The Circle
    • Construction of a circle when its radius is known
    • A-Line Segment
    • Construction of a line segment of a given length
    • Constructing a copy of a given line segment
    • Perpendiculars
    • Perpendicular to a line through a point on it
    • Perpendicular to a line through a point not on it
    • The perpendicular bisector of a line segment
    • Angles
    • Constructing an angle of a given measure
    • Constructing a copy of an angle of an unknown measure
    • Bisector of an angle
    • Angles of special measures

    Also Check

    NCERT Books and NCERT Syllabus

    NCERT Solutions for Class 6 Mathematics Chapter Wise

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

    Key Features of NCERT Solutions for Class 6 Maths Chapter 14

    Expertly Crafted Solutions: The NCERT Solutions for Chapter 14 are meticulously developed to offer clear and precise explanations for all the problems and concepts discussed in the chapter.

    Conceptual Clarity: The solutions aim to improve students' conceptual understanding by simplifying complex topics and presenting them in a more comprehensible manner.

    Comprehensive Coverage: The solutions cover all the essential topics and subtopics of Chapter 14, ensuring that students gain a thorough understanding of the entire chapter.

    NCERT Solutions for Class 6 Subject Wise

    Benefits of NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry-

    • You will get NCERT Solutions for Class 6 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
    • It is going to help you with your homework as well.

    Happy learning!!!

    Frequently Asked Question (FAQs)

    1. How to draw geometrical shapes in accurate sizes and accurate angle

    We need some instruments to draw geometrical shapes in accurate size and accurate angle here are some given below;

    • The compasses
    • The Ruler 
    • The Divider 
    • Scale 
    • Protractor
    2. What is line segment?

    The line segment is the least possible distance between the two given points. above in this article we have discussed some problems based on this concept. Practice them to get in-depth understanding the concepts. These concepts also gives drawing ideas for class 6 which are useful for upcoming classes. 

    3. What is a circle?

    In a circle the points on its boundary is at an equal distance to its center such a geometrical shape is called a circle. practice problems based on the circle given with detailed and comprehensive solution to get command in the concepts. these concepts are comprehensively covered in class 6 chapter 14 maths.

    4. What are the topics are covered in NCERT solution Class 6 Chapter Practical Geometry?

    The following are covered in NCERT solution for class 6 maths chapter 14: 

    • The Circle   
    • Construction of a circle when its radius is known   
    • A-Line Segment   
    • Construction of a line segment of a given length   
    • Constructing a copy of a given line segment   
    • Perpendiculars   
    • Perpendicular to a line through a point on it   
    • Perpendicular to a line through a point not on it   
    • The perpendicular bisector of a line segment   
    • Angles   
    • Constructing an angle of a given measure   
    • Constructing a copy of an angle of an unknown measure   
    • Bisector of an angle   
    • Angles of special measures  

    Articles

    Get answers from students and experts

    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

    Option 3)

    1.00\; J

    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

    Option 4)

    12.89×10−3 kg

     

    An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

    200 \, \, J - 500 \, \, J

    Option 3)

    2\times 10^{5}J-3\times 10^{5}J

    Option 4)

    20,000 \, \, J - 50,000 \, \, J

    A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

    Option 1)

    K/2\,

    Option 2)

    \; K\;

    Option 3)

    zero\;

    Option 4)

    K/4

    In the reaction,

    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

    6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

    Option 3)

    33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

    Option 4)

    67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

    If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

    Option 1)

    decrease twice

    Option 2)

    increase two fold

    Option 3)

    remain unchanged

    Option 4)

    be a function of the molecular mass of the substance.

    With increase of temperature, which of these changes?

    Option 1)

    Molality

    Option 2)

    Weight fraction of solute

    Option 3)

    Fraction of solute present in water

    Option 4)

    Mole fraction.

    Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

    Option 1)

    twice that in 60 g carbon

    Option 2)

    6.023 × 1022

    Option 3)

    half that in 8 g He

    Option 4)

    558.5 × 6.023 × 1023

    A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

    Option 1)

    less than 3

    Option 2)

    more than 3 but less than 6

    Option 3)

    more than 6 but less than 9

    Option 4)

    more than 9

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    A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying. 

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    Procurement Manager

    The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

    2 Jobs Available
    Production Planner

    Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner. 

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    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    .NET Developer

    .NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

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    DevOps Architect

    A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties. 

    2 Jobs Available
    Cloud Solution Architect

    Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

    2 Jobs Available
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