NCERT Solutions For Class 6 Maths Chapter 14

NCERT Solutions Class 6 Maths Practical Geometry

Ncert Solutions For Class 6 Maths Chapter 14 PDF Download

Exercise 14.1

Question 1:

Draw a circle with circle of radius 3.2 and write down it’s procedures.

Answer:

Procedure:

(a) For the required radius of 3 cm open the compass.

(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 0.

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

Hence, it is the required circle.

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Question 2:

With the same centre 0, draw two circles of radii 4 cm and 2.5 cm

Answer:

Steps of construction:

(a) Marks a point ‘0’ 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 C.

(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.

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Question 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?  

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 \(\angle\) A = \(\angle\)B= \(\angle\)C= \(\angle\)D= 90  i.e. each angle is of 90°. 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 \(\angle\)A= \(\angle\) B= \(\angle\)C= L \(\angle\)D= 90′, i.e. each angle is of 90′. Hence, it is a square.

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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:

(1)Point O is marked by using the pencil and then circle’s centre is drawn

(2) Compasses pointer is placed at O after that circle is drawn slowly using the compasses

A is on the circle.

B is in the interior of the circle.

C is in the exterior of the circle.

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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 AB and CD 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 \(\angle\) COB is 90.

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Exercise 14.2

Question 1:

A line segment was drawn of length 10cm, with the help of the ruler.

Answer:

Method of construction:

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(i)At the point X make place the zero mark of the ruler

(ii)And pick appoint Y at the exact distance of 10cm from the point X

(iii) Now join XY

 

Question 2:

Draw a line segment of length 8cm with the help of ruler and compasses

Answer:

(1)Make a line ‘l’. Note a point X on that line

(2)Compasses pointer is placed on the zero mark of that ruler. Use it to mark the pencil point till 8cm mark

(3)Not making any in the compasses. At the point A place the pointer an then with the compasses cut arc l at Y

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Question 3:

Construct \(\overline{XY}\) of certain distance 6cm .from that cut off \(\overline{XO}\) of 4cm.Measure \(\overline{OY}\)

Answer:

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(1)Keep the mark zero of this ruler at X

(2)Then Y point is marked at a distance of 6 cm from X

(3)Then again Point O is marked at a distance of 4 cm from X

Hence, measuring \(\overline{OY}\) we can find the OY=4cm

 

Question 4:

Given  \(\overline{PQ}\) of length 5cm , Draw \(\overline{AB}\) such that the length

\(\overline{XB}\) also it has the length of \(\overline{PQ}\)

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Answer:

Construction step:

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(1)Line is drawn ‘l’

(2)Draw \(\overline{AX}\) such that its length of \(\overline{AX}\)= \(\overline{PQ}\) length

(3)Cut of \(\overline{XB}\) so that \(\overline{XB}\) will have the same length of \(\overline{PQ}\)

(4) \(\overline{AX}\) length and the \(\overline{XB}\) length added together doubles the \(\overline{PQ}\)

 

Check:

Now, measurement can find that AB=10cm

=5cm+5cm

= \(\overline{PQ}\)+ \(\overline{PQ}\)=2* \(\overline{PQ}\)

 

Question 5:

Note PQ of length 8 cm and RS of length 3.4 cm, Draw a segment of line AB so that the length of AB is equal to the difference between the lengths of  PQ and RS . Check the measurement

Answer:

Steps of construction:

(1) A line is drawn  ‘l’ and take a point A on it

(2) Draw AC such that its length AC = length of PQ = 8 cm

(3) Then erase  BC = length of RS = 5 cm

(4) Thus the length of AB = length of PQ – length of RS

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Thus the measurement we can easily found that the length of PQ=3 cm

PQ-RS=8-5

 

 

Exercise 14.3

Question 1:

Construct a line segment of \(\overline{AB}\). Not measuring \(\overline{AB}\).Draw a duplicate of \(\overline{AB}\)

Answer:

Method of construction.

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(1)Now the mentioned \(\overline{AB}\) that length is not known

(2)Point A is fixed  by the compasses pointer and the B is fixed by pencil end.The instrument opening gives the \(\overline{AB}\) length accurately

(3)Now construct a line of length ‘l’ .Without making any change in the setting of the compass ,Point P is marked

(4) Now cut an arc ‘l’ at a point ,say Q

Now  \(\overline{PQ}\) is the copy of \(\overline{AB}\)

 

Question 2.

Mention some segment \(\overline{PQ}\)  ,the length of it is unknown, Draw \(\overline{AB}\) the length of \(\overline{AB}\) is double the times of \(\overline{PQ}\)

Answer:

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Method of construction:

(1)Now that \(\overline{PQ}\) whose certain length is unknown

(2)The compasses pointer on point P is fixed and then pencil end on Q .The instrument opening gives the

\(\overline{PQ}\).

(3) Line ‘l’ is drawn .Point A is chosen by ‘l’ .Not changing the compasses setup, the pointer is placed at B

(4)Now an arc is cut at the point O of length ’l’

(5)Pointer O is placed and by not changing the setting of the compass ,Another arc is drawn at a point B ‘l’

 

 

Exercise 14.4

Question 1:

A line segment  \(\overline{PQ}\) .Point O is marked on it any where.On the point  O perpendicular is drawn(by using compasses and ruler)

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Construction step:

(1)Keep O as midpoint and with some  radius,  an  intersecting arc is drawn the line PQ at 2 points X and Y.

(ii) With X and Y as centers as well as radius more than OX, draw 2 arcs, they cut together at L.

(iii) Join LO. Then LO is perpendicular to PQ through O point.

 

Question 2:

A line segment is drawn AB. X  as a point taken not on it By  R, a perpendicular  line is drawn

to AB. (Use set-square  and ruler )

 Answer: 

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construction method:

(i)A set-square is placed on \(\overline{AB}\)  such that one arm of its right angle aligns along \(\overline{AB}\)

(ii) A ruler is placed along the edge opposite to the right angle of the set-square.

(iii) Hold the ruler fixed.  Set square slide the along the ruler till the point X touches set square’s other arm

(iv) Join OX along the edge through X meeting AB at O. Then XO is perpendicular to AB.

 

Question 3:

 Draw a line l and a point X on it. Through X, draw a line segment XY perpendicular to l. Now draw a perpendicular to XY to Y. (use ruler and compasses)

Answer:

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Construction step:

(1)Make a line ‘l’ and take point O on it.

(2)With O as centre and a certain radius, an arc is draw at intersecting the line ‘l’ at 2 points P and Q.

(3) With P and Q as centers and a radius greater than OP, draw 2 arcs, which cut each other at R.

(4) Join PR and produce it to M. Then OM  is perpendicular to  ‘l’

(5) With S as centre and a certain radius, draw an arc intersecting OM  at two points R and S.

(6) With R and S as centers and radius greater than MS, draw 2 arcs and they cut each other at L.

(7) Join ML, then ML is perpendicular to OM at M

 

Exercise 14.5

Question 1:

Draw \(\overline{XY}\)  of length 10 cm and What is the axis of symmetry

Answer:

Axis of symmetry of line segment \(\overline{XY}\) will be the perpendicular bisector of \(\overline{XY}\). So, draw the perpendicular bisector of XY.

Steps of construction:

(i)A line segment is drawn XY = 10 cm

(ii) Assume  X and Y as centers and radius more than half of XY, construct 2 arcs which intersect each other at P and  Q .

(iii) Join PQ. Then PQ is the symmetry axis of the segment line XY.

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Question 2:

Construct a segment line of length 8 cm and draw its perpendicular bisector.

 Answer: 

Construction step:

(i) A line segment \(\overline{XY}\)  is drawn= 8 cm

(ii) Taking X and Y as radius and centre more than XY half, draw 2 arcs at P they intersect each other’s and Q.

(iii) PQ is then joined. Then PQ is the perpendicular bisector of \(\overline{XY}\).

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Question 3:

\(\overline{AB}\) bisector is drawn perpendicular that length is 8 cm.

(a) A point P is taken according to the drawn bisector. Prove OA = OB.

(b)If X is the \(\overline{AB}\)  midpoint, the lengths XA and XB explain about it?

Answer:

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Construction steps:

(1)A line segment is drawn \(\overline{AB}\) = 8 cm

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

(3) Join PQ. Then PQ is the required perpendicular bisector of \(\overline{AB}\).

Now,

(1)Point O is taken on the drawn bisector. By using the divider and we can check that that  \(\overline{OA}\)= \(\overline{OB}\)

(2) If    X is the midpoint of \(\overline{AB}\), then \(\overline{XA}\) =1/2 \(\overline{XB}\)

 

Question 4:

Draw a line segment of length 10 cm. Using compasses; separate it into 4 equal parts. Verify by original measurement.

Answer:

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Construction steps:

(1)A line segment is drawn PQ = 10 cm

(2) \(\overline{PQ}\) of perpendicular bisector is drawn which cuts it at a point X. Thus, \(\overline{PQ}\)  is the bisector of X.

(3) Draw \(\overline{PX}\) perpendicular bisector which cuts it at P. Thus P is the midpoint of.

(4) Again, \(\overline{XQ}\) perpendicular bisector is drawn which cuts it at B . Thus, B is the mid-point of  \(\overline{XQ}\).

(5) Now, point X, A and B separates the line segment \(\overline{PQ}\)  in the 4 equal parts.

(6) By actual measurement, we find that \(\overline{PA}\) = \(\overline{AX}\) = \(\overline{XB}\) = \(\overline{BQ}\) =2.5cm

 

Question 5:

Draw a circle with \(\overline{AB}\) of length  5 cm as diameter,.

Answer:

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construction steps:

(1) \(\overline{AB}\)  a line segment is drawn = 5 cm.

(2) \(\overline{AB}\) perpendicular bisector is drawn  which cuts, it at X. Thus X is the mid-point of AB.

(3) Assuming  X as centre and XA or XB as radius a circle is drawn  where diameter is the line segment  \(\overline{AB}\)

 

Question 6:

Draw a circle with centre O and radius 5 cm. Draw any chord \(\overline{PQ}\). Construct the perpendicular bisector  \(\overline{PQ}\)  and examine if it passes through O.

Answer:

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Steps of construction:

(i) With centre O draw a circle and radius 5 cm.

(ii)Chord \(\overline{PQ}\) is drawn.

(iii) Taking P and Q as centers and radius more than half of \(\overline{PQ}\), Make 2  arcs which cut each other at Y and X.

(iv)  XY is joined. Then XY is the perpendicular bisector of \(\overline{PQ}\).

(v) Through the centre O of the circle this perpendicular bisector of \(\overline{PQ}\)  passes.

 

Question 7:

Question 6 is repeated, if \(\overline{PQ}\)  be a diameter

Answer:        

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Construction steps:

(1)Draw a circle with centre O and radius 5 cm.

(2) \(\overline{PQ}\) diameter is drawn

(3)P and Q is taken as centers and radius more than half of it, Make 2 arcs which intersect each other at X and Y.

(4)Join XY . Then XY is the perpendicular bisector of \(\overline{PQ}\).

(5)We observe that this perpendicular bisector of \(\overline{PQ}\) passes through the centre O 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:

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Steps of construction:

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

(ii) Draw any two chords AB and CD in this circle.

(iii) Taking A and B as centers 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 CD.

(v) Similarly draw GH the perpendicular bisector of chord CD.

(vi) 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 OA and OB. Let them meet at P. Is PA = PB?

Answer:

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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 OA = OB.

(iii) Draw perpendicular bisector of OA and OB.

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

(v) With the help of divider, we check that PA PB.

 

 

Exercise 14.6

Question 1:

Draw \(\angle 75^{\circ}\) and find its line of symmetry.

Answer:

Follow the steps of construction:

(a) Construct a line l 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 l at A.

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

(d) Join OB, then \(\angle BOA=60^{\circ}\) .

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

(f) Draw bisector of \(\angle BOC\) . The angle is of \(\angle BOC\). Mark it at D. Thus, \(\angle DOA=90^{\circ}\)

(g) Draw \(\overline{OP}\) as bisector of \(\angle DOB\).

Thus, \(\angle POA=75^{\circ}\)

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Question 2:

Draw an angle of measure \(147^{\circ}\) and construct its bisector.

Answer:

Follow the steps of construction:

(a) Draw a line \(\overline{OA}\) .

(b) Using protractor, construct \(\angle AOB=147^{\circ}\) .

(c) Taking the centre as O and any suitable radius, draw an arc which cuts the arms \(\overline{OA}\) and \(\overline{OB}\) at P and Q respectively.

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

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

(f) Join OR and produce it.

(g) Thus, \(\overline{OR}\) is the required bisector of \(\angle AOB\).

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Question 3:

Construct a right angle and construct its bisector.

Answer:

Follow the steps of construction:

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

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

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

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

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

(f) Join OD. Thus, \(\overline{OD}\) is the required bisector of \(\angle COQ\) .

\(\angle COQ\) .

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Question 4:

Draw an angle of measure \(153^{\circ}\) and divide it into four equal parts.

Answer:

Follow the steps of construction:

(a) Draw a ray \(\overline{OA}\) .

(b) At O, with the help of a protractor, construct \(\angle AOB=153^{\circ}\) .

(c) Draw \(\overline{OC}\) as the bisector of \(\angle AOB\) .

(d) Again, draw \(\overline{OD}\) as bisector of \(\angle AOC\)

(e) Again, draw \(\overline{OE}\) as bisector of \(\angle BOC\) .

(f) Thus, \(\overline{OC}\) , \(\overline{OD}\) and \(\overline{OE}\) divide \(\angle AOB\) in four equal parts.

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Question 5:

Construct with ruler and compasses, angles of following measures:

(1) 60 (2) 30 (3) 90 (4) 120 (5) 45 (6) 135

Answer:

Steps of construction:

 

(1)\(60^{\circ}\)

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(i) Draw a ray \(\overline{OL}\).

(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{OL}\) at P.

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

(iv) Join OQ.

Thus, \(\angle\)MOL is required angle of 60 .

 

(2) \(30^{\circ}\) 

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(i) Draw a ray \(\overline{OL}\).

(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{OL}\) at P.

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

(iv) Join OQ. Thus, \(\angle\)MOL is required angle of \(60^{\circ}\).

(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 N.

Thus, \(\angle\)NOL is required angle of \(30^{\circ}\) .

 

(3) \(90^{\circ}\) 

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(i) Draw a ray \(\overline{OL}\).

(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{OL}\) 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 OM.

Thus, \(\angle\)MOL is required angle of \(90^{\circ}\) .

 

(4) \(120^{\circ}\) 

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(i) Draw a ray \(\overline{OI}\).

(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{OI}\) 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, \(\angle\) IOL is required angle of \(120^{\circ}\).

 

(5) \(45^{\circ}\) 

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(i) Draw a ray \(\overline{OI}\).

(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{OI}\) 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 OJ. Thus, \(\angle\) JOI is required angle of \(90^{\circ}\) .

(vii) Draw the bisector of \(\angle\)JOI.

Thus, \(\angle\)MOI is required angle of \(45^{\circ}\) .

 

(6) \(135^{\circ}\) 

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

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

(iii) Taking I and J as centres and radius more than half of IJ, draw two arcs intersecting each other at R.

(iv) Join OR. Thus, \(\angle\)QOR = \(\angle\)POQ = \(90^{\circ}\).

(v) Draw \(\overline{OL}\) the bisector of \(\angle\)POR.

Thus, \(\angle\)QOL is required angle of \(135^{\circ}\) .

 

Question 6:

Draw an angle of measure \(45^{\circ}\) and bisect it.

Answer:

Steps of construction:

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

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

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(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. Then \(\angle\)COQ is an angle of 90

(v) Draw \(\overline{OE}\) as the bisector of \(\angle\)COE. Thus, \(\angle\)QOE =\(90^{\circ}\)

(vi) Again draw \(\overline{OG}\) as the bisector of \(\angle\)QOE.

Thus, \(\angle\)QOG = \(\angle\)EOG = \(22\frac{1}{2}^{\circ}\)

 

Question 7:

Draw an angle of measure \(135^{\circ}\) and bisect it.

Answer:

Steps of construction:

(i) Draw a line PQ 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.

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

(iv) Join OR. Thus, \(\angle\)QOR = \(\angle\)POQ = \(90^{\circ}\) .

(v) Draw \(\overline{OD}\) the bisector of \(\angle\)POR. Thus, \(\angle\)QOD is required angle of \(135^{\circ}\) .

(vi) Now, draw \(\overline{OE}\) as the bisector of \(\angle\)QOD.

Thus, \(\angle\)QOE = \(\angle\)DOE = \(67\frac{1}{2}^{\circ}\)

 

Question 8:

Draw an angle of \(70^{\circ}\) . Make a copy of it using only a straight edge and compasses.

Answer:

Steps of construction:

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(i) Draw an angle \(70^{\circ}\) with protractor, i.e., \(\angle\)POQ = \(70^{\circ}\)

(ii) Draw a ray \(\overline{AB}\).

(iii) Place the compasses at O and draw an arc to cut the rays of \(\angle\)POQ at L and M.

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

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

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

(vii) Join AY.

Thus, \(\angle\)YAX = \(70^{\circ}\)

 

Question 9:

Draw an angle of \(40^{\circ}\). Copy its supplementary angle.

Answer:

Steps of construction:

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(i) Draw an angle of \(40^{\circ}\) with the help of protractor, naming \(\angle\)AOB.

(ii) Draw a line PQ.

(iii) Take any point M on PQ.

(iv) Place the compasses at O and draw an arc to cut the rays of \(\angle\)AOB at L and N.

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

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

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

(viii)Join MY. Thus, \(\angle\)QMY = \(40^{\circ}\) and \(\angle\)PMY is supplementary of it