Home » Class 6 Maths » NCERT Solution Class 6 Maths Chapter – 14 Practical Geometry

# NCERT Solution Class 6 Maths Chapter – 14 Practical Geometry

Exercise 14.1

Question 1:Draw a circle with circle of radius 3.2 and write down it’s procedures.
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. Question 2: With the same centre 0, draw two circles of radii 4 cm and 2.5 cm
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. 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?

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

(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. 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.

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 ∠ COB is 90. Exercise 14.2

Question 1:A line segment was drawn of length 10cm, with the help of the ruler. (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
(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 Question 3: Construct XY⎯ of certain distance 6cm .from that cut off XO⎯of 4cm.Measure OY⎯ (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 OY⎯ we can find the OY=4cm

Question 4:Given PQ⎯of length 5cm , Draw AB⎯such that the length
XB⎯also it has the length of PQ⎯ Construction step: (1)Line is drawn ‘l’

(2)Draw AX⎯ such that its length of AX⎯= PQ⎯⎯ length

(3)Cut of XB⎯ so that XB⎯ will have the same length of PQ⎯

(4) AX⎯length and the XB⎯ length added together doubles the PQ⎯

Check:

Now, measurement can find that AB=10cm

=5cm+5cm

= PQ⎯ + PQ⎯ = 2* 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

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 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 AB⎯. Not measuring AB⎯.Draw a duplicate of AB⎯
Method of construction. (1)Now the mentioned 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 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 PQ⎯ is the copy of AB⎯

Question 2. Mention some segment PQ⎯ ,the length of it is unknown, Draw AB⎯ the length of AB⎯ is double the times of PQ⎯ Method of construction:

(1)Now that PQ⎯ whose certain length is unknown
(2)The compasses pointer on point P is fixed and” width=”163″ height=”242″ class=”alignnone size-full” /> then pencil end on Q .The instrument opening gives the
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 PQ⎯ .Point O is marked on it any where.On the point O perpendicular is drawn(by using compasses and ruler) 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 ) construction method:

(i)A set-square is placed on AB⎯ such that one arm of its right angle aligns along 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) 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 XY⎯ of length 10 cm and What is the axis of symmetry

Axis of symmetry of line segment XY⎯ will be the perpendicular bisector of 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. Question 2:Construct a segment line of length 8 cm and draw its perpendicular bisector.

Construction step:

(i) A line segment 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 XY⎯. Question 3:
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 AB⎯ midpoint, the lengths XA and XB explain about it? Construction steps:

(1)A line segment is drawn 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 AB⎯.

Now,

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

(2) If X is the midpoint of AB⎯ , then XA⎯ =1/2 XB⎯

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

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

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

(3) Draw PX⎯ perpendicular bisector which cuts it at P. Thus P is the midpoint of.

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

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

(6) By actual measurement, we find that PA⎯ = AX⎯ = XB⎯ = BQ⎯ =2.5cm

Question 5:Draw a circle with AB⎯ of length 5 cm as diameter,. construction steps:

(1) AB⎯ a line segment is drawn = 5 cm.

(2) 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 AB⎯

Question 6:Draw a circle with centre O and radius 5 cm. Draw any chord PQ⎯. Construct the perpendicular bisector PQ⎯ and examine if it passes through O. Steps of construction:

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

(ii)Chord PQ⎯ is drawn.

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

(iv) XY is joined. Then XY is the perpendicular bisector of PQ⎯.

(v) Through the centre O of the circle this perpendicular bisector of PQ⎯ passes.

Question 7:

Question 6 is repeated, if PQ⎯ be a diameter Construction steps:

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

(2) 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 PQ⎯.

(5)We observe that this perpendicular bisector of 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? 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? 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 ∠750 and find its line of symmetry.

(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 ∠BOA=60∘ .

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

(f) Draw bisector of ∠BOC . The angle is of ∠BOC. Mark it at D. Thus, ∠DOA=900

(g) Draw OP⎯ as bisector of ∠DOB.

Thus, ∠POA=750 Question 2: Draw an angle of measure 1470
and construct its bisector.

(a) Draw a line OA⎯.

(b) Using protractor, construct ∠AOB=1470
.

(c) Taking the centre as O and any suitable radius, draw an arc which cuts the arms OA⎯ and 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, OR⎯is the required bisector of ∠AOB. Question 3: Construct a right angle and construct its bisector.

(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, ∠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, OD⎯ is the required bisector of ∠COQ .

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

(a) Draw a ray OA⎯ .

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

(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 in four equal parts. Question 5: Construct with ruler and compasses, angles of following measures:

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

Steps of construction:
(1)600 (i) Draw a ray OL⎯.

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

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

(iv) Join OQ.

Thus, ∠MOL is required angle of 60 .

(2) 30 0 (i) Draw a ray OL⎯⎯.

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

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

(iv) Join OQ. Thus, ∠MOL is required angle of 600
.

(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, ∠NOL is required angle of 300
.

(3) 900 (i) Draw a ray OL⎯.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects 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, ∠MOL is required angle of 900
.

(4) 1200 (i) Draw a ray OI⎯.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects 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, ∠ IOL is required angle of 120∘.

(5) 450 (i) Draw a ray OI⎯⎯.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects 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, ∠ JOI is required angle of 90∘ .

(vii) Draw the bisector of ∠JOI.

Thus, ∠MOI is required angle of 450 .

(6) 1350 (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, ∠QOR = ∠POQ = 900.

(v) Draw OL⎯ the bisector of ∠POR.

Thus, ∠QOL is required angle of 1350

Question 6: Draw an angle of measure 450 and bisect it.

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. (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 ∠COQ is an angle of 90

(v) Draw OE⎯ as the bisector of ∠COE. Thus, ∠QOE =900

(vi) Again draw OG⎯ as the bisector of ∠QOE.

Thus, ∠QOG = ∠EOG = 22120

Question 7:Draw an angle of measure 1350 and bisect it.

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. (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, ∠QOR = ∠POQ = 900 .

(v) Draw OD⎯the bisector of ∠POR. Thus, ∠QOD is required angle of 1350.

(vi) Now, draw OE⎯as the bisector of ∠QOD.

Thus, ∠QOE = ∠DOE = 67120

Question 8:Draw an angle of 700 . Make a copy of it using only a straight edge and compasses.

Steps of construction: (i) Draw an angle 700 with protractor, i.e., ∠POQ = 700

(ii) Draw a ray AB⎯.

(iii) Place the compasses at O and draw an arc to cut the rays of ∠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, ∠YAX = 700

Question 9: Draw an angle of 400. Copy its supplementary angle.

Steps of construction: (i) Draw an angle of 40∘ with the help of protractor, naming ∠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 ∠AOB at L and N.

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