NCERT Solutions Class 6 Mathematics Chapter 14 Practical Geometry

Practical Geometry

 

Exercise 14.1

Q.1) Draw a circle of radius 3.2 𝑐𝑚.
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 𝑐𝑚 and 2.5 𝑐𝑚.
Sol.2) (a) Marks a point ‘O’ 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 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 = ∠𝐷 = 90° , i.e. each angle is of 90°.
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 ‘O’ with sharp pencil where we want centre of the circle.
(ii) Place the pointer of the compasses at ‘O’. 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 ∠𝐶𝑂𝐵 is 90°.

Exercise 14.2

Q.1) Draw a line segment of length 7.3 𝑐𝑚 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 𝑐𝑚 from A.
(iii) Join AB.
AB is the required line segment of length 7.3 𝑐𝑚.

Q.2) Construct a line segment of length 5.6 𝑐𝑚 using ruler and compasses
Sol.2) Steps of construction:
(i) Draw a line 𝑙 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 𝑐𝑚 mark.

(iii) Without changing the opening of the compasses. Place the pointer on A and cut an arc 𝑙 at B.
AB is the required line segment of length 5.6 𝑐𝑚

Q.3) Construct AB of length 7.8 𝑐𝑚. From this cut off AC of length 4.7 𝑐𝑚. 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 𝑐𝑚 from A.
(iii) Again, mark a point C at a distance 4.7 from A.
By measuring BC, we find that 𝐵𝐶 = 3.1 𝑐𝑚

Q.4) Given AB of length 3.9 𝑐𝑚, construct PQ such that the length PQ is twice that of AB.
Verify by measurement.
Sol.4) Steps of construction:
(i) Draw a line 𝑙.
(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 𝑃𝑄 = 7.8 𝑐𝑚
= 3.9 𝑐𝑚 + 3.9 𝑐𝑚 = AB + AB = 2 × AB

Q.5) Given AB of length 7.3 cm and CD of length 3.4 𝑐𝑚, 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 𝑙 and take a point X on it.
(ii) Construct XZ such that length XZ = length of AB = 7.3𝑐𝑚
(iii) Then cut off ZY = length of CD = 3.4 𝑐𝑚
(iv) Thus the length of XY = length of AB − length of CD

Verification:
By measurement we find that length of XY = 3.9 𝑐𝑚
= 7.3 𝑐𝑚 – 3.4 𝑐𝑚 = AB − 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 𝑃 and the pencil end on 𝑄. The opening of the instrument now gives the length of 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 𝑙 at a point, say 𝐵. 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 𝐴 and the pencil end on 𝐵. The opening of the instrument now gives the length of 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 PQ is the required line segment whose length is twice that of 𝐴𝐵.

Exercise 14.4

Q.1) Draw any line segment AB. Mark any point 𝑀 on it. Through 𝑀, draw a perpendicular to AB. (Use ruler and compasses)
Sol.1) Steps of construction:

(i) With 𝑀 as 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 the point 𝑀.

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 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.

Q.3) Draw a line 𝑙 and a point 𝑋 on it. Through 𝑋, draw a line segment ̅𝑋̅̅𝑌̅ perpendicular to 𝑙.
Now draw a perpendicular to ̅𝑋̅̅𝑌̅ to 𝑌. (use ruler and compasses)
Sol.3) Steps of construction:
(i) Draw a line 𝑙 and take point 𝑋 on it.
(ii) With X as 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 D as 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.

Exercise 14.5

Q.1) Draw AB of length 7.3 𝑐𝑚 and find its axis of symmetry.
Sol.1) The below given steps will be followed to construct a line segment 𝐴𝐵 of length 7.3 𝑐𝑚 & to find its axis of symmetry

i) Draw a lone segment of 7.3 𝑐𝑚
ii) Taking 𝐴 as centre, draw a circle by using compasses. The radius of circle should be more than half the length of 𝐴𝐵

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

Q.3) Draw the perpendicular bisector of XY whose length is 10.3 𝑐𝑚.
(a) Take any point 𝑃 on the bisector drawn. Examine whether PX = PY.
(b) If 𝑀 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 𝑐𝑚

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 𝐴𝐵 = 12.8 𝑐𝑚
(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 𝑐𝑚

Q.5) With PQ of length 6.1 𝑐𝑚 as diameter, draw a circle.
Sol.5) Steps of construction:

(i) Draw a line segment 𝑃𝑄 = 6.1 𝑐𝑚.
(ii) Draw the perpendicular bisector of 𝑃𝑄 which cuts, it at O. Thus O is the mid-point of 𝑃𝑄.
Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment 𝑃𝑄.

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 𝑐𝑚.
(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 𝑐𝑚.
(ii) Draw its diameter 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 𝐴𝐵.
We observe that this perpendicular bisector of 𝐴𝐵 passes through the centre C of the circle.

Q.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?
Sol.8) 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 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 𝐶𝐷.
(v) Similarly draw 𝐺𝐻 the perpendicular bisector of chord 𝐶𝐷.
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 𝑂𝐴 = 𝑂𝐵. Draw the perpendicular bisectors of 𝑂𝐴 and 𝑂𝐵. Let them meet at P. Is 𝑃𝐴 = 𝑃𝐵?
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 𝑂𝐴 and 𝑂𝐵
(iv) Let them meet at P. Join 𝑃𝐴 and 𝑃𝐵.
With the help of divider, we check that 𝑃𝐴 = 𝑃𝐵

Exercise 14.6

Q.1) Draw ∠ POQ of measure 75° and find its line of symmetry.
Sol.1) Steps of construction:

(a) Draw 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 90°. Mark it at D. Thus, ∠𝐷𝑂𝐴 = 90°
(g) Draw 𝑂𝑃 as bisector of ∠𝐷𝑂𝐵.
Thus, ∠𝑃𝑂𝐴 = 75°

Q.2) Draw an angle of measure 147° and construct its bisector.
Sol.2) Steps of construction:

(a) Draw a ray OA
(b) With the help of protractor, construct ∠𝐴𝑂𝐵 = 147°.
(c) Taking centre O and any convenient radius, draw an arc which intersects the arms 𝑂𝐴 and 𝑂𝐵¯ at P and Q respectively.
(d) Taking P as 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 intersects the previous at R.
(f) Join 𝑂𝑅 and produce it.
Thus, 𝑂𝑅 is the required bisector of ∠𝐴𝑂𝐵.

Q.3) Draw a right angle and construct its bisector.
Sol.3) 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 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 which intersect each other at the point D.
(f) Join 𝑂𝐷.
Thus, 𝑂𝐷 is the required bisector of ∠𝐶𝑂𝑄.

Q.4) Draw an angle of measure 153° 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 ∠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.

Q.5) Construct with ruler and compasses, angles of following measures:
(a) 60° (b) 30° (c) 90° (d) 120° (e) 45° (f) 135°
Sol.5) Steps of construction:

(a) 60°
(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, ∠𝐵𝑂𝐴 is required angle of 60°.
(b) 30°

(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, ∠𝐵𝑂𝐴 is required angle of 60°.
(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, ∠𝐶𝑂𝐴 is required angle of 60°.
(c) 90°

(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, ∠𝐵𝑂𝐴 is required angle of 90°
(d) 120°

(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, ∠𝐴𝑂𝐷 is required angle of 120°
(e) 45°

(i) Draw a ray 𝑂𝐴
(ii) Taking O as centre and convenient radius, mark an arc, which intersects 𝑂 𝐴 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, ∠𝐵𝑂𝐴 is required angle of 90°
(vii) Draw the bisector of ∠𝐵𝑂𝐴.
Thus, ∠𝑀𝑂𝐴 is required angle of 45°
(f) 135°

(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 = 90°
(v) Draw 𝑂 𝐷 the bisector of ∠𝑃𝑂𝑅.
thus, ∠𝑄𝑂𝐷 is required angle of 135°

Q.6) Draw an angle of measure 45° 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 𝑂𝐶. Then ∠𝐶𝑂𝑄 is an angle of 90°
(e) Draw 𝑂 𝑒 as the bisector of ∠𝐶𝑂𝐸. Thus, ∠𝑄𝑂𝐸 = 45°
(f) Again draw 𝑂 𝐺 as the bisector of ∠𝑄𝑂𝐸.
Thus, ∠𝑄𝑂𝐺 = ∠𝐸𝑂𝐺 = 22(1°/2)°

Q.7) Draw an angle of measure 135° 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 𝐴𝐵, draw two arcs intersecting each other at R.
(d) Join 𝑂𝑅. Thus, ∠𝑄𝑂𝑅 = ∠𝑃𝑂𝑄 = 90°
(e) Draw 𝑂 𝐷 the bisector of ∠𝑃𝑂𝑅. Thus, ∠𝑄𝑂𝐷 is required angle of 135°
(f) Now, draw 𝑂 𝐸 as the bisector of ∠∠QOD.
Thus, ∠𝑄𝑂𝐸 = ∠𝐷𝑂𝐸 = 67(1°/2)

Q.8) Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.
Sol.8) Steps of construction:

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

Q.9) Draw an angle of 40°. Copy its supplementary angle.
Sol.9) Steps of construction:

(a) Draw an angle of 40° with the help of protractor, naming ∠𝐴𝑂𝐵.
(b) Draw a line PQ.
(c) Take any point M on PQ.
(d) Place the compasses at O and draw an arc to cut the rays of ∠𝐴𝑂𝐵 at L and N.
(e) Use the same compasses setting to draw an arc O as 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, ∠𝑄𝑀𝑌 = 40° and ∠𝑃𝑀𝑌 is supplementary of it.