# Practical geometry

**Class –VI Mathematics**

**(Ex. 14.1)**

**Question 1.**Draw a circle of radius 3.2 cm.

**Answer: 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.

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

**Answer: Steps of construction**:

(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.

NCERT Solutions for Class 6 Maths Exercise 14.1

**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 ∠∠A = ∠∠B = ∠∠C = ∠∠D = 90∘,90∘, , i.e. each angle is of 90∘.90∘.

Henc,e 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∘,90∘, , i.e. each angle is of 90∘.90∘.

Hence, it is a square.

NCERT Solutions for Class 6 Maths Exercise 14.1

**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:**

(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.

NCERT Solutions for Class 6 Maths Exercise 14.1

**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¯¯¯¯¯¯¯AB¯ and CD¯¯¯¯¯¯¯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 ∠∠COB is 90∘.90∘.

**(Ex. 14.2)**

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

**Answer: 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¯¯¯¯¯¯¯AB¯ is the required line segment of length 7.3 cm.

NCERT Solutions for Class 6 Maths Exercise 14.2

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

**Answer: Steps of construction**:

(i) Draw a line ′l′.′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′′l′ at B.

AB¯¯¯¯¯¯¯AB¯ is the required line segment of length 5.6 cm.

NCERT Solutions for Class 6 Maths Exercise 14.2

**Question 3.**Construct AB¯¯¯¯¯¯¯AB¯ of length 7.8 cm. From this cut off AC¯¯¯¯¯¯¯AC¯ of length 4.7 cm. Measure BC¯¯¯¯¯¯¯.BC¯.

**Answer: 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¯¯¯¯¯¯¯BC¯, we find that BC = 3.1 cm

NCERT Solutions for Class 6 Maths Exercise 14.2

**Question 4.**Given AB¯¯¯¯¯¯¯AB¯ of length 3.9 cm, construct PQ¯¯¯¯¯¯¯PQ¯ such that the length PQ¯¯¯¯¯¯¯PQ¯ is twice that of AB¯¯¯¯¯¯¯AB¯. Verify by measurement.

(**Hint**: Construct PX¯¯¯¯¯¯¯PX¯ such that length of PX¯¯¯¯¯¯¯PX¯ = length of AB¯¯¯¯¯¯¯AB¯; then cut off XQ¯¯¯¯¯¯¯¯XQ¯ such that XQ¯¯¯¯¯¯¯¯XQ¯ also has the length of AB¯¯¯¯¯¯¯AB¯.

**Answer: Steps of construction**:

(i) Draw a line ′l′.′l′.

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

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

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

**Verification**:

By measurement we find that PQ = 7.8 cm

= 3.9 cm + 3.9 cm = AB¯¯¯¯¯¯¯AB¯ + AB¯¯¯¯¯¯¯AB¯ = 2 x AB¯¯¯¯¯¯¯AB¯

NCERT Solutions for Class 6 Maths Exercise 14.2

**Question 5.**Given AB¯¯¯¯¯¯¯AB¯ of length 7.3 cm and CD¯¯¯¯¯¯¯CD¯ of length 3.4 cm, construct a line segment XY¯¯¯¯¯¯¯¯XY¯ such that the length of XY¯¯¯¯¯¯¯¯XY¯ is equal to the difference between the lengths of AB¯¯¯¯¯¯¯AB¯ and CD¯¯¯¯¯¯¯.CD¯. Verify by measurement.

**Answer: Steps of construction**:

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

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

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

(iv) Thus the length of XY¯¯¯¯¯¯¯¯XY¯ = length of AB¯¯¯¯¯¯¯AB¯ – length of CD¯¯¯¯¯¯¯CD¯

**Verification**:

By measurement we find that length of XY¯¯¯¯¯¯¯¯XY¯ = 3.9 cm

= 73. Cm – 3.4 cm = AB¯¯¯¯¯¯¯–CD¯¯¯¯¯¯¯

**(Ex. 14.3)**

**Question 1.**Draw any line segment PQ¯¯¯¯¯¯¯.PQ¯. Without measuring PQ¯¯¯¯¯¯¯,PQ¯, construct a copy of PQ¯¯¯¯¯¯¯.PQ¯.

**Answer: Steps of construction**:

(i) Given PQ¯¯¯¯¯¯¯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¯¯¯¯¯¯¯PQ¯.

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

Draw an arc that cuts ′l′′l′ at a point, say B. Now AB¯¯¯¯¯¯¯AB¯ is a copy of PQ¯¯¯¯¯¯¯PQ¯.

NCERT Solutions for Class 6 Maths Exercise 14.3

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

**Answer:** Steps of construction:

(i) Given AB¯¯¯¯¯¯¯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¯¯¯¯¯¯¯AB¯.

(iii) Draw any line ′l′.′l′. Choose a point P on ′l′.′l′. Without changing the compasses setting, place the pointer on Q.

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

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

Thus PQ¯¯¯¯¯¯¯PQ¯ is the required line segment whose length is twice that of AB.

**(Ex. 14.4)**

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

**Answer: 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.

NCERT Solutions for Class 6 Maths Exercise 14.4

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

**Answer: Steps of construction**:

(i) Place a set-square on PQ←→PQ↔ such that one arm of its right angle aligns along PQ←→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←→PQ↔ at M. Then RM ⊥⊥ PQ.

NCERT Solutions for Class 6 Maths Exercise 14.4

**Question 3.**Draw a line ll and a point X on it. Through X, draw a line segment XY¯¯¯¯¯¯¯¯XY¯ perpendicular to l.l.

Now draw a perpendicular to XY¯¯¯¯¯¯¯¯XY¯ to Y. (use ruler and compasses)

**Answer:** Steps of construction:

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

(ii) With X as centre and a convenient radius, draw an arc intersecting the line ′l′′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′′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.

**(Ex. 14.5)**

**Question 1.**Draw AB¯¯¯¯¯¯¯AB¯ of length 7.3 cm and find its axis of symmetry.

**Answer:** Axis of symmetry of line segment AB¯¯¯¯¯¯¯AB¯ will be the perpendicular bisector of AB¯¯¯¯¯¯¯.AB¯. So, draw the perpendicular bisector of AB.

**Steps of construction**:

(i) Draw a line segment AB¯¯¯¯¯¯¯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: Steps of construction**:

(i) Draw a line segment AB¯¯¯¯¯¯¯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 AB¯¯¯¯¯¯¯AB¯.

NCERT Solutions for Class 6 Maths Exercise 14.5

**Question 3.**Draw the perpendicular bisector of XY¯¯¯¯¯¯¯¯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¯¯¯¯¯¯¯¯XY¯, what can you say about the lengths MX and XY?

**Answer: Steps of construction**:

(i) Draw a line segment XY¯¯¯¯¯¯¯¯XY¯ = 10.3 cm

(ii) Taking X and Y 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 required perpendicular bisector of XY¯¯¯¯¯¯¯¯XY¯.

Now:

(a) Take any point P on the bisector drawn. With the help of divider we can check that PX¯¯¯¯¯¯¯=PY¯¯¯¯¯¯¯PX¯=PY¯.

(b) If M is the mid-point of XY¯¯¯¯¯¯¯¯,XY¯, then MX¯¯¯¯¯¯¯¯=12XY¯¯¯¯¯¯¯¯MX¯=12XY¯.

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

**Answer: Steps of construction**:

(i) Draw a line segment AB = 12.8 cm

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

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

(iv) Again, draw the perpendicular bisector of CB¯¯¯¯¯¯¯CB¯ which cuts it at E. Thus, E is the mid-point of CB¯¯¯¯¯¯¯CB¯.

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

(vi) By actual measurement, we find that

AD¯¯¯¯¯¯¯¯=DC¯¯¯¯¯¯¯=CE¯¯¯¯¯¯¯=EB¯¯¯¯¯¯¯=3.2AD¯=DC¯=CE¯=EB¯=3.2 cm

NCERT Solutions for Class 6 Maths Exercise 14.5

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

**Answer: Steps of construction**:

(i) Draw a line segment PQ¯¯¯¯¯¯¯PQ¯ = 6.1 cm.

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

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

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

**Answer: Steps of construction**:

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

(ii) Draw any chord AB¯¯¯¯¯¯¯.AB¯.

(iii) Taking A and B as centers and radius more than half of AB¯¯¯¯¯¯¯,AB¯, draw two arcs which cut each other at P and Q.

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

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

NCERT Solutions for Class 6 Maths Exercise 14.5

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

**Answer: Steps of construction**:

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

(ii) Draw its diameter AB¯¯¯¯¯¯¯AB¯.

(iii) Taking A and B as centers 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¯¯¯¯¯¯¯AB¯.

We observe that this perpendicular bisector of AB¯¯¯¯¯¯¯AB¯ passes through the centre C of the circle.

NCERT Solutions for Class 6 Maths Exercise 14.5

**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: Steps of construction**:

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

(ii) Draw any two chords AB¯¯¯¯¯¯¯AB¯ and CD¯¯¯¯¯¯¯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¯¯¯¯¯¯¯CD¯.

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

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

NCERT Solutions for Class 6 Maths Exercise 14.5

**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¯¯¯¯¯¯¯¯OA¯ and OB¯¯¯¯¯¯¯.OB¯. Let them meet at P. Is PA = PB?

**Answer: 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¯¯¯¯¯¯¯¯OA¯ and OB¯¯¯¯¯¯¯.OB¯.

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

With the help of divider, we check that PA¯¯¯¯¯¯¯=PB¯¯¯¯¯¯¯.

**(Ex. 14.6)**

**Question 1.**Draw ∠∠POQ of measure 75∘75∘ and find its line of symmetry.

**Answer: Steps of construction**:

(a) Draw a line ll 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 ll at A.

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

(d) Join OB, then ∠∠BOA = 60∘.60∘.

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

(f) Draw bisector of ∠∠BOC. The angle is of 90∘.90∘. Mark it at D. Thus, ∠∠DOA = 90∘90∘

(g) Draw OP¯¯¯¯¯¯¯OP¯ as bisector of ∠∠DOB.

Thus, ∠∠POA = 75∘75∘

**Question 2.**Draw an angle of measure 147∘147∘ and construct its bisector.

**Answer: Steps of construction**:

(a) Draw a ray OA←→.OA↔.

(b) With the help of protractor, construct ∠∠AOB = 147∘.147∘.

(c) Taking centre O and any convenient radius, draw an arc which intersects the arms OA¯¯¯¯¯¯¯¯OA¯ and OB¯¯¯¯¯¯¯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¯¯¯¯¯¯¯¯OR¯ is the required bisector of ∠∠AOB.

NCERT Solutions for Class 6 Maths Exercise 14.6

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

**Answer: 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, ∠∠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¯¯¯¯¯¯¯¯OD¯ is the required bisector of ∠∠COQ.

NCERT Solutions for Class 6 Maths Exercise 14.6

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

**Answer: Steps of construction**:

(a) Draw a ray OA←→.OA↔.

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

(c) Draw OC¯¯¯¯¯¯¯¯OC¯ as the bisector of ∠∠AOB.

(d) Again, draw OD¯¯¯¯¯¯¯¯OD¯ as bisector of ∠∠AOC.

(e) Again, draw OE¯¯¯¯¯¯¯OE¯ as bisector of ∠∠BOC.

(f) Thus, OC¯¯¯¯¯¯¯¯OC¯, OD¯¯¯¯¯¯¯¯OD¯ and OE¯¯¯¯¯¯¯OE¯ divide ∠∠AOB in four equal arts.

NCERT Solutions for Class 6 Maths Exercise 14.6

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

(a) 60∘60∘

(b) 30∘30∘

(c) 90∘90∘

(d) 120∘120∘

(e) 45∘45∘

(f) 135∘135∘

**Answer:**

**Steps of construction**:

(a) 60∘60∘

(i) Draw a ray OA←→.OA↔.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA←→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∘.60∘.

(b) 30∘30∘

(i) Draw a ray OA←→.OA↔.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA←→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∘.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 30∘.30∘.

(c)90∘90∘

(i) Draw a ray OA←→.OA↔.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA←→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 centers 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∘.90∘.

(d)120∘120∘

(i) Draw a ray OA←→.OA↔.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA←→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∘.120∘.

(e)45o45o

(i) Draw a ray OA−→−OA→

(ii) Taking O as centre and convenient radius, mark an arc, which intersects OA−→−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 centers 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 90o90o

(vii) Draw the bisector of ∠∠BOA.

Thus, ∠∠MOA is required angle of 45o45o

(f)135o135o

(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 centers and radius more than half of AB, draw two arcs intersecting each other at R.

(iv) Join OR. Thus, ∠∠QOR = ∠∠POQ = 90o90o

(v) Draw OD−→−OD→ the bisector of ∠∠POR.

thus, ∠∠QOD is required angle of 135o135o

NCERT Solutions for Class 6 Maths Exercise 14.6

**Question 6.**Draw an angle of measure 45∘45∘ and bisect it.

**Answer: 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 centers 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 90o90o

(e)Draw OE−→−OE→ as the bisector of ∠∠COE. Thus, ∠∠QOE = 45o45o

(f)Again draw OG−→−OG→ as the bisector of ∠∠QOE.

Thus, ∠∠QOG = ∠∠EOG = 2212o2212o

NCERT Solutions for Class 6 Maths Exercise 14.6

**Question 7.**Draw an angle of measure 135∘135∘ and bisect it.

**Answer: 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 centers and radius more than half of AB, draw two arcs intersecting each other at R.

(d) Join OR. Thus, ∠∠QOR = ∠∠POQ = 90o90o

(e) Draw OD−→−OD→ the bisector of ∠∠POR. Thus, ∠∠QOD is required angle of 135o135o

(f) Now, draw OE−→−OE→ as the bisector of ∠∠QOD.

Thus, ∠∠QOE = ∠∠DOE = 6712o6712o

NCERT Solutions for Class 6 Maths Exercise 14.6

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

**Answer: Steps of construction**:

(a) Draw an angle 70o70o with protractor, i.e., ∠POQ=70o∠POQ=70o

(b) Draw a ray AB−→−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=70o∠YAX=70o

NCERT Solutions for Class 6 Maths Exercise 14.6

**Question 9.**Draw an angle of 40∘.40∘. Copy its supplementary angle.

**Answer:** Steps of construction:

(a) Draw an angle of 40o40o 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 = 40o40o and ∠∠PMY is supplementary of it.