Practical Geometry-Solutions 14.6

CBSE Class –VI Mathematics
NCERT Solutions
Chaper 14 Practical Geometry (Ex. 14.6)

---

Question 1. Draw ∠$\mathrm{\angle }$POA of measure 75∘${75}^{\circ }$ and find its line of symmetry.

Answer: Steps of construction:

(a) Draw a line l$l$ 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$l$ at A.

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

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

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

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

(g) Draw OP¯¯¯¯¯¯¯$\overline{\text{OP}}$ as bisector of ∠$\mathrm{\angle }$DOB.

Thus, ∠$\mathrm{\angle }$POA = 75∘${75}^{\circ }$

Question 2. Draw an angle of measure 147∘${147}^{\circ }$ and construct its bisector.

Answer: Steps of construction:

(a) Draw a ray OA←→.$\overleftrightarrow{\text{OA}}.$

(b) With the help of protractor, construct ∠$\mathrm{\angle }$AOB = 147∘.${147}^{\circ }.$

(c) Taking centre O and any convenient radius, draw an arc which intersects the arms OA¯¯¯¯¯¯¯¯$\overline{\text{OA}}$ and OB¯¯¯¯¯¯¯$\overline{\text{OB}}$ at P and Q respectively.

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

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

(f) Join OR and produce it.

Thus, OR¯¯¯¯¯¯¯¯$\overline{\text{OR}}$ is the required bisector of ∠$\mathrm{\angle }$AOB.

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 the 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, ∠$\mathrm{\angle }$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, OD¯¯¯¯¯¯¯¯$\overline{\text{OD}}$ is the required bisector of ∠$\mathrm{\angle }$COQ.

Question 4. Draw an angle of measure 153∘${153}^{\circ }$ and divide it into four equal parts.

Answer: Steps of construction:

(a) Draw a ray OA←→.$\overleftrightarrow{\text{OA}}.$

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

(c) Draw OC¯¯¯¯¯¯¯¯$\overline{\text{OC}}$ as the bisector of ∠$\mathrm{\angle }$AOB.

(d) Again, draw OD¯¯¯¯¯¯¯¯$\overline{\text{OD}}$ as bisector of ∠$\mathrm{\angle }$AOC.

(e) Again, draw OE¯¯¯¯¯¯¯$\overline{\text{OE}}$ as bisector of ∠$\mathrm{\angle }$BOC.

(f) Thus, OC¯¯¯¯¯¯¯¯$\overline{\text{OC}}$, OD¯¯¯¯¯¯¯¯$\overline{\text{OD}}$ and OE¯¯¯¯¯¯¯$\overline{\text{OE}}$ divide ∠$\mathrm{\angle }$AOB in four equal arts.

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

(a) 60∘${60}^{\circ }$

(b) 30∘${30}^{\circ }$

(c) 90∘${90}^{\circ }$

(d) 120∘${120}^{\circ }$

(e) 45∘${45}^{\circ }$

(f) 135∘${135}^{\circ }$

Answer:

Steps of construction:

(a) 60∘${60}^{\circ }$

(i) Draw a ray OA←→.$\overleftrightarrow{\text{OA}}.$

(ii) Taking O as the centre and convenient radius, mark an arc, which intersects OA←→$\overleftrightarrow{\text{OA}}$ at P.

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

(iv) Join OQ and extend to B 

Thus, ∠$\mathrm{\angle }$BOA is required angle of 60∘.${60}^{\circ }.$

(b) 30∘${30}^{\circ }$

(i) Draw a ray OA←→.$\overleftrightarrow{\text{OA}}.$

(ii) Taking O as the centre and convenient radius, mark an arc, which intersects OA←→$\overleftrightarrow{\text{OA}}$ at P.

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

(iv) Join OQ. Thus, ∠$\mathrm{\angle }$BOA is the required angle of 60∘.${60}^{\circ }.$

(v) Put the pointer on P and mark an arc.

(vi) Put the pointer on Q and with the same radius, cut the previous arc at C.

Thus, ∠$\mathrm{\angle }$COA is required angle of 30∘.${30}^{\circ }.$

(c) 90∘${90}^{\circ }$

(i) Draw a ray OA←→.$\overleftrightarrow{\text{OA}}.$

(ii) Taking O as the centre and convenient radius, mark an arc, which intersects OA←→$\overleftrightarrow{\text{OA}}$ at X.

(iii) Taking X as the centre and the same radius, cut the previous arc at Y.

(iv) Taking Y as the centre and the same radius, draw another arc intersecting the same arc at Z.

(v) Taking Y and Z as centres and the same radius, draw two arcs intersecting each other at S.

(vi) Join OS and produce it to form a ray OB.

Thus, ∠$\mathrm{\angle }$BOA is required angle of 90∘.${90}^{\circ }.$

(d) 120∘${120}^{\circ }$

(i) Draw a ray OA←→.$\overleftrightarrow{\text{OA}}.$

(ii) Taking O as the centre and convenient radius, mark an arc, which intersects OA←→$\overleftrightarrow{\text{OA}}$ at P.

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

(iv) Taking Q as the centre and the same radius cut the arc at S.

(v) Join OS.

Thus, ∠$\mathrm{\angle }$AOD is required angle of 120∘.${120}^{\circ }.$

(e) 45o${45}^o$

 points y and z are not marked in the figure

(i) Draw a ray OA−→−$\overrightarrow{OA}$

(ii) Taking O as the centre and convenient radius, mark an arc, which intersects OA−→−$\overrightarrow{OA}$ at X.

(iii) Taking X as the centre and the same radius, cut the previous arc at Y.

(iv) Taking Y as the centre and the same radius, draw another arc intersecting the same arc at Z.

(v) Taking Y and Z as centres and the same radius, draw two arcs intersecting each other at S.

(vi) Join OS and produce it to form a ray OB. Thus, ∠$\mathrm{\angle }$BOA is required angle of 90o${90}^o$

(vii) Draw the bisector of ∠$\mathrm{\angle }$BOA.

Thus, ∠$\mathrm{\angle }$MOA is required angle of 45o${45}^o$

(f) 135o${135}^o$

 the angle 135 degrees should be shown in between QOD

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

(ii) Taking O as the 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, ∠$\mathrm{\angle }$QOR = ∠$\mathrm{\angle }$POR = 90o${90}^o$

(v) Draw OD−→−$\overrightarrow{OD}$ the bisector of ∠$\mathrm{\angle }$POR.

thus, ∠$\mathrm{\angle }$QOD is the required angle of 135o${135}^o$

Question 6.Draw an angle of measure 45∘${45}^{\circ }$ and bisect it.

Answer: Steps of construction:

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

(b)Taking O as the 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 ∠$\mathrm{\angle }$COQ is an angle of 90o${90}^o$

(e)Draw OE−→−$\overrightarrow{OE}$ as the bisector of ∠$\mathrm{\angle }$COQ. Thus, ∠$\mathrm{\angle }$QOE = 45o${45}^o$

(f)Again draw OG−→−$\overrightarrow{OG}$ as the bisector of ∠$\mathrm{\angle }$QOE.

Thus, ∠$\mathrm{\angle }$QOG = ∠$\mathrm{\angle }$EOG = 2212o$22\frac{1}{2}{}^o$

Question 7.Draw an angle of measure 135∘${135}^{\circ }$ and bisect it.

Answer: Steps of construction:

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

(b) Taking O as a 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, ∠$\mathrm{\angle }$QOR = ∠$\mathrm{\angle }$POQ = 90o${90}^o$

(e) Draw OD−→−$\overrightarrow{OD}$ the bisector of ∠$\mathrm{\angle }$POR. Thus, ∠$\mathrm{\angle }$QOD is the required angle of 135o${135}^o$

(f) Now, draw OE−→−$\overrightarrow{OE}$ as the bisector of ∠$\mathrm{\angle }$QOD.

Thus, ∠$\mathrm{\angle }$QOE = ∠$\mathrm{\angle }$DOE = 6712o$67\frac{1}{2}{}^o$

Question 8.Draw an angle of 70∘.${70}^{\circ }.$ Make a copy of it using only a straight edge and compasses.

Answer: Steps of construction:

(a) Draw an angle 70o${70}^o$ with protractor, i.e., ∠POQ=70o$\mathrm{\angle }POQ={70}^o$

(b) Draw a ray AB−→−$\overrightarrow{AB}$

(c) Place the compasses at O and draw an arc to cut the rays of ∠$\mathrm{\angle }$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=70o$\mathrm{\angle }YAX={70}^o$

Question 9.Draw an angle of 40∘.${40}^{\circ }.$ Copy its supplementary angle.

Answer: Steps of construction:

(a) Draw an angle of 40o${40}^o$ with the help of protractor, naming ∠$\mathrm{\angle }$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 ∠$\mathrm{\angle }$AOB at L and N.

(e) Use the same compasses setting to draw an arc M 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, ∠$\mathrm{\angle }$QMY = 40o${40}^o$ and ∠$\mathrm{\angle }$PMY =100 degrees is supplementary of it.