NCERT Solutions for Class 6 Maths Chapter 14: Practical Geometry - Exercise 14.6

Exercise 14.6

1. Draw \[\angle POQ\] of measure ${75^ \circ }$and find its line of symmetry.

Ans: (a) At first draw a line P and mark a point O on it.

(b )Then place the pointer of the compass at O and draw an arc on the line. The arc intersects the line P at T.

(c) Now considering the same radius, with centre T, cut the previous arc at S.

(d) Considering the same radius, with centre S, cut the previous arc at R.

(e) Then draw a bisector of $\angle SOR$. This is an angle of ${90^ \circ }$. Denote it as N.

(f) Now draw bisector of $\angle SON$, which will intersect arc at X.

Thus, $\angle POQ = {75^ \circ }$

Now we will find the line of symmetry of ${75^ \circ }$

We will draw two arcs from X and T with a fixed radius, intersecting at point Y.

Join the line OY which is the line of symmetry of the angle POQ.

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

Ans:

(a) Draw a line OP.

(b) With the help of protractor, construct $\angle POQ = {147^ \circ }$

(c) Consider centre O and draw an arc which intersects the arms $\overline {OP} $ and $\overline {OQ} $ at S and T respectively.

(d) Consider S as the centre and radius more than half of ST and draw an arc.

(e) Considering T as centre and with the same radius, draw another arc which intersects the previous at U.

(f) Join OU and produce it.

(g) OU is the required bisector of $\angle POQ = {147^ \circ }$.

3. Draw a right angle and construct its bisector.

Ans.

(a) Draw a line AB and mark a point called O on it.

(b) Considering O as the centre and considering a convenient radius, draw an arc intersecting AB at C and D.

(c) Considering C and D as centres and considering radius as more than half of CD.

(d) Draw two arcs which intersect each other at E.

(e) Join OE. 

Thus, $\angle EOB$ is the required right-angle.

(f) Considering D and G as centre and radius more than half of DG.

(g) Draw two arcs which intersect each other at the point F.

(h) Join OF.

(i) OF is the required bisector of$\angle EOB$.

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

Ans:

(a) Draw a line OP.

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

(c) Draw $\overline {OR} $ as the bisector of$\angle POQ$.

(d) Draw $\overline {OS} $ as bisector of $\angle POR$

(e) Draw $\overline {OT} $ as bisector of $\angle QOR$

(f) OR, OS and OT divide$\angle POQ$ into four equal parts.

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

(a) $60^\circ $ 

Ans: 

(i) Draw a line OL.

(ii) Considering O as the centre, draw an arc which will intersect OL at A.

(iii) Consider A as the centre and take the same radius and cut the previous arc at B.

(iv) Join OB.

(v) $\angle MOL$ is the required angle of ${60^ \circ }$.

(b) ${30^ \circ }$

Ans: 

(i) Draw a line OL.

(ii) Considering O as centre, draw an arc, which intersects OL at Y.

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

(iv) Join OZ.

(v) $\angle LOM$ is the required angle of ${60^ \circ }$

(vi) Put the pointer on Y and draw an arc.

(vii) Put the pointer on Z and with the same radius, cut the previous arc at N.

(viii) $\angle NOL$ is the required angle of ${30^ \circ }$

(c) ${90^ \circ }$

Ans: 

(i) Draw a line

(ii) Considering O as centre, draw an arc, which intersects OP at A.

(iii) Now, Considering A as centre and same radius, cut the previous arc at the point B.

(iv) Considering B as centre and same radius, draw another arc that intersects the same arc at the point C.

(v) Considering B and C as centres and the same radius, draw two arcs which intersect each other at R.

(vi) Join OR and produce it to form a line OQ.

(vii) $\angle QOP$ is the required angle of ${90^ \circ }$.

(d) ${120^ \circ }$

Ans: 

(i) Draw a ray OP.

(ii) Considering O as centre, Draw an arc, which intersects OP at X.

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

(iv) Considering Y as centre and same radius cut the arc at Z.

(v) Join OZ.

(vi)$\angle ROP$ is the required angle of ${120^ \circ }$.

(e) ${45^ \circ }$

Ans: 

(i) Draw a ray OP.

(ii) Considering O as centre, draw an arc, which intersects OP at M.

(iii) Considering M as centre and same radius, cut the previous arc at S.

(iv) ConsideringS as centre and same radius, draw another arc intersecting the same arc at T.

(v) Considering S and T as centres and the same radius, draw two arcs intersecting each other at K.

(vi) Join OK and produce it to form a ray OQ.

(vii) $\angle QOP$ is the required angle of ${90^ \circ }$.

(viii) Draw the bisector of $\angle QOP$

(ix) $\angle ROP$ is the required angle of ${45^ \circ }$.

(f) ${135^ \circ }$

Ans: 

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

(ii) Considering O as centre, draw an arc, which intersects XY at P and Q.

(iii) Considering P and Q as centres and considering the radius more than half of PQ, draw two arcs intersecting each other at S.

(iv) Join OS. 

Thus,

$\angle SOX = \angle SOY$

$\angle SOX = {90^ \circ }$

(v) Now, draw OR the bisector of$\angle SOX$.

(vi)  $\angle ROY$ is the required angle of ${135^ \circ }$.

6. Draw an angle of measure 45 and bisect it.

Ans: 

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

(b) Considering O as centre, draw an arc which intersects XY at two points P and Q.

(c) Considering P and Q as centres and radius more than half of PQ, draw two arcs which intersect each other at N.

(d) Join ON. Then $\angle NOY = {90^ \circ }$

(e) Draw OM as the bisector of$\angle NOY$. 

Thus, $\angle YOM = {45^ \circ }$

(f) Again, draw OL as the bisector of$\angle YOM$.

(g) Hence, $\angle MOL = \angle LOY$

$\angle MOL = 22{\frac{1}{2}^ \circ }$.

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

Ans: 

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

(b) Considering O as centre, draw an arc, which intersects XY at P and Q.

(c) Considering P and Q as centres and considering radius more than half of PQ and draw two arcs intersecting each other at S.

(d) Join OS. 

Thus,

\[\angle SOX = \angle SOY\]

\[\angle SOX = {90^ \circ }\].

(e) Now, draw OT the bisector of $\angle SOX$

(f) $\angle TOY$is the required angle of ${135^ \circ }$.

(g) Now, draw OU as the bisector of $\angle TOY$.

(h)Hence.

\[\angle TOU = \angle TOY\]

\[\angle TOU = 67{\frac{1}{2}^ \circ }\]

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

Ans: 

(a) Draw $\angle ABC = {70^ \circ }$ using a protractor.

(b) Draw a ray XY.

(c) Place the compass at B and draw an arc to cut the rays of $\angle ABC$ at T and S. (i.e. BC at T and BA at S)

(d) Use the same compass radius to draw an arc with X as centre, cutting XY at P.

(e) Now place the compass pointer at point T and pencil(other leg of compass) at point S, thus setting your compass setting to the length ST.

(f) Place the compass pointer at P and draw the arc of the same radius as set in the previous step to cut the arc drawn earlier at Q.

(g) Join XQ and produce it to form a ray XZ.

(h) $\angle ZXY = {70^ \circ }$.

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

Ans: 

(a) Draw $\angle XOY = {40^ \circ }$

(b) Draw a line AB.

(c) Take any point R on AB.

(d) Place the compass at O and draw an arc to cut the rays of $\angle XOY$ at L and M.

(e) Use the same compass setting to draw an arc O as centre, cutting RB at S.

(f) Set the compass to length LM.

(g) Place the compass at S and draw the arc of the radius set up in the previous step to cut the arc drawn earlier T.

(h) Join RT.

(i) $\angle TRB = {40^ \circ }$

and $\angle TRA$ is supplementary of it.

Thus, 

$\angle TRA = {180^ \circ } - {40^ \circ }$

$\angle TRA = {140^ \circ }$

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6

Opting for the NCERT solutions for Ex 14.6 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 14.6 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 14 Exercise 14.6 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 6 Maths Chapter 14 Exercise 14.6, there are plenty of exercises in this chapter that contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 6 Maths Chapter 14 Exercise 14.6 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.