NCERT Solutions for Class 6 Maths Exercise 14.6

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NCERT solutions for Maths Practical geometry Download as PDF

NCERT Solutions for Class 6 Maths Exercise 14.6

NCERT Solutions for Class 6 Maths Practical geometry

Class –VI Mathematics
(Ex. 14.6)
Question 1.Draw {tex}\angle {/tex}POQ of measure {tex}75^\circ {/tex} and find its line of symmetry.

Answer: Steps of construction:

(a) Draw a line {tex}l{/tex} 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 {tex}l{/tex} at A.

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

(d) Join OB, then {tex}\angle {/tex}BOA = {tex}{60^ \circ }.{/tex}

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

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

(g) Draw {tex}\overline {{\text{OP}}} {/tex} as bisector of {tex}\angle {/tex}DOB.

Thus, {tex}\angle {/tex}POA = {tex}75^\circ {/tex}


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

Answer: Steps of construction:

(a) Draw a ray {tex}\overleftrightarrow {{\text{OA}}}.{/tex}

(b) With the help of protractor, construct {tex}\angle {/tex}AOB = {tex}147^\circ .{/tex}

(c) Taking centre O and any convenient radius, draw an arc which intersects the arms {tex}\overline {{\text{OA}}} {/tex} and {tex}\overline {{\text{OB}}} {/tex} 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, {tex}\overline {{\text{OR}}} {/tex} is the required bisector of {tex}\angle {/tex}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, {tex}\angle {/tex}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, {tex}\overline {{\text{OD}}} {/tex} is the required bisector of {tex}\angle {/tex}COQ.


NCERT Solutions for Class 6 Maths Exercise 14.6

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

Answer: Steps of construction:

(a) Draw a ray {tex}\overleftrightarrow {{\text{OA}}}.{/tex}

(b) At O, with the help of a protractor, construct {tex}\angle {/tex}AOB = {tex}153^\circ .{/tex}

(c) Draw {tex}\overline {{\text{OC}}} {/tex} as the bisector of {tex}\angle {/tex}AOB.

(d) Again, draw {tex}\overline {{\text{OD}}} {/tex} as bisector of {tex}\angle {/tex}AOC.

(e) Again, draw {tex}\overline {{\text{OE}}} {/tex} as bisector of {tex}\angle {/tex}BOC.

(f) Thus, {tex}\overline {{\text{OC}}} {/tex}, {tex}\overline {{\text{OD}}} {/tex} and {tex}\overline {{\text{OE}}} {/tex} divide {tex}\angle {/tex}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) {tex}{60^ \circ }{/tex}

(b) {tex}{30^ \circ }{/tex}

(c) {tex}{90^ \circ }{/tex}

(d) {tex}{120^ \circ }{/tex}

(e) {tex}{45^ \circ }{/tex}

(f) {tex}{135^ \circ }{/tex}

Answer:

Steps of construction:

(a) {tex}{60^ \circ }{/tex}

(i) Draw a ray {tex}\overleftrightarrow {{\text{OA}}}.{/tex}

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

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

(iv) Join OQ.

Thus, {tex}\angle {/tex}BOA is required angle of {tex}{60^ \circ }.{/tex}

(b) {tex}{30^ \circ }{/tex}

(i) Draw a ray {tex}\overleftrightarrow {{\text{OA}}}.{/tex}

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

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

(iv) Join OQ. Thus, {tex}\angle {/tex}BOA is required angle of {tex}{60^ \circ }.{/tex}

(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, {tex}\angle {/tex}COA is required angle of {tex}{30^ \circ }.{/tex}

(c){tex}{90^ \circ }{/tex}

(i) Draw a ray {tex}\overleftrightarrow {{\text{OA}}}.{/tex}

(ii) Taking O as centre and convenient radius, mark an arc, which intersects {tex}\overleftrightarrow {{\text{OA}}}{/tex} 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, {tex}\angle {/tex}BOA is required angle of {tex}{90^ \circ }.{/tex}

(d){tex}{120^ \circ }{/tex}

(i) Draw a ray {tex}\overleftrightarrow {{\text{OA}}}.{/tex}

(ii) Taking O as centre and convenient radius, mark an arc, which intersects {tex}\overleftrightarrow {{\text{OA}}}{/tex} 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, {tex}\angle {/tex}AOD is required angle of {tex}{120^ \circ }.{/tex}

(e){tex}{45^o}{/tex}

(i) Draw a ray {tex}\overrightarrow {OA} {/tex}

(ii) Taking O as centre and convenient radius, mark an arc, which intersects {tex}\overrightarrow {OA} {/tex} 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, {tex}\angle {/tex}BOA is required angle of {tex}{90^o}{/tex}

(vii) Draw the bisector of {tex}\angle {/tex}BOA.

Thus, {tex}\angle {/tex}MOA is required angle of {tex}{45^o}{/tex}

(f){tex}{135^o}{/tex}

(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, {tex}\angle {/tex}QOR = {tex}\angle {/tex}POQ = {tex}{90^o}{/tex}

(v) Draw {tex}\overrightarrow {OD} {/tex} the bisector of {tex}\angle {/tex}POR.

thus, {tex}\angle {/tex}QOD is required angle of {tex}{135^o}{/tex}


NCERT Solutions for Class 6 Maths Exercise 14.6

Question 6.Draw an angle of measure {tex}{45^ \circ }{/tex} 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 {tex}\angle {/tex}COQ is an angle of {tex}{90^o}{/tex}

(e)Draw {tex}\overrightarrow {OE} {/tex} as the bisector of {tex}\angle {/tex}COE. Thus, {tex}\angle {/tex}QOE = {tex}{45^o}{/tex}

(f)Again draw {tex}\overrightarrow {OG} {/tex} as the bisector of {tex}\angle {/tex}QOE.

Thus, {tex}\angle {/tex}QOG = {tex}\angle {/tex}EOG = {tex}22\frac{1}{2}{\,^o}{/tex}


NCERT Solutions for Class 6 Maths Exercise 14.6

Question 7.Draw an angle of measure {tex}{135^ \circ }{/tex} 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, {tex}\angle {/tex}QOR = {tex}\angle {/tex}POQ = {tex}{90^o}{/tex}

(e) Draw {tex}\overrightarrow {OD} {/tex} the bisector of {tex}\angle {/tex}POR. Thus, {tex}\angle {/tex}QOD is required angle of {tex}{135^o}{/tex}

(f) Now, draw {tex}\overrightarrow {OE} {/tex} as the bisector of {tex}\angle {/tex}QOD.

Thus, {tex}\angle {/tex}QOE = {tex}\angle {/tex}DOE = {tex}67\frac{1}{2}{\,^o}{/tex}


NCERT Solutions for Class 6 Maths Exercise 14.6

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

Answer: Steps of construction:

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

(b) Draw a ray {tex}\overrightarrow {AB} {/tex}

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


NCERT Solutions for Class 6 Maths Exercise 14.6

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

Answer: Steps of construction:

(a) Draw an angle of {tex}{40^o}{/tex} with the help of protractor, naming {tex}\angle {/tex}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 {tex}\angle {/tex}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, {tex}\angle {/tex}QMY = {tex}{40^o}{/tex} and {tex}\angle {/tex}PMY is supplementary of it.

NCERT Solutions for Class 6 Maths Exercise 14.6

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