Practical Geometry (Mathematics) Class 6 - NCERT Questions

Draw a circle of radius 3.2 cm.

SOLUTION: Steps of construction :

(i) Mark a point '*O*' with a sharp pencil where we want the centre of the circle.

(ii) Open the compasses for the required radius of 3.2 cm.

(iii) Place the pointer of compasses on *O*.

(iv) Turn the compass slowly to draw the circle.

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

Steps of construction :

(i) Mark a point '*O*' with a sharp pencil where we want the centre of the circle.

(ii) Open the compass for 4 cm.

(iii) Place the pointer of the compass on *O*.

(iv) Turn the compass slowly to draw the circle.

(v) Again open the compass for 2.5 cm and place the pointer of the compass on *O*.

(vi) Turn the compass slowly to draw the second circle.

Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?

SOLUTION: Steps of construction:

(i) Mark a point '*O*' with sharp pencil where we want the centre of the circle.

(ii) Place the pointer of the compasses on '*O*'.

(iii) Turn the compasses slowly to draw the circle.

(A) By joining the ends of two diameters, we get a rectangle. By measuring, we find*AB* *=* *CD* and *BC* *=* *AD*, *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.

(B) 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*, *i.e.,* all four sides are equal.

Also ∠*A* = ∠*B* = ∠*C* = ∠*D* = 90°, *i.e.*, each angle is 90°.

Hence, it is a square.

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.

Steps of construction:

(i) Mark a point '*O*' with sharp pencil where we want centre of the circle.

(ii) Place the pointer of the compasses on '*O*'.

(iii) Turn the compasses slowly to draw a circle.

(A) Point *A* is on the circle.

(B) Point *B* is in the interior of the circle.

(C) Point *C* is in the exterior of the circle.

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 and are at right angles.

Steps of construction :

(i) Mark a point '*A*' with sharp pencil.

(ii) Place the pointer of the compasses on *A*.

(iii) Turn the compasses slowly to draw a circle with centre *A*.

(iv) Take a point '*B*' on the circle with centre *A*.

(v) Place the pointer of the compasses on *B*.

(vi) Draw the circle with the radius same as radius of circle with centre *A.*

Now, both circles intersect at *C* and *D*.

Let *AB* and *CD* intersect at *O*.

Now, measuring ∠*COB*, we get

∠*COB* = 90°

∴ and are perpendicular.

Draw a line segment of length 7.3 cm using a ruler.

SOLUTION: 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*.

is the required line segment of length7.3 cm.

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

SOLUTION: 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 upto 5.6 cm mark.

(iii) Without changing the opening of the compasses, place the pointer on *A* and draw an arc that cuts *l* at *B*.

is the required line segment of length 5.6 cm.

Construct of length 7.8 cm. From this, cut off of length 4.7 cm. Measure .

SOLUTION: 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*.

(iv) By measuring , we find that *BC* = 3.1 cm.

Given of length 3.9 cm, construct such that the length of is twice that of . Verify by measurement.

(Hint: Construct such that length of = length of ; then cut off such that also has the length of .)

Steps of construction :

(i) Draw a line '*l*'.

(ii) Construct such that length of = lenght of = 3.9 cm

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

(iv) Thus the length of and the length of added together make twice the length of
**Verification:**

By measurement we find that
*PQ* = 7.8 cm = 3.9 cm + 3.9 cm

Given of length 7.3 cm and of length 3.4 cm, construct a line segment such that the length of is equal to the difference between the lengths of and . Verify by measurement.

SOLUTION: Steps of construction :

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

(ii) Construct such that

length = length of

(iii) Then cut off = length of = 3.4 cm

(iv) Thus the length of

= length of - length of

**Verification:**

By measurement, we find that length of

Draw any line segment . Without measuring , construct a copy of .

SOLUTION: Steps of construction :

(i) Draw whose length is not known.

(ii) Fix the compasses pointer on *P* and the pencil end on *Q*. The opening of the compasses now gives the length of .

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

(iv) Draw an arc that cuts '*l*' at a point, say *B*. Now, is a copy of .

Given some line segment , whose length you do not know, construct such that the length of is twice that of .

SOLUTION: Steps of construction :

(i) Draw whose length is not known.

(ii) Fix the compasses pointer on *A* and the pencil end on *B*. The opening of the compasses now gives the length of

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

(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*.

(vi) Thus is the required line segment whose length is twice that of *AB*.

Draw any line segment . Mark any point *M* on it. Through *M*, draw a perpendicular to . (use ruler and compasses)

Steps of construction :

(i) Draw .

(ii) With *M* as centre and a convenient radius, draw an arc intersecting at two points *C* and *D*.

(iii) With *C* and *D* as centres and a radius greater than *MC*, draw two arcs, which cut each other at *P*.

(iv) Join *PM*. Then *PM* is perpendicular to *AB* through the point *M*.

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

Steps of construction :

(i) Draw any line segment .

(ii) Place a set-square on such that one arm of its right angle aligns along .

(iii) Place a ruler along the edge opposite to the right angle of the set-square.

(iv) Hold the ruler fixed. Slide the set square along the ruler till the point *R* touches the other arm of the set square.

(v) Join *RM* along the edge through *R* meeting at *M*. Then *RM* ┴ *PQ*.

Draw a line *l* and a point *X* on it. Through *X*, draw a line segment perpendicular to *l*. Now draw a perpendicular to at *Y*. (use ruler and compasses)

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 *E*.

(iv) Join *XE* and produce it to *Y*. Then *XY* is perpendicular to '*l*'.

(v) With *Y* as centre and a convenient radius, draw an arc intersecting *XY* at two points *C* and *D*.

(vi) With *C* and *D* as centre 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*.

Draw of length 7.3 cm and find its axis of symmetry.

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

Steps of construction :

(i) Draw = 7.3cm.

(ii) With *A* as centre and radius more than half of *AB*, draw two arcs, one on each side of *AB*.

(iii) With *B* as a centre and the same radius as in step (ii), draw arcs cutting the arcs drawn in the previous step at *C* and *D*.

(iv) Join *CD*. Then *CD* is the axis of symmetry of the line segment *AB*.

Draw a line segment of length 9.5 cm and construct its perpendicular bisector.

SOLUTION: Steps of construction :

(i) Draw = 9.5cm.

(ii) With *A* as centre and radius more than half of *AB*, draw two arcs one on each side of *AB*.

(iii) With *B* as a centre and the same radius as in step (ii), draw arcs cutting the arcs drawn in the previous step at *C* and *D*.

(iv) Join *CD*. Then *CD* is the perpendicular bisector of .

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

Steps of construction :

(i) Draw = 10.3cm.

(ii) With *X* as centre and radius more than half of *XY*, draw two arcs one on each side of *XY*.

(iii) With *Y* as centre and the same radius as in step (ii), draw two arcs cutting the arcs drawn in the previous step at *C* and *D*.

(iv) Join *CD*. Then *CD* is the required perpendicular bisector of .

Now

(A) Take any point *P* on the bisector drawn. With the help of divider we can check that if *P* is the point of intersection of *XY* and *CD*.

(B) If *M* is the mid-point of , then

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

SOLUTION: Steps of construction :

(i) Draw *AB* = 12.8 cm.

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

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

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

(v) Now, point *C, D* and *E* divide in four equal parts.

(vi) By actual measurement, we find that

With of length 6.1 cm as diameter, draw a circle.

SOLUTION: Steps of construction :

(i) Draw a line segment = 6.1cm.

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

(iii) Taking *O* as centre and *OP* or *OQ* as radius draw a circle where is the diameter.

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

Steps of construction :

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

(ii) Draw any chord .

(iii) With *A* as center and radius more than half of draw two arcs one on each side of *AB*.

(iv) With *B* as a centre and the radius same as in step (iii), draw two arcs cutting the arcs drawn in the previous step at *P* and *Q*.

(v) Join *PQ*. Then *PQ* is the perpendicular bisector of .

(vi) This perpendicular bisector of passes through the centre *C* of the circle.

Repeat Question 6, if happens to be a diameter.

SOLUTION: Steps of construction :

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

(ii) Draw its diameter *AB*.

(iii) With *A* as center and radius more than half of *AB*, draw two arcs one on each side of *AB*.

(iv) With *B* as a centre and the radius same as in step (iii), draw two arcs cutting the arcs drawn in the previous step at *P* and *Q*.

(v) Join *PQ*. Then *PQ* is the perpendicular bisector of .

(vi) We observe that this perpendicular bisector of intersect it at the centre *C* of the circle.

Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?

SOLUTION: Steps of construction :

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

(ii) Draw any two chords *AB* and *CD* in this circle.

(iii) With *A* as center and radius more than half *AB*, draw two arcs one on each side of *AB*.

(iv) With *B* as centre and radius same as in step (ii), draw two arcs cutting the arcs drawn in previous step at *E* and *F*.

(v) Join *EF*. Thus *EF* is the perpendicular bisector of chord *AB.*

(vi) Similarly draw *GH* the perpendicular bisector of chord *CD.*

(vii) These two perpendicular bisectors meet at *O*, the centre of the circle.

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

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 *OA* = *OB*.

(iii) Draw perpendicular bisector of *OA* and *OB*.

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

(v) With the help of divider, we check that *PA* = *PB*.

Draw ∠*POQ* of measure 75° and find its line of symmetry.

Steps of construction :

(i) Draw a line *l* and mark a point *O* on it.

(ii) Place the pointer of the compasses at *O* and draw an arc of any radius which intersects the line *l* at *Q*.

(iii) Taking same radius, with centre *Q*, draw an arc which cuts the previous arc at *B*.

(iv) Join *OB*, then ∠*BOQ* = 60°.

(v) Taking same radius, with centre *B*, draw an arc which cuts the arc drawn in step (ii) at *C*.

(vi) Draw bisector of ∠*BOC* which cuts the at *D*. Thus, ∠*DOQ* = 90°.

(vii) Draw *OP* as bisector of ∠*DOB*. Thus, ∠*POQ* = 75°.

Draw an angle of measure 147° and construct its bisector.

SOLUTION: Steps of construction :

(i) Draw .

(ii) With the help of protractor, draw ∠*AOB* = 147°.

(iii) Taking centre *O* and any convenient radius, draw an arc which intersects at *P* and *Q* respectively.

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

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

(vi) Join *OR* and produce it.

Thus, is the required bisector of ∠*AOB*.

Draw a right angle and construct its bisector.

SOLUTION: Steps of construction :

(i) Draw .

(ii) With the help of protractor draw ∠*QOP* = 90°

(iii) Taking *O* as centre and any convenient radius, draw an arc which intersect the at *A* and *B* respectively.

(iv) Taking *B* and *A* as centre and radius more than half of* BA*, draw two arcs which intersect each other at the point *D*.

(v) Join *OD*. Thus, is the required bisector of ∠*QOP*.

Draw an angle of measure 153° and divide it into four equal parts.

SOLUTION: Steps of construction :

(i) Draw .

(ii) At *O*, with the help of a protractor, draw ∠*AOB* = 153°.

(iii) Draw as the bisector of ∠*AOB*.

(iv) Again, draw as the bisector of ∠*AOC*.

(v) Again, draw as the bisector of ∠*BOC*.

(vi) Thus, divide ∠*AOB* in four equal parts.

Construct with ruler and compasses, angles of following measures:

(A) 60°

(B) 30°

(C) 90°

(D) 120°

(E) 45°

(F) 135°

Steps of construction :

(A) 60°

(i) Draw .

(ii) Taking *O* as centre and convenient radius, draw an arc, which intersects at *P*.

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

(iv) Join* OQ* and produce it to *B*. Thus, ∠*BOA* is the required angle of 60°.

(B) 30°

(i) Draw .

(ii) Taking *O* as centre and convenient radius, draw an arc, which intersects at *P*.

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

(iv) Join* OQ* and produce it to *B*. Thus, ∠*BOA* is the angle of 60°.

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

(vi) Taking *Q* as centre and with same radius, draw an arc which cut the previous arc at *C*.

(vii) Join *OC*. Thus ∠*COA* is the required angle of 30°.

(C) 90°

(i) Draw .

(ii) Taking *O* as centre and convenient radius, draw an arc, which intersects at *X*.

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

(iv) Taking *Y* as centre and same radius, draw another arc intersecting the arc drawn in step (ii) 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 .

(ii) Taking *O* as centre and convenient radius, draw an arc, which intersects at *P*.

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

(iv) Taking *Q* as centre and same radius draw another arc intersecting the arc drawn in step (ii) at *S*.

(v) Join *OS* and produce it to *D*.

Thus, ∠*AOD* is the required angle of 120°.

(E) 45°

(i) Draw .

(ii) Taking *O* as centre and convenient radius, draw an arc, which intersects at *X*.

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

(iv) Taking *Y* as centre and same radius, draw another arc intersecting the arc drawn in step (ii) 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 *B*. Thus, ∠*BOA* is the angle of 90°.

(vii) Draw *OM* the bisector of ∠*BOA*.

Thus, ∠*MOA* is the required angle of 45°.

(F) 135°

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

(ii) Taking *O* as centre and convenient radius, draw an arc, which intersects
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* = ∠*POR* = 90°.

(v) Draw the bisector of ∠*POR*.

Thus, ∠*QOD* is the required angle of 135°.

Draw an angle of measure 45° and bisect it.

SOLUTION: Steps of construction :

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

(ii) Taking *O* as centre and a convenient radius, draw an arc which intersects at two points *A* and *B*.

(iii) Taking *A* and *B* as centres and radius more than half of *AB*, draw two arcs which intersect each other at *C*.

(iv) Join *OC*. Then ∠*COQ* is an angle of 90°.

(v) Draw as the bisector of ∠*COQ*. Thus ∠*QOE* = 45°.

(vi) Again draw as the bisector of ∠*QOE*.

Draw an angle of measure 135° and bisect it.

SOLUTION: **Steps of construction :**

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

(ii) Taking *O* as centre and convenient radius, draw an arc, which intersects 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* = ∠*POR* = 90°.

(v) Draw the bisector of ∠*POR*. Thus, ∠*QOD* is the required angle of 135°.

(vi) Now, draw as the bisector of ∠*QOD*.

.

Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.

SOLUTION: Steps of construction :

(i) Draw an angle of 70° with the help of protractor, *i.e.*, ∠*POQ* = 70°.

(ii) Draw .

(iii) Place the compasses at *O* and draw an arc to cut the rays of ∠*POQ* at *L* and *M*.

(iv) Use the same compasses setting to draw an arc with *A* as centre, cutting *AB* at *X*.

(v) Set your compasses setting to the length *LM* with the same radius.

(vi) Place the compasses pointer at *X* and draw the arc to cut the arc drawn earlier at *Y*.

(vii) Join *AY*.

Thus, ∠*YAX* = 70°.

Draw an angle of 40°. Copy its supplementary angle.

SOLUTION: Steps of construction :

(i) Draw an angle of 40° with the help of protractor, *i.e.*, ∠*AOB* = 40°.

(ii) Draw a line .

(iii) Take any point *M* on *PQ*.

(iv) Place the compasses at* O* and draw an arc to cut the rays of ∠*AOB* at *l* and *N*.

(v) Use the same compasses setting to draw an arc *M* as centre, cutting *MQ* at *X*.

(vi) Set your compasses to length *LN* with the same radius.

(vii) Place the compasses at *X* and draw the arc to cut the arc drawn earlier at *Y*.

(viii) Join *MY*.

(ix) Thus, ∠*QMY* = 40° and ∠*PMY* is supplementary of it.