Class
6 Maths NCERT Solutions
Chapter
14 Practical Geometry
Exercise 14.1
Q.1: Draw a circle of radius 3.2 cm.
Ans :
The required circle can be drawn as follows.
Step 1
First, open the compasses for the
required radius 3.2 cm.
Step 2
Mark a point ‘O’ where we want the
centre of the circle to be.
Step 3
Place the pointer of compasses on O.
Step 4
Turn the compasses slowly to draw the
circle.
Q.2: With the same centre O, draw two circles
of radii 4 cm and 2.5 cm.
Ans :
The required circle can be drawn as follows.
Step 1
First, open the compasses for the
required radius 4 cm.
Step 2
Mark a point ‘O’ where we want the
centre of the circle to be.
Step 3
Place the pointer of compasses on O.
Step 4
Turn the compasses slowly to draw the
circle.
Step 5
Now, open the compasses for 2.5 cm.
Step 6
Again
put the pointer of the compasses on point ‘O’ and turn the compasses slowly to
draw the circle.
Q.3: 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?
Ans :
A circle can be drawn of any convenient radius,
also having its centre as O. Let AB and CD be two diameters of this circle.
When we join the ends of these diameters, a quadrilateral ACBD is formed.
As we know that the diameters of a circle are equal
in length, therefore, the quadrilateral so formed will have its diagonals of
equal length.
Also, OA = OB = OC = OD = radius r and
if a quadrilateral has its diagonals of same length which are bisecting each
other, then it will be a rectangle.
Let DE and FG be two diameters of this circle such
that these are perpendicular to each other. A quadrilateral is formed by
joining the ends of these diameters.
Here, OD = OE = OF = OG = radius r
In this quadrilateral DFEG, the diagonals are equal
and perpendicular to each other. Also, since these are bisecting each other, it
will be a square.
The length of the sides of the quadrilateral so
formed can be measured to check our answers.
Q.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.
Ans :
A circle and three required points A, B, C can be
drawn as follows.
Q.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 
and 
are at right angles.
Ans :
Let us draw two circles of same radius which are
passing through the centres of the other circle.
Here, point A and B are the centres of these
circles and these circles are intersecting each other at point C and D.
In quadrilateral ADBC,
AD = AC (Radius of circle centered at A)
BC = BD (Radius of circle centered at B)
As radius of both circles are equal, therefore, AD
= AC = BC = BD
Hence, 
is a rhombus and in a rhombus,
the diagonals bisect each other at 90°. Hence, and are at right angles.
Exercise 14.2
Q.1: Draw a line segment of length 7.3 cm
using a ruler.
Ans :
A line segment of length 7.3 cm can be drawn using
a ruler as follows.
(1) Mark a point A on the sheet.
(2) Put 0 mark of ruler at point A.
(3) Mark a point B on the sheet at 7.3
cm on ruler.
(4) Join A and B.
is the required line segment.
Q.2: Construct a line segment of length 5.6
cm using ruler and compasses.
Ans :
A line segment of length 5.6 cm can be drawn using
a ruler and compasses as follows.
(1) Draw a line l and
mark a point A on this line.
(2) Place the compasses on the zero
mark of the ruler. Open it to place the pencil up to the 5.6 cm mark.
(3) Place the pointer of compasses on
point A and draw an arc to cut l at B. AB is the line segment
of 5.6 cm length.
Q.3: Construct of length 7.8 cm. From this, cut
off 
of length 4.7 cm. Measure
.
Ans :
(1) Draw a line l and mark a point
A on it.
(2) By adjusting the compasses up to 7.8 cm, draw
an arc to cut l on B, while putting the pointer of compasses
on point A.
is the line segment of 7.8 cm.
(3) By adjusting the compasses up to 4.7 cm, draw
an arc to cut l on C, while putting the pointer of compasses on
point A. is the line segment of 4.7 cm.
(4) Now, put the ruler along with this line such
that 0 mark of the ruler will match with point C.
On reading the position of point B, it comes to 3.1
cm. is 3.1 cm.
Q.4: 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 thatalso has the length of.)
Ans :
A line segment can be drawn such that the
length of is twice that of as follows.
(1) Draw a line l and
mark a point P on it and let AB be the given line segment of 3.9 cm.
(2) By adjusting the compasses up to the length of
AB, draw an arc to cut
the line at X, while taking the pointer of
compasses at point P.
(3) Again put the pointer
on point X and draw an arc to cut line l again at
Q.
is the required line segment. By
ruler, the length of can be measured which comes
to 7.8 cm.
Q.5: 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.
Ans :
(1) Given that, =7.3 cm and = 3.4 cm
(2) Adjust the compasses up to the length of CD and
put the pointer of the compasses at A. Draw an arc to cut AB at P.
(3) Adjust the compasses up to the length of PB.
Now draw a line l and mark a point X on it.
(4) Now, putting the pointer of compasses at point
X, draw an arc to cut the line at Y.
is the required line segment.
Exercise 14.3
Q.1: Draw any line segment. Without measuring, construct a copy of.
Ans :
The following steps will be followed to draw the
given line segment and to construct a copy of.
(1) Let be the given line segment.
(2) Adjust the compasses up to the length of .
(3) Draw any line l and mark a
point A on it.
(4) Put the pointer on point A, and without
changing the setting of compasses, draw an arc to cut the line segment at point
B.
is the required line segment.
Q.2: Given some line segment, whose length you do not know,
construct such that the length
of is twice that of .
Ans :
The following steps will be followed to construct a
line segment such that the length of is twice that of .
(1) Let be the given line segment.
(2) Adjust the compasses up to the length of .
(3) Draw any line l and mark a
point P on it.
(4) Put the pointer on P and without changing the
setting of compasses, draw an arc to cut the line segment at point X.
(5) Now, put the pointer on point X and again draw an
arc with the same radius as before, to cut the line l at point
Q.
is the required line segment.
Exercise 14.4
Q.1: Draw any line segment. Mark any point M on it. Through
M, draw a perpendicular to. (Use ruler and compasses)
Ans :
(1) Draw the given line segment and mark any point M on it.
(2) With M as centre and a convenient radius,
construct an arc intersecting the line segmentat two points C and D.
(3) With C and D as centres and a radius greater
than CM, construct two arcs. Let these be intersecting each other at E.
(4) Join EM. 
is perpendicular to .
Q.2: Draw any line segment. Take any point R not on it.
Through R, draw a perpendicular to. (Use ruler and set-square)
Ans :
(1) Take the given line segment and mark any point R
outside.
(2) Place a set square on such that one arm of its
right angle aligns along.
(3) Place the ruler along the edge opposite to the
right angle of the set square.
(4) Hold the ruler fixed. Slide the set square
along the ruler till the point R touches the other arm of the set square.
(5) Draw a line along this edge of the set square
which will be passing through R. It is the required line, which is
perpendicular to.
Q.3: Draw a line l and
point X on it. Through X, draw a line segment perpendicular to l.
Now draw a perpendicular to at Y. (use ruler and
compasses)
Ans :
(1) Draw a line l and mark a point
X on it.
(2) Taking X as centre and with a convenient
radius, draw an arc intersecting line l at two points A and B.
(3) With A and B as centres and a radius more than
AX, construct two arcs intersecting each other at Y.
(4) Join XY. is perpendicular to l.
Similarly, a perpendicular to at the point Y can be
drawn. The line 
is perpendicular to at Y.
Exercise 14.5
Q.1: Draw of length 7.3 cm and find its
axis of symmetry.
Ans :
The below given steps will be followed to
construct of length 7.3 cm and to find its
axis of symmetry.
(1) Draw a line segment
of 7.3 cm.
(2) Taking A as centre, draw a circle by using
compasses. The radius of circle should be more than half the length of.
(3) With the same radius as before, draw another
circle using compasses while taking point B as centre. Let it cut the previous
circle at C and D.
(4) Join
. is the axis of
symmetry.
Q.2: Draw a line segment of length 9.5 cm
and construct its perpendicular bisector.
Ans :
The below given steps will be followed to construct
a line segment of length 9.5 cm and its perpendicular bisector.
(1) Draw a line segment
of 9.5 cm.
(2) Taking P as centre, draw a circle by using
compasses. The radius of circle should be more than half the length of .
(3) With the same radius as before, draw another
circle using compasses while taking point Q as centre. Let it cut the previous
circle at R and S.
(4) Join RS. 
is the axis of symmetry
i.e., the perpendicular bisector of line .
Q.3: Draw the perpendicular bisector ofwhose 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?
Ans :
(1) Draw a line segment
of 10.3 cm.
(2) Taking point X as centre, draw a circle by
using compasses. The radius of circle should be more than half the length
of .
(3) With the same radius as before, draw another
circle using compasses while taking point Y as centre. Let it cut the previous
circle at A and B.
(4) Join. is the axis of symmetry.
(a) Take any point P on . We will find that the measures
of the lengths of PX and PY are same.
It is because is the axis of symmetry. Hence,
any point lying on will be at the same distance from
both the ends of.
(b) M is the mid-point of. Perpendicular bisector will be passing through point M.
Hence, length ofis just double of
.
Or, 2MX = XY
Q.4: Draw a line segment of length 12.8 cm.
Using compasses; divide it into four equal parts. Verify by actual measurement.
Ans :
(1) Draw a line segmentof 12.8 cm.
(2) Draw a circle, while taking point X as centre
and radius more than half of XY.
(3) With same radius and taking centre as Y, again
draw arcs to cut the circle at A and B. Join AB which intersectsat M.
(4) Taking X and Y as centres, draw two circles
with radius more than half of
.
(5) With same radius and taking M as centre, draw
arcs to intersect these circles at P, Q and R, S.
(6) Join PQ and RS. These are intersectingat T and U.
(7) Now, 
. These are 4 equal parts of.
By measuring these line segments with the help of
ruler, we will find that each is of 3.2 cm.
Q.5: With of length 6.1 cm as diameter draw
a circle.
Ans :
(1) Draw a line segmentof 6.1 cm.
(2) Taking point P as centre and radius more than
half of, draw a circle.
(3) With same radius and taking Q as centre, draw
arcs to intersect this circle at points R and S.
(4) Join RS which intersects at T.
(5) Taking T as centre and with radius TP, draw a
circle which will also pass through Q. It is the required circle.
Q.6: 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.
Ans :
(1) Mark any point C on the sheet.
(2) By adjusting the compasses up to 3.4 cm and by
putting the pointer of the compasses at point C, turn the compasses slowly to
draw the circle. It is the required circle of 3.4 cm radius.
(3) Now, mark any chordin the circle.
(4) Taking A and B as centres, draw arcs on both
sides of. Let these intersect each other
at D and E.
(5) Join DE, which is the perpendicular bisector of
AB.
When
is extended, it will pass through
point C.
Q.7: Repeat question 6, if happens to be a diameter.
Ans :
(1) Mark any point C on the sheet.
(2) By adjusting the compasses up to 3.4 cm and by
putting the pointer of the compasses at point C, turn the compasses slowly to
draw the circle. It is the required circle of 3.4 cm radius.
(3) Mark any diameterin the circle.
(4) Now, taking A and B as centres, draw arcs on
both sides of taking radius more than. Let these intersect each other
at D and E.
(5) Join DE, which is the perpendicular bisector of
AB.
It can be observed that is passing through the centre C of
the circle.
Q.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?
Ans :
(1) Mark any point C on the sheet. Now, by adjusting
the compasses up to
4 cm and by putting the pointer of compasses at
point C, turn the compasses slowly to draw the circle. It is the required
circle of 4 cm radius.
(2) Take any two chordsandin the circle.
(3) Taking A and B as centres and with radius more
than half of, draw arcs on both sides of AB,
intersecting each other at E, F. Join EF which is the perpendicular bisector of
AB.
(4) Taking C and D as centres and with radius more
than half of, draw arcs on both sides of CD,
intersecting each other at G, H. Join GH which is the perpendicular bisector of
CD.
Now, we will find that when EF and GH are extended,
they meet at the centre of the circle i.e., point O.
Q.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 
and
.
Let them meet at P. Is PA = PB?
Ans :
(1)Draw any angle whose vertex is O.
(2) With a convenient radius, draw arcs on both
rays of this angle while taking O as centre. Let these points be A and B.
(3) Taking O and A as centres and with radius more
than half of OA, draw arcs on both sides of OA. Let these be intersecting at C
and D. Join CD.
(4) Similarly, we can find the perpendicular
bisector
of
. These perpendicular bisectorsandwill intersect each other at P.
Now, PA and PB can be measured. These are equal in
length.
Exercise 14.6
Q.1: Draw ∆POQ of measure 75° and find
its line of symmetry.
Ans :
The below given steps will be followed to construct
an angle of 75° and its line of symmetry.
(1) Draw a line l and mark two
points O and Q on it, as shown in the figure. Draw an arc of convenient radius,
while taking point O as centre. Let it intersect line l at R.
(2) Taking R as centre and with the same radius as
before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as
before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same
radius to intersect each other at U.
(5) Join OU. Let it intersect the arc at V. Now,
taking S and V as centres, draw arcs with radius more than 
SV. Let those intersect each other
at P. Join OP, which is the ray making 75° with the line l.
(6) Let this ray be intersecting our major arc at
point W. Now, taking R and W as centres, draw arcs
with radius more than RW in the interior of angle of
75º. Let these be intersecting each other at X. Join OX.
OX is the line of symmetry for ∆POQ = 75°.
Q.2: Draw an angle of measure 147° and
construct its bisector.
Ans :
The below given steps will be followed to construct
an angle of 147º measure and its bisector.
(1) Draw a line l and mark a point
O on it. Place the centre of the protractor at point O and the zero edge along
line l.
(2) Mark a point A at 147º. Join OA. OA is the
required ray making 147º with line l.
(3) Draw an arc of convenient radius, while taking
point O as centre. Let it intersect both rays of angle 147º at point A and B.
(4) Taking A and B as centres, draw arcs of radius
more than AB in the interior of angle of
147º. Let those intersect each other at C. Join OC.
OC is the required bisector of 147º angle.
Q.3: Draw a right angle and construct its
bisector.
Ans :
The below given steps will be followed to construct
a right angle and its bisector.
(1) Draw a line l and mark a point
P on it. Draw an arc of convenient radius, while taking point P as centre. Let
it intersect line l at R.
(2) Taking R as centre and with the same radius as
before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as
before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centres, draw arcs of same
radius to intersect each other at U.
(5) Join PU. PU is the required ray making 90º with
line l. Let it intersect the major arc at point V.
(6) Now, taking R and V as centres, draw arcs with
radius more than RV to intersect each other at W.
Join PW.
PW is the required bisector of this right angle.
Q.4: Draw an angle of measure153° and divide
it into four equal parts.
Ans :
The below given steps will be followed to construct
an angle of 153º measure and its bisector.
(1) Draw a line l and mark a point
O on it. Place the centre of the protractor at point O and the zero edge along
line l.
(2) Mark a point A at 153º. Join OA. OA is the
required ray making 153º with line l.
(3) Draw an arc of convenient radius, while taking
point O as centre. Let it intersect both rays of angle 153º at point A and B.
(4) Taking A and B as centres, draw arcs of radius
more than AB in the interior of angle of
153º. Let those intersect each other at C. Join OC.
(5) Let OC intersect the major arc at point D. Now,
with radius more than AD, draw arcs while taking A and D
as centres, and D and B as centres. Let these be intersecting each other at
point E and F respectively. Join OE, OF.
OF, OC, OE are the rays dividing 153º angle in 4
equal parts.
Q.5: Construct with ruler and compasses,
angles of following measures:
(a) 60° (b) 30° (c) 90°
(d) 120° (e) 45° (f) 135°
Ans :
(a) 60°
The below given steps will be followed to construct
an angle of 60°.
(1) Draw a line l and mark a point
P on it. Now, taking P as centre and with a convenient radius, draw an arc of a
circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as
before, draw an arc intersecting the previously drawn arc at point R.
(3) Join PR which is the required ray making 60°
with line l.
(b) 30°
The below given steps will be followed to construct
an angle of 30°.
(1) Draw a line l and mark a point
P on it. Now taking P as centre and with convenient radius, draw an arc of a
circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as
before, draw an arc intersecting the previously drawn arc at point R.
(3) Now, taking Q and R as centre and with radius
more than RQ, draw arcs to intersect each
other at S. Join PS which is the required ray making 30° with line l.
(c) 90º
The below given steps will be followed to construct
an angle of 90°.
(1) Draw a line l and mark a point
P on it. Now taking P as centre and with a convenient radius, draw an arc of a
circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as
before, draw an arc intersecting the previously drawn arc at R.
(3) Taking R as centre and with the same radius as
before, draw an arc intersecting the arc at S (see figure).
(4) Taking R and S as centre, draw an arc of same
radius to intersect each other at T.
(5) Join PT, which is the required ray making 90°
with line l.
(d) 120º
The below given steps will be followed to construct
an angle of 120°.
(1) Draw a line l and mark a point
P on it. Now taking P as centre and with a convenient radius, draw an arc of a
circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as
before, draw an arc intersecting the previously drawn arc at R.
(3) Taking R as centre and with the same radius as
before, draw an arc intersecting the arc at S.
(4) Join PS, which is the required ray making 120°
with line l.
(e) 45º
The below given steps will be followed to construct
an angle of 45°.
(1) Draw a line l and mark a point
P on it. Now taking P as centre and with a convenient radius, draw an arc of a
circle which intersects line l at Q.
(2) Taking Q as centre and with the same radius as
before, draw an arc intersecting the previously drawn arc at R.
(3) Taking R as centre and with the same radius as
before, draw an arc intersecting the arc at S (see figure).
(4) Taking R and S as centres, draw arcs of same
radius to intersect each other at T.
(5) Join PT. Let it intersect the major arc at
point U.
(6) Taking Q and U as centres, draw arcs with
radius more than QU to intersect each other at V.
Join PV.
PV is the required ray making 45° with the given
line l.
(f) 135°
The below given steps will be followed to construct
an angle of 135°.
(1) Draw a line l and mark a point
P on it. Now taking P as centre and with a convenient radius, draw a
semi-circle which intersects line l at Q and R.
(2) Taking R as centre and with the same radius as
before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as
before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw arcs of same
radius to intersect each other at U.
(5) Join PU. Let it intersect the arc at V. Now
taking Q and V as centres and with radius more than QV, draw arcs to intersect each
other at W.
(6) Join PW which is the required ray making 135°
with line l.
Q.6: Draw an angle of measure 45° and bisect
it.
Ans :
The below given steps will be followed to construct
an angle of 45º and its
bisector.
(1) ∠POQ
of 45º measure can be formed on a line l by using the
protractor.
(2) Draw an arc of a convenient radius, while
taking point O as centre. Let it intersect both rays of angle 45º at point A
and B.
(3) Taking A and B as centres, draw arcs of radius
more than AB in the interior of angle of 45º.
Let those intersect each other at C. Join OC.
OC is the required bisector of 45º angle.
Q.7: Draw an angle of measure 135° and
bisect it.
Ans :
The below given steps will be followed to construct
an angle of 135° and its bisector.
(1) ∆POQ of 135º measure can be formed on a
line l by using the protractor.
(2) Draw an arc of a convenient radius, while taking
point O as centre. Let it intersect both rays of angle 135º at point A and B.
(3) Taking A and B as centres, draw arcs of radius
more than AB in the interior of angle of
135º. Let those intersect each other at C. Join OC.
OC is the required bisector of 135º angle.
Q.8: Draw an angle of 70°. Make a copy of it
using only a straight edge and compasses.
Ans :
The below given steps will be followed to construct
an angle of 70º measure and its copy.
(1) Draw a line l and mark a point
O on it. Place the centre of the protractor at point O and the zero edge along
line l.
(2) Mark a point A at 70º. Join OA. OA is the ray
making 70º with line l. Draw an arc of convenient radius in the
interior of 70º angle, while taking point O as centre. Let it intersect both
rays of angle 70º at point B and C.
(3) Draw a line m and mark a point
P on it. With the same radius as used before, again draw an arc while taking
point P as centre. Let it cut the line m at point D.
(4) Now, adjust the compasses up to the length of
BC. With this radius, draw an arc while taking D as centre, which will
intersect the previously drawn arc at point E.
(5) Join PE. PE is the required ray which makes the
same angle (i.e. 70º) with line m.
Q.9: Draw an angle of 40°. Copy its
supplementary angle.
Ans :
The below given steps will be followed to construct
an angle of 40º measure and the copy of its supplementary angle.
(1) Draw a line segment and mark a point O on it.
Place the centre of the protractor at point O and the zero edge along line
segment.
(2) Mark a point A at 40º. Join OA. OA is the
required ray making 40º with. ∆POA is the supplementary
angle of 40º.
(3) Draw an arc of convenient radius in the
interior of ∆POA, while taking point O as centre. Let it intersect both
rays of ∆POA at point B and C.
(4) Draw a line m and mark a point
S on it. With the same radius as used before, again draw an arc while taking
point S as centre. Let it cut the line m at point T.
(5) Now, adjust the compasses up to the length of
BC. With this radius, draw an arc while taking T as centre, which will
intersect the previously drawn arc at point R.
(6) Join RS. RS is the required ray which makes the
same angle with line m, as the supplementary of 40º is 140º.
End of Questions