NCERT Solutions for Class 8 Maths Practical Geometry [Old Book]
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Page 1
NCERT
SOLUTIONS
CLASS - 8TH
aglase .co
SOLUTIONS
CLASS - 8TH
aglase .co
Page 2
Book : Mathematics Ncert Solutions | Chapter-4 Maths
Class : 8th
Subject : Maths
Chapter : 4
Chapter Name : Practical Geometry
Exercise 4.1
Q1 Construct the following quadrilaterals.
(i) Quadrilateral ABCD.
AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AD = 6 cm
AC = 7 cm
(ii) Quadrilateral JUMP
JU = 3.5 cm
UM = 4 cm
MP = 5 cm
PJ = 4.5 cm
PU = 6.5 cm
(iii) Parallelogram MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm
(iv) Rhombus BEST
BE = 4.5 cm
ET = 6 cm
Answer. (i) Firstly, a rough sketch of this quadrilateral can be drawn as follows.
(1) △ABC can be constructed by using the given measurements as follows.
(2) Vertex D is 6 cm away from vertex A. Therefore, while taking Aas centre, draw an arc of radius 6
Page 1 of 20 Aglasem Schools
Class : 8th
Subject : Maths
Chapter : 4
Chapter Name : Practical Geometry
Exercise 4.1
Q1 Construct the following quadrilaterals.
(i) Quadrilateral ABCD.
AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AD = 6 cm
AC = 7 cm
(ii) Quadrilateral JUMP
JU = 3.5 cm
UM = 4 cm
MP = 5 cm
PJ = 4.5 cm
PU = 6.5 cm
(iii) Parallelogram MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm
(iv) Rhombus BEST
BE = 4.5 cm
ET = 6 cm
Answer. (i) Firstly, a rough sketch of this quadrilateral can be drawn as follows.
(1) △ABC can be constructed by using the given measurements as follows.
(2) Vertex D is 6 cm away from vertex A. Therefore, while taking Aas centre, draw an arc of radius 6
Page 1 of 20 Aglasem Schools
Page 3
Book : Mathematics Ncert Solutions | Chapter-4 Maths
cm.
(3) Taking C as centre, draw an arc of radius 4 cm, cutting the previous arc at point D. Join D to A and
C.
ABCD is the required quadrilateral.
(ii) Firstly, a rough sketch of this quadrilateral can be drawn as follows.
(1) ΔJUP can be constructed by using the given measurements as follows.
(2) Vertex M is 5 cm away from vertex P and 4 cm away from vertex U. Taking P and U as centres, draw
arcs of radii 5 cm and 4 cm respectively. Let the point of intersection be M.
(3) Join M to P and U.
Page 2 of 20 Aglasem Schools
cm.
(3) Taking C as centre, draw an arc of radius 4 cm, cutting the previous arc at point D. Join D to A and
C.
ABCD is the required quadrilateral.
(ii) Firstly, a rough sketch of this quadrilateral can be drawn as follows.
(1) ΔJUP can be constructed by using the given measurements as follows.
(2) Vertex M is 5 cm away from vertex P and 4 cm away from vertex U. Taking P and U as centres, draw
arcs of radii 5 cm and 4 cm respectively. Let the point of intersection be M.
(3) Join M to P and U.
Page 2 of 20 Aglasem Schools
Page 4
Book : Mathematics Ncert Solutions | Chapter-4 Maths
JUMP is the required quadrilateral.
(iii) We know that opposite sides of a parallelogram are equal in length and also these are parallel to
each other.
Hence, ME = OR, MO = ER
A rough sketch of this parallelogram can be drawn as follows.
(1) △EOR can be constructed by using the given measurements as follows.
(2) Vertex M is 4.5 cm away from vertex O and 6 cm away from vertex E. Therefore, while taking O and
E as centres, draw arcs of 4.5 cm radius and 6 cm radius respectively. These will intersect each other at
point M.
(3) Join M to O and E.
Page 3 of 20 Aglasem Schools
JUMP is the required quadrilateral.
(iii) We know that opposite sides of a parallelogram are equal in length and also these are parallel to
each other.
Hence, ME = OR, MO = ER
A rough sketch of this parallelogram can be drawn as follows.
(1) △EOR can be constructed by using the given measurements as follows.
(2) Vertex M is 4.5 cm away from vertex O and 6 cm away from vertex E. Therefore, while taking O and
E as centres, draw arcs of 4.5 cm radius and 6 cm radius respectively. These will intersect each other at
point M.
(3) Join M to O and E.
Page 3 of 20 Aglasem Schools
Page 5
Book : Mathematics Ncert Solutions | Chapter-4 Maths
MORE is the required parallelogram.
(iv) We know that all sides of a rhombus are of the same measure.
Hence, BE = ES = ST = TB
A rough sketch of this rhombus can be drawn as follows.
(1) ΔBET can be obstructed by using the given measurements as follows.
(2) Vertex S is 4.5 cm away from vertex E and also from vertex T. Therefore, while taking E and T as
centres, draw arcs of 4.5 cm radius, which will be intersecting each other at point S.
(3) Join S to E and T.
BEST is the required rhombus.
Page : 60 , Block Name : Exercise 4.1
Exercise 4.2
Page 4 of 20 Aglasem Schools
MORE is the required parallelogram.
(iv) We know that all sides of a rhombus are of the same measure.
Hence, BE = ES = ST = TB
A rough sketch of this rhombus can be drawn as follows.
(1) ΔBET can be obstructed by using the given measurements as follows.
(2) Vertex S is 4.5 cm away from vertex E and also from vertex T. Therefore, while taking E and T as
centres, draw arcs of 4.5 cm radius, which will be intersecting each other at point S.
(3) Join S to E and T.
BEST is the required rhombus.
Page : 60 , Block Name : Exercise 4.1
Exercise 4.2
Page 4 of 20 Aglasem Schools
Page 6
Book : Mathematics Ncert Solutions | Chapter-4 Maths
Q1 Construct the following quadrilaterals.
(i) quadrilateral LIFT
LI = 4 cm
IF = 3 cm
TL = 2.5 cm
LF = 4.5 cm
IT = 4 cm
(ii) Quadrilateral GOLD
OL = 7.5 cm
GL = 6 cm
GD = 6 cm
LD = 5 cm
OD = 10 cm
(iii) Rhombus BEND
BN = 5.6 cm
DE = 6.5 cm
Answer. (i) A rough sketch of this quadrilateral can be drawn as follows.
(1) ΔITL can be constructed by using the given measurements as follows.
(2) Vertex F is 4.5 cm away from vertex L and 3 cm away from vertex I. Therefore, while taking L and I
as centres, draw arec of 4.5 cm radius and 3 cm radius respectively, which will be intersecting each other
at point F.
Page 5 of 20 Aglasem Schools
Q1 Construct the following quadrilaterals.
(i) quadrilateral LIFT
LI = 4 cm
IF = 3 cm
TL = 2.5 cm
LF = 4.5 cm
IT = 4 cm
(ii) Quadrilateral GOLD
OL = 7.5 cm
GL = 6 cm
GD = 6 cm
LD = 5 cm
OD = 10 cm
(iii) Rhombus BEND
BN = 5.6 cm
DE = 6.5 cm
Answer. (i) A rough sketch of this quadrilateral can be drawn as follows.
(1) ΔITL can be constructed by using the given measurements as follows.
(2) Vertex F is 4.5 cm away from vertex L and 3 cm away from vertex I. Therefore, while taking L and I
as centres, draw arec of 4.5 cm radius and 3 cm radius respectively, which will be intersecting each other
at point F.
Page 5 of 20 Aglasem Schools
Page 7
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(3) Join F to T and F to I.
LIFT is the required quadrilateral.
(ii) A rough sketch of this quadrilateral can be drawn as follows.
(1) ΔGDL can be constructed by using the given measurements as follows.
(2) Vertex O is 10 cm away from vertex D and 7.5 cm away from vertex L. Therefore, while taking D
and L as centres, draw arcs of 10 cm radius and 7.5 cm radius respectively. These will intersect each
other at point O.
(3) Join O to G and L.
Page 6 of 20 Aglasem Schools
(3) Join F to T and F to I.
LIFT is the required quadrilateral.
(ii) A rough sketch of this quadrilateral can be drawn as follows.
(1) ΔGDL can be constructed by using the given measurements as follows.
(2) Vertex O is 10 cm away from vertex D and 7.5 cm away from vertex L. Therefore, while taking D
and L as centres, draw arcs of 10 cm radius and 7.5 cm radius respectively. These will intersect each
other at point O.
(3) Join O to G and L.
Page 6 of 20 Aglasem Schools
Page 8
Book : Mathematics Ncert Solutions | Chapter-4 Maths
GOLD is the required quadrilateral.
(iii) We know that the diagonals of a rhombus always bisect each other at 90 . Let us assume that these
∘
are intersecting each other at point O in this rhombus.
Hence, EO = OD = 3.25 CM
A rough sketch of this rhombus can be drawn as follows.
(1) Draw a line segment BN of 5.6 cm and also draw its perpendicular bisector. Let it
intersect the line segment BN at point O.
(2) Taking O as centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at point D and
E.
Page 7 of 20 Aglasem Schools
GOLD is the required quadrilateral.
(iii) We know that the diagonals of a rhombus always bisect each other at 90 . Let us assume that these
∘
are intersecting each other at point O in this rhombus.
Hence, EO = OD = 3.25 CM
A rough sketch of this rhombus can be drawn as follows.
(1) Draw a line segment BN of 5.6 cm and also draw its perpendicular bisector. Let it
intersect the line segment BN at point O.
(2) Taking O as centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at point D and
E.
Page 7 of 20 Aglasem Schools
Page 9
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(3) Join points D and E to points 3 and N.
BEND is the required quadrilateral.
Page : 62 , Block Name : Exercise 4.2
Exercise 4.3
Page 8 of 20 Aglasem Schools
(3) Join points D and E to points 3 and N.
BEND is the required quadrilateral.
Page : 62 , Block Name : Exercise 4.2
Exercise 4.3
Page 8 of 20 Aglasem Schools
Page 10
Book : Mathematics Ncert Solutions | Chapter-4 Maths
Q1 Construct the following quadrilaterals.
(i) Quadrilateral MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°
(ii) Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°
(iii) Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°
(iv) Rectangle OKAY
OK = 7 cm
KA = 5 cm
Answer. (i) (1) A rough sketch of this quadrilateral can be drawn as follows.
(2) Draw a line segment MO of 6 cm and an angle of 105 at point O. As vertex R is 4.5 cm away from
∘
the vertex O, cut a line segment OR of 4.5 cm from this ray.
Page 9 of 20 Aglasem Schools
Q1 Construct the following quadrilaterals.
(i) Quadrilateral MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°
(ii) Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°
(iii) Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°
(iv) Rectangle OKAY
OK = 7 cm
KA = 5 cm
Answer. (i) (1) A rough sketch of this quadrilateral can be drawn as follows.
(2) Draw a line segment MO of 6 cm and an angle of 105 at point O. As vertex R is 4.5 cm away from
∘
the vertex O, cut a line segment OR of 4.5 cm from this ray.
Page 9 of 20 Aglasem Schools
Page 11
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(3) Again, draw an angle of 105 at point R.
∘
(4) Draw an angle of 60 at point M. Let this ray meet the previously drawn ray from R at point E.
∘
MORE is the required quadrilateral.
(ii) (1) The sum of the angles of a quadrilateral is 360 ∘
In quadrilateral PLAN, ∠P + ∠L + ∠A + ∠N = 360 ∘
∘ ∘ ∘ ∘
90 + ∠L + 110 + 85 = 360
∘ ∘
285 + ∠L = 360
∘ ∘ ∘
∠L = 360 − 285 = 75
(2) A rough sketch of this quadrilateral is as follows.
Page 10 of 20 Aglasem Schools
(3) Again, draw an angle of 105 at point R.
∘
(4) Draw an angle of 60 at point M. Let this ray meet the previously drawn ray from R at point E.
∘
MORE is the required quadrilateral.
(ii) (1) The sum of the angles of a quadrilateral is 360 ∘
In quadrilateral PLAN, ∠P + ∠L + ∠A + ∠N = 360 ∘
∘ ∘ ∘ ∘
90 + ∠L + 110 + 85 = 360
∘ ∘
285 + ∠L = 360
∘ ∘ ∘
∠L = 360 − 285 = 75
(2) A rough sketch of this quadrilateral is as follows.
Page 10 of 20 Aglasem Schools
Page 12
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(3) Draw a line segment PL of 4 cm and draw an angle of 75 at point L. As vertex A is 6.5 cm away
∘
from vertex L, cut a line segment LA of 6.5 cm from this ray.
(4) Again draw an angle of 110 at point A.
∘
(5) Draw an angle of 90 at point P. This ray will meet the previously drawn ray from A at point N.
∘
PLAN is the required quadrilateral.
(iii) (1) Firstly, a rough sketch of this quadrilateral is as follows.
Page 11 of 20 Aglasem Schools
(3) Draw a line segment PL of 4 cm and draw an angle of 75 at point L. As vertex A is 6.5 cm away
∘
from vertex L, cut a line segment LA of 6.5 cm from this ray.
(4) Again draw an angle of 110 at point A.
∘
(5) Draw an angle of 90 at point P. This ray will meet the previously drawn ray from A at point N.
∘
PLAN is the required quadrilateral.
(iii) (1) Firstly, a rough sketch of this quadrilateral is as follows.
Page 11 of 20 Aglasem Schools
Page 13
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(2) Draw a line segment HE of 5 cm and an angle of 85 at point E. As vertex A is 6 cm away from
∘
vertex E, cut a line segment EA of 6 cm from this ray.
(3) Vertex R is 6 cm and 5 cm away from vertex H and A respectively. By taking radius as 6 cm and 5
cm, draw arcs from point H and A respectively. These will be intersecting each other at point R.
HEAR is the required quadrilateral.
Page 12 of 20 Aglasem Schools
(2) Draw a line segment HE of 5 cm and an angle of 85 at point E. As vertex A is 6 cm away from
∘
vertex E, cut a line segment EA of 6 cm from this ray.
(3) Vertex R is 6 cm and 5 cm away from vertex H and A respectively. By taking radius as 6 cm and 5
cm, draw arcs from point H and A respectively. These will be intersecting each other at point R.
HEAR is the required quadrilateral.
Page 12 of 20 Aglasem Schools
Page 14
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(iv) (1) A rough sketch of this quadrilateral is drawn as follows.
(2) Draw a line segment OK of 7 cm and angle of 90 at point K. As vertex A is 5 cm away from vertex
∘
K, cut a line segment KA of 5 cm from this ray.
(3) Vertex Y is 5 cm and 7 cm away from vertex O and A respectively. By taking radius as 5 cm and 7
cm, draw arcs from point O and A respectively. These will be intersecting each other at point Y.
(4) Join Y to A and O.
OKAY is the required quadrilateral.
Page : 64 , Block Name : Exercise 4.3
Page 13 of 20 Aglasem Schools
(iv) (1) A rough sketch of this quadrilateral is drawn as follows.
(2) Draw a line segment OK of 7 cm and angle of 90 at point K. As vertex A is 5 cm away from vertex
∘
K, cut a line segment KA of 5 cm from this ray.
(3) Vertex Y is 5 cm and 7 cm away from vertex O and A respectively. By taking radius as 5 cm and 7
cm, draw arcs from point O and A respectively. These will be intersecting each other at point Y.
(4) Join Y to A and O.
OKAY is the required quadrilateral.
Page : 64 , Block Name : Exercise 4.3
Page 13 of 20 Aglasem Schools
Page 15
Book : Mathematics Ncert Solutions | Chapter-4 Maths
Exercise 4.4
Q1 Construct the following quadrilaterals.
(i) Quadrilateral DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°
(ii) Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U = 120°
Answer. (i) (1) A rough sketch of this quadrilateral can be drawn as follows.
(2) Draw a line segment DE of 4 cm and an angle of 60 at point E. As vertex A is 5 cm away from
∘
vertex E, cut a line segment EA of 5 cm from this ray.
(3) Again draw an angle of 90 at point A. As vertex R is 4.5 cm away from vertex A, cut a line segment
∘
RA of 4.5 cm from this ray.
Page 14 of 20 Aglasem Schools
Exercise 4.4
Q1 Construct the following quadrilaterals.
(i) Quadrilateral DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°
(ii) Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U = 120°
Answer. (i) (1) A rough sketch of this quadrilateral can be drawn as follows.
(2) Draw a line segment DE of 4 cm and an angle of 60 at point E. As vertex A is 5 cm away from
∘
vertex E, cut a line segment EA of 5 cm from this ray.
(3) Again draw an angle of 90 at point A. As vertex R is 4.5 cm away from vertex A, cut a line segment
∘
RA of 4.5 cm from this ray.
Page 14 of 20 Aglasem Schools
Page 16
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(4) Join D to R.
DEAR is the required quadrilateral.
(ii) (1) A rough sketch of this quadrilateral can be drawn as follows.
(2) Draw a line segment RU of 3 cm and angle of 120 at point U. As vertex E is 4 cm away from vertex
∘
U, cut line segment UE of 4 cm from this ray.
(3) Next, draw an angle of 75 at point R. As vertex T is 3.5 cm away from vertex R, cut a line segment
∘
RT of 3.5 cm from this ray.
Page 15 of 20 Aglasem Schools
(4) Join D to R.
DEAR is the required quadrilateral.
(ii) (1) A rough sketch of this quadrilateral can be drawn as follows.
(2) Draw a line segment RU of 3 cm and angle of 120 at point U. As vertex E is 4 cm away from vertex
∘
U, cut line segment UE of 4 cm from this ray.
(3) Next, draw an angle of 75 at point R. As vertex T is 3.5 cm away from vertex R, cut a line segment
∘
RT of 3.5 cm from this ray.
Page 15 of 20 Aglasem Schools
Page 17
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(4) Join T to E.
Page : 67 , Block Name : Exercise 4.4
Exercise 4.5
Q1 Draw the following.
The square READ with RE = 5.1 cm.
Answer. All the sides of a square of the same measure and also all the interior angles of a square are of
90 measure. Therefore, the given square READ can be drawn as follows.
∘
(1) A rough sketch of this square READ can be drawn as follows.
(2) Draw a line segment RE of 5.1 cm and an angle of 90 at point R and E.
∘
Page 16 of 20 Aglasem Schools
(4) Join T to E.
Page : 67 , Block Name : Exercise 4.4
Exercise 4.5
Q1 Draw the following.
The square READ with RE = 5.1 cm.
Answer. All the sides of a square of the same measure and also all the interior angles of a square are of
90 measure. Therefore, the given square READ can be drawn as follows.
∘
(1) A rough sketch of this square READ can be drawn as follows.
(2) Draw a line segment RE of 5.1 cm and an angle of 90 at point R and E.
∘
Page 16 of 20 Aglasem Schools
Page 18
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(3) As vertex A and D are 5.1 cm away from vertex E and R respectively, cut line segments EA and RD,
each of 5.1 cm from these rays.
(4) Join D to A.
READ is the required square.
Page : 68 , Block Name : Exercise 4.5
Q2 Draw the following.
A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
Answer. In a rhombus, diagonals bisect each other at 900. Therefore, the given rhombus
ABCD can be drawn as follows.
(1) A rough sketch of this rhombus ABCD is as follows.
Page 17 of 20 Aglasem Schools
(3) As vertex A and D are 5.1 cm away from vertex E and R respectively, cut line segments EA and RD,
each of 5.1 cm from these rays.
(4) Join D to A.
READ is the required square.
Page : 68 , Block Name : Exercise 4.5
Q2 Draw the following.
A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
Answer. In a rhombus, diagonals bisect each other at 900. Therefore, the given rhombus
ABCD can be drawn as follows.
(1) A rough sketch of this rhombus ABCD is as follows.
Page 17 of 20 Aglasem Schools
Page 19
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(2) Draw a line segment AC Of 5.2 cm and draw its perpendicular bisector. Let it intersect the line
segment AC at point O.
6.4cm
(3) Draw arcs of 2
= 3.2cm on both sides of this perpendicular bisector. Let the arcs intersect the
perpendicular bisector at point B and D.
(4)Join points B and D with points A and C.
ABCD is the required rhombus.
Page : 68 , Block Name : Exercise 4.5
Q3 Draw the following.
A rectangle with adjacent sides of lengths 5 cm and 4 cm.
Answer. Opposite sides of a rectangle have their lengths os same measure and also. All the interior
Page 18 of 20 Aglasem Schools
(2) Draw a line segment AC Of 5.2 cm and draw its perpendicular bisector. Let it intersect the line
segment AC at point O.
6.4cm
(3) Draw arcs of 2
= 3.2cm on both sides of this perpendicular bisector. Let the arcs intersect the
perpendicular bisector at point B and D.
(4)Join points B and D with points A and C.
ABCD is the required rhombus.
Page : 68 , Block Name : Exercise 4.5
Q3 Draw the following.
A rectangle with adjacent sides of lengths 5 cm and 4 cm.
Answer. Opposite sides of a rectangle have their lengths os same measure and also. All the interior
Page 18 of 20 Aglasem Schools
Page 20
Book : Mathematics Ncert Solutions | Chapter-4 Maths
angles of a rectangle are of 90 measure. The given rectangle ABCD may be drawn as follows.
∘
(1) A rough sketch of this rectangle ABCD can be drawn as follows.
(2) Draw a line segment AB of 5 cm and an angle of 90 at point A and B.
∘
(3) As vertex C and D are 4 cm away from vertex B and A respectively, cut the segments AD and BC,
each of 4 cm, from thee rays.
(4) Join D to C.
ABCD is the required rectangle.
Page : 68 , Block Name : Exercise 4.5
Q4 Draw the following.
A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?
Answer. Opposite sides of a parallelogram are equal and parallel to each other. The given parallelogram
OKAY can be drawn as follows.
(1) A rough sketch of this parallelogram OKAY is drawn as follows.
Page 19 of 20 Aglasem Schools
angles of a rectangle are of 90 measure. The given rectangle ABCD may be drawn as follows.
∘
(1) A rough sketch of this rectangle ABCD can be drawn as follows.
(2) Draw a line segment AB of 5 cm and an angle of 90 at point A and B.
∘
(3) As vertex C and D are 4 cm away from vertex B and A respectively, cut the segments AD and BC,
each of 4 cm, from thee rays.
(4) Join D to C.
ABCD is the required rectangle.
Page : 68 , Block Name : Exercise 4.5
Q4 Draw the following.
A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?
Answer. Opposite sides of a parallelogram are equal and parallel to each other. The given parallelogram
OKAY can be drawn as follows.
(1) A rough sketch of this parallelogram OKAY is drawn as follows.
Page 19 of 20 Aglasem Schools
Page 21
Book : Mathematics Ncert Solutions | Chapter-4 Maths
(2) Draw a line segment OK of 5.5 cm and a ray at point K at a convenient angle.
(3) Draw a ray at point O parallel to the ray at K. As the vertices, A and Y, are 4.2 cm away from the
vertices K and O respectively, cut the segments KA and OY, each of 4.2 cm, from these rays.
OKAY is the required rectangle.
Page : 68 , Block Name : Exercise 4.5
Page 20 of 20 Aglasem Schools
(2) Draw a line segment OK of 5.5 cm and a ray at point K at a convenient angle.
(3) Draw a ray at point O parallel to the ray at K. As the vertices, A and Y, are 4.2 cm away from the
vertices K and O respectively, cut the segments KA and OY, each of 4.2 cm, from these rays.
OKAY is the required rectangle.
Page : 68 , Block Name : Exercise 4.5
Page 20 of 20 Aglasem Schools
Document Details
| Board / Org | NCERT |
| Exam | Class 8 |
| Type | Solution |
| Pages | 21 |
| Language | english |
| Updated | 22 Jul 2026 |