NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry, are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1

Question 1.
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
Solution.
(i) Steps of Construction

1. Draw AB 4.5 cm
2. With A as centre and radius AC = 7 cm, draw an arc.
   
3. With B as center and radius BC = 5.5 cm, draw another arc to intersect the arc drawn in step (2) at C.
4. With A as center and radius AD = 6 cm, draw an arc on the side of AC, opposite to that of B.
5. With C as center and radius CD = 4 cm, draw another arc to intersect the arc drawn in step (4) at D.
6. Join BC, CD, DA, and AC.

Then, ABCD is the required quadrilateral.

(ii) Steps of Construction

1. Draw JU = 3.5 cm
2. With J as center and radius JP = 4.5 cm, draw an arc.
3. With U as center and radius UP = 6.5 cm, draw another arc to intersect the arc drawn in step 2 at P.
   
4. With U as center and radius UM = 4 cm, draw an arc on the side of PU opposite to that of J.
5. With P as center and radius PM = 5 cm, draw another arc to intersect the arc drawn in step 4 at M.
6. Join UM, MP, PJ, and UP.

Then, JUMP is the required quadrilateral.

(iii) Steps of Construction
[We know that in a parallelogram, opposite sides are equal in length.
∴ MO = ER = 4.5 cm and ME – OR = 6 cm]

1. Draw MO = 4.5 cm
2. With M as center and radius ME = 6 cm, draw an arc.
   
3. With O as center and radius OE = 7.5 cm, draw an arc to intersect the arc drawn in step 2 at E.
4. With O as center and radius OR = 6 cm, draw an arc on the side of OE opposite to that of M.
5. With E as center and radius ER = 4.5 cm, draw another arc to intersect the arc drawn in step 4 at R.
6. Join OR, RE, EM, and EO.

Then, MORE is the required parallelogram.

(iv) Steps of Construction
[We know that in a rhombus, all the sides are equal in length.
∴ BE = ES = ST = TB = 4.5 cm]

1. Draw BE = 4.5 cm
2. With B as centre and radius BT = 4.5 cm, draw an arc.
   
3. With E as center and radius
   ET = 6 cm, draw another arc to intersect the arc drawn in step 2 at T.
4. With E as center and radius
   ES = 4.5 cm, draw an arc on the side of ET opposite to that of B.
5. With T as center and radius
   TS = 4.5 cm, draw another arc to
   intersect the arc drawn in step 4 at S.
6. Join ES, ST, TB, and TE.

Then, BEST is the required rhombus.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2

Question 1.
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
Solution.
(i) Steps of Construction

1. Draw LI = 4 cm.
2. With L as center and radius LT = 2.5 cm, draw an arc.
   
3. With I as center and radius, IT = 4 cm, draw another arc to intersect the arc drawn in step 2 at T.
4. With I as center and radius IF = 3 cm, draw an arc.
5. With L as center and radius LF = 4.5 cm, draw another arc to intersect the arc drawn in step 4 at F.
6. Join IF, FT, TL, LF and IT.

Then, LIFT is the required quadrilateral.

(ii) Steps of Construction

1. Draw LD = 5 cm.
2. With L as center and radius LG = 6 cm, draw an arc.
   
3. With D as center and radius DG = 6 cm, draw another arc to intersect the arc drawn in step 2 at G.
4. With L as center and radius LO = 7.5 cm, draw an arc.
5. With D as center and radius DO = 10 cm, draw another arc to intersect the arc drawn in step 4 at O.
6. Join DG, GO, OL, LG and DO.

Then GOLD is the required quadrilateral.

(iii) Steps of Construction

1. Draw DE = 6.5 cm.
2. Draw perpendicular bisector PQ of DE so as to intersect DE at M. Then M is the mid-point of DE.
3. With M as centre and radius
   \(=\frac { 1 }{ 2 } \times \left( 5.6 \right) =2.8 cm\)
   opposite sides of DE to intersect MP at N and MQ at B.
   
4. Join DN, NE, EB, and BD.

Then, BEND is the required rhombus.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3

Question 1.
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.
Solution.
(i) Steps of Construction

1. Draw MO = 6 cm.
2. At 0, draw ray OX such that Z MOX = 105°.
   
3. Cut OR = 4.5 cm from ray OX.
4. At M, draw ray MY such that ∠OMY = 60°.
5. At R, draw ray RZ such that ∠ORZ = 105°.
6. Let the rays MY and RZ meet at E.

Then, MORE is the required quadrilateral.

(ii) Steps of Construction

1. Draw PL = 4 cm.
2. At L, draw ray LX such that ∠PLX = 75°.
   
   By Angle-sum property of a quadrilateral,
   ∠P + ∠A + ∠N + ∠L = 360°
   ⇒ 90° + 110° + 85° + Z L = 360°
   ⇒ 285° + ∠ L = 360°
   ⇒ ∠L = 360° – 285°
   ⇒ ∠L = 75°.
3. Cut LA = 6.5 cm from ray LX.
4. At A, draw ray AY such that ∠LAY = 110°.
5. At P, draw ray PZ such that ∠LPZ = 90°.
   Let the rays AY and PZ meet at N.

Then, PLAN is the required quadrilateral.

(iii) Steps of Construction

1. Draw HE = 5 cm.
2. At E, draw ray EX such that ∠HEX = 85°
   ∴ Opposite angles of a parallelogram are equal.
   ∵ ∠E = ∠R = 85°
3. Cut EA = 6 cm from the ray EX.
   
4. With A as centre and radius AR = 5 cm, draw an arc.
5. With H as centre and radius HR = 6 cm; draw another arc to intersect the arc drawn in step 4 at R.
   ∴ opposite sides of a parallelogram are equal in length
   ∵ AR = EH = 5 cm
   and HR = EA = 6 cm
6. Join AR and HR.

Then, HEAR is the required parallelogram.

(iv) Steps of Construction
[We know that each angle of a rectangle
is 90°.
∴ ∠O=∠K=∠A=∠Y= 90°.
Also, opposite sides of a rectangle are equal in length.
∴ OY = KA = 5 cm and AY = KO = 7 cm]

1. Draw OK = 7 cm.
2. At K, draw ray KX such that ∠OKX = 90°.
3. Cut KA – 5 cm from ray KX.
   
4. Taking A as centre and radius AY = 7 cm, draw an arc.
5. Taking O as centre and radius OY = 5 cm, draw another arc to intersect the arc drawn in step 4 at Y.
6. Join AY and OY.

Then OKAY is the required rectangle.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4

Question 1.
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°
Solution.
(i) Steps of Construction

1. Draw DE = 4 cm.
2. At E, draw ray EX such that ∠DEX = 60°.
3. From ray EX, cut EA = 5 cm.
   
4. At A, draw ray AY such that ∠EAY = 90°.
5. Cut AR = 4.5 cm from ray AY.
6. Join RD.

Then, DEAR is the required quadrilateral.

(ii) Steps of Construction

1. Draw TR = 3.5 cm.
2. At R, draw ray RX such that ∠TRX = 75°.
   
3. Cut RU = 3 cm from ray RX.
4. At U, draw ray UY such that ∠RUY = 120°.
5. Cut UE = 4 cm from ray UY.
6. Join ET.

Then, TRUE is the required quadrilateral.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5

Question 1.
Draw the following:
1. The square READ with RE = 5.1 cm.
2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
3. A rectangle with adjacent sides of lengths 5 cm and 4 cm.
4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?
Solution.
1. Steps of Construction

1. Draw RE = 5.1 cm.
2. At R, draw a ray RX such that ∠ERX
   
3. From ray RX, cut RD = 5.1 cm.
4. At E, draw a ray EY such that ∠REY = 90°.
5. From ray EY, cut EA = 5.1 cm.
6. Join AD.

Then, READ is the required square.

2. Steps of Construction
[We know that the diagonals of a rhombus bisect each other at right angles. So in rhombus ABCD, the diagonals AC and BD will bisect each other at right angles.]

1. Draw AC = 5.2 cm.
2. Construct its perpendicular bisector. Let it intersect AC at O.
   
3. Cut off \(\frac { 6.4 }{ 2 } \)= 3.2 cm lengths on either side of the bisector drawn in step 2, we get B and D.
4. Join AB, BC, CD, and DA.

Then, ABCD is the required rhombus.

3. Steps of Construction
[We know that each angle of a rectangle is 90°. So, in rectangle PQRS,
∠P=∠Q=∠R=∠S= 90°.
Also, opposite sides of a rectangle are parallel.
So, in rectangle PQRS,
PQ || SR and PS || QR]

1. Draw PQ = 5 cm.
2. At Q, draw a ray QX such that ∠PQX = 90°.
   
3. From ray QX, cut QR = 4 cm.
4. At P, draw a ray PY parallel to QR.
5. At R, draw a ray RZ parallel to QP to meet the ray drawn in step 4 at S.

Then, PQRS is the required rectangle.

4. Steps of Construction
[We know that in a parallelogram, opposite sides are parallel and equal. So,
OK = YA and OK || YA;
KA = OY and KA || OY]

1. Draw OK = 5.5 cm.
   
2. At K, draw a ray KX at any suitable angle from OK.
3. From ray KX, cut KA = 4.2 cm.
4. A, draw a ray AT parallel to KO.
5. At O, draw a ray OZ parallel to KA to cut the ray drawn in step 4 at Y.

Then, OKAY is the required parallelogram.
This is not unique.
Note: We can construct countless parallelograms with these dimensions by varying ∠OKA

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