NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry

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

Answer :

(i)Draw a rough sketch first as per the given dimensions.

Construction:

Step 1: Draw a line segment LI = 4 cm.

Step 2: Draw an arc with centre at I and radius 3 cm.

Step 3: Draw another arc with centre L and radius 4.5 cm to meet the former arc at F and Join LF and IF.

Step 4: Draw an arc with centre L and radius 2.5 cm.

Step 5: Draw another arc with centre I and radius 4 cm to meet the previous arc at T and Join LT and IT.

Hence we have LIFT as the required quadrilateral with given dimensions.

(ii)Draw a rough sketch first as per the given dimensions.


Construction:

Step 1: Draw a line segment OL = 7.5 cm

Step 2: Draw an arc with centre at O and of radius 10 cm.

Step 3: Draw another arc with centre at L and of radius 5 cm to meet the previous arc at D and join OD and LD.

Step 4: Draw an arc with centre L and D with equal radii of 6 cm to meet each other at G and join LG and DG.

Hence we have GOLD as the required quadrilateral with given dimensions.

(iii)Draw a rough sketch first as per the given dimensions and we know that the diagonals of a rhombus bisect each other at the right angle.


Construction:

Step 1: Draw a line segment BN = 5.6 cm.

Step 2: Draw the right bisector of BN at O.

Step 3: Draw two arcs with centre O and radius \(\frac { 1 }{ 2 } \times DE\), i.e., \(\frac { 1 }{ 2 } \times 6.5 = 3.25 cm\) to meet the right bisector at D and E.

Step 4: Join BE, EN, ND and BD.

Hence we have, BEND as the required rhombus with given dimensions.

1.Construct the following quadrilaterals:

(i) Quadrilateral MORE

MO = 6 cm ∠R = \(105^{\circ}\), OR = 4.5 cm, ∠M = \(60^{\circ}\), ∠O = \(105^{\circ}\)

(ii) Quadrilateral PLAN

PL = 4 cm, LA = 6.5 cm, ∠P = \(90^{\circ}\), ∠A = \(110^{\circ}\), ∠N = \(85^{\circ}\)

(iii) Parallelogram HEAR

HE = 5 cm, EA = 6 cm, ∠R = \(85^{\circ}\)

(iv) Rectangle OKAY

OK = 7 cm, KA = 5 cm

Answer :

(i) Draw a rough sketch first as per the given dimensions.


Construction:

Step 1: Draw a line segment OR = 4.5 cm

Step 2: Draw an angle of \(105^{\circ}\) at O and also an angle of \(105^{\circ}\) at R with the help of protractor.

Step 3: Make an arc OM = 6 cm.

Step 4: Draw an angle of \(60^{\circ}\) at M to meet the line through R at E.

Hence we have, MORE as the required quadrilateral.

(ii)Draw a rough sketch first as per the given dimensions.


Construction:

Step 1: Draw a line segment LA = 6.5 cm

Step 2: Draw an angle of 75° at L and \(110^{\circ}\) at A with the help of a protractor. [\(∵ 360^{\circ} – (110^{\circ} + 90^{\circ} + 85^{\circ}) = 75^{\circ}\)]

Step 3: Cut LP = 4 cm.

Step 4: Draw an angle of 90° at P which meets the line through A at N.

Hence we have PLAN as the required quadrilateral.

(iii)Draw a rough sketch first as per the given dimensions and we know that opposite sides of a parallelogram are equal.


Construction:

Step 1: Draw a line segment HE = 5 cm.

Step 2: Draw an angle of \(85^{\circ}\) at E and cut an arc of EA = 6 cm.

Step 3: Draw an arc with centre A and radius 5 cm.

Step 4: Draw another arc with centre H and radius 6 cm to meet the previous arc at R and join HR and AR.

Hence we have, HEAR as the required parallelogram.

(iv)Draw a rough sketch first as per the given dimensions and we know that each angle of a rectangle is \(90^{\circ}\) and opposite sides are equal.


Construction:

Step 1: Draw a line segment OK = 7 cm.

Step 2: Draw the angle of \(90^{\circ}\) at K and cut an arc of KA = 5 cm.

Step 3: Draw an arc with centre at O and radius 5 cm.

Step 4: Draw another arc with centre at A and radius 7 cm to meet the previous arc at Y and join OY and AY.

Hence we have OKAY as the required rectangle.

1.Construct the following quadrilaterals:
(i) Quadrilateral DEAR

DE = 4 cm, EA = 5 cm, AR = 4.5 cm, ∠E = \(60^{\circ}\), ∠A =\( 90^{\circ}\)

(ii) Quadrilateral TRUE

TR = 3.5 cm, RU = 3 cm, UE = 4.5 cm, ∠R = \(75^{\circ}\), ∠U =\( 120^{\circ}\)

Answer :

(i)Draw a rough sketch first as per the given dimensions


Construction:

Step 1: Draw a line segment DE = 4 cm.

Step 2: Draw an angle of \(60^{\circ}\) at E.

Step 3: Draw an arc with centre E and radius 5 cm to meet the angle line at A.

Step 4: Draw an angle of \(90^{\circ}\) at A and cut an arc AR = 4.5 cm and join DR.

Hence we have, DEAR as the required quadrilateral.

(ii)Draw a rough sketch first as per the given dimensions


Construction:

Step 1: Draw a line segment TR = 3.5 cm

Step 2: Draw an angle of \(75^{\circ}\) at R and cut RU = 3 cm.

Step 3: Draw an angle of \(120^{\circ}\) at U and cut UE = 4.5 cm.

Step 4: Join TE.

Hence we have TRUE as the required quadrilateral.

1.The square READ with RE = 5.1 cm.

Answer :

Construction:



Step 1: Draw RE = 5.1 cm.

Step 2: Draw an angle of \(90^{\circ}\) at E and cut an arc EA = 5.1 cm.

Step 3: Draw two arcs from A and R with radius 5.1 cm to cut each other at D.

Step 4: Join RD and AD.

Hence we have, READ as the required square.

2.A rhombus whose diagonals are 5.2 cm and 6.4 cm long.

Answer :

Construction:


Step 1: Draw a line segment AC = 6.4 cm.

Step 2: Draw the right angle bisector of AC at E.

Step 3: Draw two arcs with centre at E and radius = \(\frac { 5.2 }{ 2 }\) = 2.6 cm to cut the previous diagonal at B and D.

Step 4: Join AD, AB, BC and DC.

Hence we have ABCD as the required rhombus.

3.A rectangle with adjacent sides of lengths 5 cm and 4 cm.

Answer :

Construction:


Step 1: Draw PQ = 5 cm.

Step 2: Draw an angle of \(90^{\circ}\) at Q and cut an arc QR = 4 cm.

Step 3: Draw an arc with centre R and radius 5 cm.

Step 4: Draw another arc with centre P and radius 4 cm to meet the previous arc at S.

Step 5: Join RS and PS.

Hence we have PQRS as the required rectangle.

4.A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?

Answer :

Construction:


Step 1: Draw a line segment OK = 5.5 cm.

Step 2: Draw an angle of any measure (let's say \(60^{\circ}\)) at K and cut an arc KA = 4.2 cm.

Step 3: Draw an arc with centre A and radius of 5.5 cm.

Step 4: Draw another arc with centre O and radius 4.2 cm to cut the previous arc at Y.

Step 5: Join AY and OY.

Hence we have OKAY as the required parallelogram.

No, it is not a unique parallelogram because the angle at K can be of any measure other than \(60^{\circ}\).