Class 8 Exam  >  Class 8 Notes  >  NCERT Solutions (Ex - 4.1, 4.2, 4.3) - Practical Geometry, Maths, Class 8

NCERT Solutions (Ex - 4.1, 4.2, 4.3) - Practical Geometry, Maths, Class 8 PDF Download

Exercise 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 

Answer 1:

(i) Given: AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm, AC = 7 cm

To construct: A quadrilateral ABCD

Steps of construction: 

(a) Draw AB = 4.5 cm.

(b) Draw an arc taking radius 5.5 cm from point B.

(c) Taking radius 7 cm, draw another arc from point A which intersects the first arc at point C.

(d) Join BC and AC.

(e) Draw an arc of radius 6 cm from point A and draw another arc of radius 4 cm from point C which intersects at D.

(f) Join AD and CD.

It is required quadrilateral ABCD.

CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved

(ii) Given:

JU = 3.5 cm, UM = 4 cm, MP = 5 cm, PJ = 4.5 cm, PU = 6.5 cm

To construct: A quadrilateral JUMP

Steps of construction: 

(a) Draw JU = 3.5 cm.

(b) Draw an arc of radius 4.5 cm taking centre J and then draw another arc of radius 6.5 cm taking U as centre. Both arcs intersect at P.

(c) Join PJ and PU.

(d) Draw arc of radius 5 cm and 4 cm taking P and U as centres respectively, which intersect at M.

(e) Join MP and MU.

It is required quadrilateral JUMP.

CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved

(iii) Given: OR = 6 cm, RE = 4.5 cm, EO = 7.5 cm

To construct: A parallelogram MORE.

Steps of construction: 

(a) Draw OR = 6 cm.

(b) Draw arcs of radius 7.5 cm and radius 4.5 cm taking O and R as centres respectively, which intersect at E.

(c) Join OE and RE.

(d) Draw an arc of 6 cm radius taking E as centre.

(e) Draw another arc of 4.5 cm radius taking O as centre, which intersects at M.

(f) Join OM and EM.

It is required parallelogram MORE. 

CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved

(iv) Given: BE = 4.5 cm, ET = 6 cm

To construct: A rhombus BEST.

Steps of construction: 

(a) Draw TE = 6 cm and bisect it into two equal parts.

(b) Draw up and down perpendiculars to TE.

(c) Draw two arcs of 4.5 cm taking E and T as centres, which intersect at S.

(d) Again draw two arcs of 4.5 cm taking E and T as centres, which intersects at B.

(e) Join TS, ES, BT and EB.

It is the required rhombus BEST.

CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved

 

Exercise 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 

Answer 1: 

(i) Given: LI = 4 cm, IF = 3 cm, TL = 2.5 cm, LF = 4.5 cm, IT = 4 cm

To construct: A quadrilateral LIFT

Steps of construction: 

(a) Draw a line segment LI = 4 cm.

(b) Taking radius 4.5 cm, draw an arc taking L as centre.

(c) Draw an arc of 3 cm taking I as centre which intersects the first arc at F.

(d) Join FI and FL.

(e) Draw another arc of radius 2.5 cm taking L as centre and 4 cm taking I as centre which intersect at T.

(f) Join TF, Tl and TI.

It is the required quadrilateral LIFT.

CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved

 

(ii) Given:

OL = 7.5 cm, GL = 6 cm, GD = 6 cm, LD = 5 cm, OD = 10 cm

To construct:  A quadrilateral GOLD

Steps of construction:

(a) Draw a line segment OL = 7.5 cm

(b) Draw an arc of radius 5 cm taking L as centre and another arc of radius 10 cm taking O as centre which intersect the first arc point at D.

(c) Join LD and OD.

(d) Draw an arc of radius 6 cm from D and draw another arc of radius 6 cm taking L as centre, which intersects at G.

(e) Join GD and GO.

It is the required quadrilateral GOLD.

CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved

 

(iii) Given: BN = 5.6 cm, DE = 6.5 cm

To construct: A rhombus BEND.

Steps of construction: 

(a) Draw DE = 6.5 cm.

(b) Draw perpendicular bisector of line segment DE.

(c) Draw two arcs of radius 2.8 cm from intersection point O, which intersects the line KN at B and N. (d)Join BE, BD as well as ND and NE.

It is the required rhombus BEND.

CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved

 

Exercise 4.3 

Question 1: 

Construct the following quadrilaterals: 

[i] Quadrilateral MORE

MO = 6 cm, OR = 4.5 cm, ∠M = 60% ∠O = 105o , ∠R = 105°

[ii] Quadrilateral PLAN

PL = 4 cm, LA = 6.5 cm, ∠P = 90o ∠A = 110o ∠N = 850

(iii) Parallelogram HEAR

HE = 5 cm, EA = 6 cm, ∠R = 850

Rectangle OKAY

OK = 7 cm, KA = 5 cm 

Answer 1: 

(i) Given: MO = 6 cm, OR = 4.5 cm, CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, SolvedM = 60o , CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, SolvedO = 105o , CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, SolvedR = 105o

To construct: A quadrilateral MORE.

Steps of construction:

(a) Draw a line segment MO = 6 cm.

(b) Construct CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved R = 105o and taking radius 4.5 cm, draw an arc taking O as centre, which intersects at R.

(c) Also construct an angle 105o at R and produce the side RE.

(d) Construct another angle of 60o at point M and produce the side ME. Both sides ME and RE intersect at E.

It is the required quadrilateral MORE.

CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, Solved

(ii) Given: PL = 4 cm, LA = 6.5 cm, CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, SolvedP = 90o , CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, SolvedA = 110o , CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, SolvedN = 85o

To construct: A quadrilateral PLAN.

To find:  ∠L = 3600 - (90 + 85+ 110°) = 360 -285= 75o

Steps of construction:

  1. Draw a line segment PL = 4 cm.
  2. Construct angle of 90o at P and produce the side PN.
  3. Construct angle of 75o at L and with L as centre, draw an arc of radius 6 cm, which intersects at A.
  4. Construct ∠A = 110° at A and produce the side AN which intersects PN at N.

It is the required quadrilateral PLAN.

NCERT Solutions (Ex - 4.1, 4.2, 4.3) - Practical Geometry, Maths, Class 8

(iii) Given: HE = 5 cm, EA = 6 cm, CBSE, NCERT, Syllabus, Important, Class VI, Mathematics, Questions and Answers, SolvedR = 85o

To construct: A parallelogram HEAR.

To find:   ∠H = 180° - 85o = 95°    [ ∵Sum of adjacent angle of ||gm is 1800]

Steps of construction:

  1. Draw a line segment HE = 5 cm.
  2. Construct ∠H = 95* and draw an arc of radius 6 cm with centre H. It intersects AR at R,
  3. Join RH.
  4. Draw ∠R = ∠E = 85° and draw an arc of radius 6 cm with E as a centre which intersects RA at A.
  5. Join RA

It is the required parallelogram HEAR.

NCERT Solutions (Ex - 4.1, 4.2, 4.3) - Practical Geometry, Maths, Class 8

(iv) Given: OK = 7 cm, KA = 5 cm

To construct: A rectangle OKAY.

Steps of construction:

  1. Draw a line segment OK = 7 cm.
  2. Construct angle 90° at both points 0 and K and produce these sides.
  3. Draw two arcs of radius 5 cm from points 0 and K respectively. These arcs intersect at Y and A
  4. Join YA.

It is the required rectangle OKAY.

NCERT Solutions (Ex - 4.1, 4.2, 4.3) - Practical Geometry, Maths, Class 8

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FAQs on NCERT Solutions (Ex - 4.1, 4.2, 4.3) - Practical Geometry, Maths, Class 8

1. What is practical geometry in Class 8 Maths?
Ans. Practical geometry is a branch of mathematics that deals with the study of shapes, sizes, and positions of objects in the real world. In Class 8 Maths, practical geometry involves constructing various geometrical figures using a ruler, compass, and protractor.
2. What are the basic tools required for practical geometry in Class 8?
Ans. The basic tools required for practical geometry in Class 8 are a ruler, compass, and protractor. A ruler is used to measure the length of a line segment, a compass is used to draw circles and arcs of circles, and a protractor is used to measure angles.
3. What are the different types of triangles in practical geometry?
Ans. In practical geometry, there are three types of triangles based on their sides and angles. They are: - Equilateral triangle: A triangle with all three sides of equal length and all three angles of equal measure (60 degrees). - Isosceles triangle: A triangle with two sides of equal length and two angles of equal measure. - Scalene triangle: A triangle with all three sides of different lengths and all three angles of different measures.
4. How can we construct a perpendicular bisector of a line segment in practical geometry?
Ans. To construct a perpendicular bisector of a line segment in practical geometry, follow these steps: - Draw a line segment AB using a ruler. - Place the compass at point A and draw an arc that intersects the line segment AB at point C. - Place the compass at point B and draw an arc that intersects the line segment AB at point D. - Draw a line passing through points C and D using a ruler. This line is the perpendicular bisector of line segment AB.
5. What is the difference between a line segment and a line in practical geometry?
Ans. In practical geometry, a line segment is a part of a line that has two endpoints. A line, on the other hand, is a straight path that extends infinitely in both directions. A line has no endpoints.
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