The NCERT Solutions for Class 8 Maths Chapter 4 on Practical Geometry offer valuable assistance to students, facilitating a comprehensive understanding of the subject and aiding them in achieving high marks in examinations. These solutions feature meticulous, step-by-step explanations for all problems found in Chapter 4 of the Class 8 NCERT Textbook, ensuring that students can grasp the intricacies of Practical Geometry.

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**

**(iv) Rhombus BEST**

**BE = 4.5 cm**

**ET = 6 cm**

Answer :

(i)

The rough sketch of the quadrilateral ABCD 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 A as the centre, draw an arc of radius 6 cm.

(3) Taking C as the centre, draw an arc of radius 4 cm, cutting the previous arc at point D. Joint D to A and C.

ABCD is the required quadrilateral.

(ii)

The rough sketch of the quadrilateral JUMP can be drawn as follows.

(1) ∆ JUP 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.

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.

i.e., ME = OR, MO = ER

The rough sketch of the parallelogram MORE 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.

MORE is the required parallelogram.

(iv)

We know that all sides of a rhombus are of the same measure. Hence, BE = ES = ST = TB

The rough sketch of the rhombus BEST can be drawn as follows.

(1) ∆ BET can be constructed 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 intersect each other at point S.

(3) Join S to E and T.

NCERT Solution For Class 8 Maths Chapter 4 Image

BEST is the required rhombus.

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 :

(i) A rough sketch of the quadrilateral LIFT can be drawn as follows.

(1) ∆ ITL 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. ∴ while taking L and I as centres, draw arcs of 4.5 cm radius and 3 cm radius, respectively, which will intersect each other at point F.

(3) Join F to T and F to I.

LIFT is the required quadrilateral.

(ii) The rough sketch of the quadrilateral GOLD can be drawn as follows.

(1) ∆ GDL 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.

GOLD is the required quadrilateral.

(iii) We know that the diagonals of a rhombus always bisect each other at 90º.

Let us assume that these are intersecting each other at point O in this rhombus. Hence, EO = OD = 3.25 cm

The rough sketch of the rhombus BEND 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 the centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at points D and E.

(3) Join points D and E to points B and N.

BEND is the required quadrilateral.

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**

Answer :

(i) Rough Figure:

(1) Draw a line segment MO of 6 cm and an angle of 105º 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.

(2) Again, draw an angle of 105º at point R.

(3) Draw an angle of 60º at point M. Let this ray meet the previously drawn ray from R at point E.

MORE is the required quadrilateral.

(ii) 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°

Rough Figure:

(1) Draw a line segment PL of 4 cm and draw an angle of 75º 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.

(2) Again, draw an angle of 110º at point A.

(3) Draw an angle of 90º at point P. This ray will meet the previously drawn ray from A at point N.

PLAN is the required quadrilateral.

(iii) Rough Figure:

(1) Draw a line segment HE of 5 cm and an angle of 85º at point E. As vertex A is 6 cm away from vertex E, cut a line segment EA of 6 cm from this ray.

(2) Vertex R is 6 cm and 5 cm away from vertex H and A, respectively. By taking radii as 6 cm and 5 cm, draw arcs from points H and A, respectively. These will intersect each other at point R.

(3) Join R to H and A.

HEAR is the required quadrilateral.

(iv) Rough Figure:

(1) Draw a line segment OK of 7 cm and an angle of 90º at point K. As vertex A is 5 cm away from vertex K, cut a line segment KA of 5 cm from this ray.

(2) Vertex Y is 5 cm and 7 cm away from vertex O and A, respectively. By taking

radii as 5 cm and 7 cm, draw arcs from points O and A, respectively. These will intersect each other at point Y.

(3) Join Y to A and O.

OKAY is the required quadrilateral.

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°**

Answer :

(i) Rough Figure:

(1) Draw a line segment DE of 4 cm and an angle of 60º at point E. As vertex A is 5 cm away from vertex E, cut a line segment EA of 5 cm from this ray.

(2) Again, draw an angle of 90º 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.

(3) Join D to R.

DEAR is the required quadrilateral.

(ii) Rough Figure:

(1) Draw a line segment RU of 3 cm and an angle of 120º at point U. As vertex E is 4 cm away from vertex U, cut a line segment UE of 4 cm from this ray.

(2) Next, draw an angle of 75º 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.

(3) Join T to E.

TRUE is the required quadrilateral.

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 length 5 cm and 4 cm**

**4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. **

Answer :

1. All the sides of a square are of the same measure, and also, all the interior angles of a square are 90º measure. Therefore, the given square READ can be drawn as follows.

Rough Figure:

(1) Draw a line segment RE of 5.1 cm and an angle of 90º at points R and E.

(2) 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.

(3) Join D to A.

READ is the required square.

2. In a rhombus, diagonals bisect each other at 90º. ∴, the given rhombus ABCD can be drawn as follows.

Rough Figure:

(1) Draw a line segment AC of 5.2 cm and draw its perpendicular bisector. Let it intersect the line segment AC at point O.

(2) Draw arcs of 6.4/2 = 3.2 on both sides of this perpendicular bisector. Let the arcs intersect the perpendicular bisector at points B and D.

(3) Join points B and D with points A and C.

ABCD is the required rhombus.

3. Opposite sides of a rectangle have lengths of the same measure, and also, all the interior angles of a rectangle are 90º measure. The given rectangle ABCD may be drawn as follows.

Rough figure:

(1) Draw a line segment AB of 5 cm and an angle of 90º at points A and B.

(2) As vertex C and D are 4 cm away from vertex B and A, respectively, cut line segments AD and BC, each of 4 cm, from these rays.

(3) Join D to C.

ABCD is the required rectangle.

4. Opposite sides of a parallelogram are equal and parallel to each other. The given parallelogram OKAY can be drawn as follows.

Rough Figure:

(1) Draw a line segment OK of 5.5 cm and a ray at point K at a convenient angle.

(2) 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 line segments KA and OY, each of 4.2 cm, from these rays.

(3) Join Y to A.

OKAY is the required rectangle.

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The NCERT solution for class 8 Chapter 4: Practical Geometry is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tacking more difficult concepts in their further education.

Yes, the NCERT solution for class 8 Chapter 4: Practical Geometry is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Practical Geometry ally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.

You can get all the NCERT solutions for class 8 Maths Chapter 4 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.

Yes, students must practice all the questions provided in the NCERT solution for class 8 Maths Chapter 4 : Practical Geometry as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.

Students can utilize the NCERT solution for class 8 Maths Chapter 4 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution. Also, you can make Practical Geometry al notes and jot down the important concepts for your understanding.

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