NCERT Solutions for Class 8 Maths Chapter 4 - Practical Geometry

Exercise (4.1)

1. Construct the following quadrilaterals.

(i) Quadrilateral  $ \text{ABCD} $ 

 $ AB=4.5cm $ 

 $ BC=5.5cm $ 

 $ CD=4cm $ 

 $ AD=6cm $ 

 $ AC=7cm $ 

Ans: Let us first draw the rough diagram of the given quadrilateral.

1. Draw  $ \text{BC=5}\text{.5 cm} $ 

2. Draw  $ \vartriangle ABC $  by using the given measurement as follows:

3. Since  $ \text{D} $  is  $ \text{6 cm} $  away from the vertex  $ \text{A} $ . So take  $ \text{A} $  as center and draw an arc of  $ \text{6 cm} $  . The vertex  $ \text{D} $  is  $ \text{4 cm} $  away from  $ \text{C} $ . So draw an arc of radius  $ \text{4 cm} $  from  $ \text{C} $ cutting the previous arc.

4. Join  $ \text{AD} $  and  $ \text{CD} $ .

Thus  $ \text{ABCD} $   is the required quadrilateral.

(ii) Quadrilateral  JUMP 

 JU=3.5 cm

 UM=4 cm

 MP=5 cm

 PJ=4.5 cm 

 PU=6.5 cm 

Ans: Let us first draw the rough diagram of the given quadrilateral  $ \text{JUMP} $ 

1. Draw  $ \text{PU=6}\text{.5 cm} $ 

2. Draw  $ \vartriangle JUP $  by using the given measurement as follows

3. Since the vertex  $ \text{M} $  is  $ \text{5 cm} $  away from the vertex  $ \text{P} $ . So take  $ \text{P} $  as center and draw an arc of radius  $ \text{5 cm} $  . The vertex  $ M $  is  $ \text{4 cm} $  away from  $ \text{U} $ . So draw an arc of radius  $ \text{4 cm} $  from  $ U $ cutting the previous arc.

4. Join  $ \text{PM} $  and  $ \text{UM} $ .

Thus  $ \text{JUMP} $  is the required quadrilateral.

(iii) Parallelogram  $ \text{MORE} $ 

OR=6 cm

RE=4.5 cm 

EO=7.5 cm

Ans: Since the opposite sides of parallelogram are equal,

 $ \text{ME=OR} $ 

 $ \text{MO=ER} $ 

Let us first draw the rough diagram of the parallelogram.

1. Draw  $ \text{OR=6 cm} $ 

2. Construct  $ \vartriangle EOR $  by using the given measurement as follows:

3. Since the vertex  $ \text{M} $  is  $ \text{4}\text{.5 cm} $  away from the vertex  $ O $ . So take  $ O $  as center and draw an arc of radius  $ \text{4}\text{.5 cm} $  . The vertex  $ M $  is  $ \text{6 cm} $  away from  $ E $ . So draw an arc of radius  $ \text{6 cm} $  from  $ E $ cutting the previous arc.

4. Join  $ \text{OM} $  and  $ \text{EM} $ .

Thus  $ \text{MORE} $  is the required parallelogram.

(iv) Rhombus  $ \text{BEST} $ 

 BE=4.5 cm 

 ET=6 cm

Ans: All sides of the rhombus are equal,

Let us first draw the rough diagram of the rhombus.

1. Draw  $ \text{ET=6 cm} $ 

2. Construct  $ \vartriangle BET $  by using the given measurement as follows:

3. Since vertex  $ \text{S} $  is  $ \text{4}\text{.5 cm} $  away from the vertex  $ \text{E} $ . So take  $ \text{E} $  as center and draw an arc of radius  $ \text{4}\text{.5 cm} $  . The vertex  $ S $  is  $ \text{4}\text{.5 cm} $  away from  $ \text{T} $ . So draw an arc of radius  $ \text{4}\text{.5 cm} $  from  $ T $ cutting the previous arc.

4. Join  $ \text{ES} $  and  $ \text{ST} $ .

Thus  $ \text{BEST} $  is the required rhombus.

Exercise ( 4.2)

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 

Ans: Let us first draw the rough diagram of the given quadrilateral.

1. Draw  $ \text{TL=2}\text{.5 cm} $ 

2. Draw  $ \vartriangle ITL $  by using the given measurement as follows:

3. Since  $ \text{F} $  is  $ \text{4}\text{.5 cm} $  away from the vertex  $ L $ . So take  $ \text{L} $  as center and draw an arc of  $ \text{4}\text{.5 cm} $  . The vertex  $ \text{F} $  is  $ \text{3 cm} $  away from  $ \text{I} $ . So draw an arc of radius  $ \text{3 cm} $  from  $ I $ cutting the previous arc.

4. Join  $ LF,TF $  and  $ \text{IF} $ .

Thus  $ \text{LIFT} $   is the required quadrilateral.

(ii) Quadrilateral  GOLD

 OL=7.5 cm 

 GL=6 cm 

 GD=6 cm 

 LD=5 cm

 OD=10 cm

Ans. Let us first draw the rough diagram of the given quadrilateral.

1. Draw  $ \text{GD=6 cm} $ 

2. Draw  $ \vartriangle GDL $  by using the given measurement as follows:

3. Since  $ \text{O} $  is  $ \text{10 cm} $  away from the vertex  $ \text{D} $ . So take  $ D $  as center and draw an arc of  $ \text{10 cm} $  . The vertex  $ \text{O} $  is  $ \text{7}\text{.5 cm} $  away from  $ \text{L} $ . So draw an arc of radius  $ \text{7}\text{.5 cm} $  from  $ \text{L} $ cutting the previous arc.

4. Join  $ \text{OD, LO} $  and  $ \text{GO} $ .

Thus  $ \text{GOLD} $   is the required quadrilateral.

(iii) Rhombus  BEND

BN=5.6 cm

DE=6.5 cm

Ans. Let us first draw the rough diagram of a given rhombus

The diagonals of the rhombus bisect each other at  $ {{90}^{{}^\circ }} $ . Let us assume that the point is intersecting at the point  $ \text{O} $  in the rhombus. 

Since  $ \text{ED=6}\text{.5 cm} $ 

 $ \text{EO=OD} $ 

      $ \text{=3}\text{.25 cm} $ 

1. Draw a line segment  $ \text{BN=5}\text{.6 cm} $ 

2. Draw a perpendicular bisector of the line  $ \text{BN} $ . Let the bisector intersect the line segment  $ \text{BN} $  at point  $ \text{O} $ .

3. Taking  $ \text{O} $  as center, draw arcs of radius  $ \text{3}\text{.25 cm} $  to intersect he perpendicular bisector at the point  $ \text{D, E} $ 

4. Join  $ \text{BD, BE, ND} $ and  $ \text{NE} $ .

Thus  $ \text{BEND} $  is the required rhombus.

Exercise (4.3)

1. Construct the following quadrilateral 

(i) Quadrilateral  MORE

MO=6 cm

OR=4.5 cm

 $ \angle M={{60}^{{}^\circ }} $ 

 $ \angle O={{105}^{{}^\circ }} $ 

 $ \angle R={{105}^{{}^\circ }} $ 

Ans: 

1. Let us first draw the rough diagram of the quadrilateral.

2. Draw a line segment  $ \text{MO=6 cm} $ 

3. Draw an angle  $ \angle M={{105}^{{}^\circ }} $  at the point  $ \text{O} $ . As vertex  $ \text{R} $  is  $ \text{4}\text{.5 cm} $ away from the vertex  $ \text{O} $ , cut a line segment  $ \text{OR=4}\text{.5 cm} $ from this ray.

4. Draw an angel  $ \angle R=105{}^\circ  $ 

5. Draw an angle of  $ \angle M=60{}^\circ  $ 

Thus  $ \text{MORE} $ is the required quadrilateral.

(ii) Quadrilateral   PLAN 

PL=4 cm 

LA=6.5 cm

 $ \angle P={{90}^{{}^\circ }} $ 

 $ \angle A={{110}^{{}^\circ }} $ 

 $ \angle N={{85}^{{}^\circ }} $ 

Ans:

1. Let us first draw the rough diagram of the quadrilateral.

2. Draw a line segment  $ \text{PL=4 cm} $ 

3. Draw an angle  $ \angle L={{75}^{{}^\circ }} $  at the point  $ L $ . As vertex  $ \text{R} $  is  $ \text{6}\text{.5 cm} $ away from the vertex  $ L $ , cut a line segment  $ \text{LA=6}\text{.5 cm} $ from this ray.

4. Draw an angel  $ \angle A=110{}^\circ  $ 

5. Draw an angle of  $ \angle P=90{}^\circ  $ 

Thus  $ \text{PLAN} $ is the required quadrilateral.

(iii) Parallelogram HEAR

HE=5 cm 

EA=6 cm 

 $ \angle R={{85}^{{}^\circ }} $ 

Ans:

1. Let us first draw the rough diagram of the parallelogram.

2. Draw a line segment  $ \text{HE=5 cm} $ 

3. Draw an angle  $ \angle E={{85}^{{}^\circ }} $  at the point  $ \text{E} $ . As vertex  $ \text{A} $  is  $ \text{6 cm} $ away from the vertex  $ \text{E} $ , cut a line segment  $ \text{EA=6 cm} $ from this ray.

4. Vertex  $ \text{R} $  is  $ \text{6 cm} $ away from  $ \text{H} $  and  $ \text{4 cm} $  away from  $ \text{A} $ . With  $ \text{H} $  and  $ \text{A} $  as centers, draw an arc of radius  $ \text{6 cm} $  and  $ \text{5 cm} $  respectively. And these will intersect at the point  $ \text{R} $ .

5. Join  $ \text{AR} $  and  $ \text{HR} $ 

Thus  $ \text{HEAR} $ is the required parallelogram.

(iv) Rectangle   OKAY

OK=7 cm 

KA=5 cm 

Ans:

1. Let us first draw the rough diagram of the rectangle.

2. Draw a line segment  $ \text{OK=7 cm} $ 

3. Draw an angle  $ \angle K={{90}^{{}^\circ }} $  at the point  $ \text{K} $ . As vertex  $ \text{A} $  is  $ \text{5 cm} $ away from the vertex  $ \text{K} $ , cut a line segment  $ \text{KA=5 cm} $ from this ray.

4. Vertex  $ \text{Y} $  is  $ \text{5 cm} $  and  $ \text{7 cm} $  away from  $ O,\text{ A} $  respectively. Draw arcs from  $ \text{O, A } $ respectively with the radius of  $ 5\text{ cm, 7 cm} $  respectively.

Thus  $ \text{OKAY} $ is the required rectangle.

Exercise (4.4)

1. Construct the following quadrilaterals.

(i) Quadrilateral   DEAR

DE=4 cm 

EA=5 cm 

AR=4.5 cm

$ \angle E={{60}^{{}^\circ }} $ 

$ \angle A={{90}^{{}^\circ }} $ 

Ans:

1. Let us first draw the rough diagram of the quadrilateral.

2. Draw a line segment  $ \text{DE=4 cm} $ 

3. Draw an angle  $ \angle E={{60}^{{}^\circ }} $ . As vertex  $ \text{R} $  is  $ \text{4}\text{.5 cm} $ away from the vertex  $ \text{A} $ , cut a line segment  $ \text{AR=4}\text{.5 cm} $ from this ray

4. Draw an angle  $ \angle A=90{}^\circ  $ . As vertex  $ \text{R} $  is  $ \text{4}\text{.5 cm} $  away from vertex  $ \text{A} $ , draw a line segment  $ \text{AR=4}\text{.5 cm} $  from the ray.

5. Join  $ \text{DR} $ 

Thus  $ \text{DEAR} $  is the required quadrilateral.

(ii) Quadrilateral   TRUE 

TR=3.5 cm 

RU=3 cm 

UE=4 cm 

 $ \angle R={{75}^{{}^\circ }} $ 

 $ \angle U={{120}^{{}^\circ }} $ 

Ans:

1. Let us first draw the rough diagram of the quadrilateral.

2. Draw a line segment  $ \text{RU=3 cm} $ 

3. Draw an angle  $ \angle U={{120}^{{}^\circ }} $ . As vertex  $ \text{E} $  is  $ \text{4 cm} $  away from the vertex  $ \text{U} $ , cut a line segment  $ \text{EU=4 cm} $ from this ray.

4. Draw an angle  $ \angle R=75{}^\circ  $ As vertex  $ \text{T} $  is  $ \text{3}\text{.5 cm} $  away from the vertex  $ \text{R} $ , cut a line segment  $ \text{RT=3}\text{.5 cm} $  from this ray.

5. Join  $ \text{ET} $ 

Thus  $ \text{TRUE} $  is the required quadrilateral.

Exercise (4.5)

1. Draw the following:

The square  $ \text{READ} $  with  $ \text{RE=5}\text{.1 cm} $ .

Ans: All sides of a square are of the same measures and all angles of a square are of  $ 90{}^\circ  $ . Therefore the given square can be drawn as follows:

1. Let us first draw the rough diagram of the square.

2. Draw a line segment  $ \text{RE=5}\text{.1 cm} $ 

3. Draw an angle at the point  $ \text{R} $  and  $ \text{E} $  each with an angle  $ 90{}^\circ  $ 

4. Draw an arc with the radius of  $ \text{5}\text{.1 cm} $ from  $ \text{R} $  and  $ \text{E} $ .

5. Join  $ \text{AD} $ 

Thus  $ \text{READ} $  is the required Square

2. Draw the following:

A rhombus whose diagonals are  $ \text{5}\text{.2 cm} $  and  $ \text{6}\text{.4 cm} $  long.

Ans: In a rhombus, diagonals bisect each other at  $ 90{}^\circ  $ . Therefore the given rhombus  $ \text{ABCD} $  can be drawn as follows:

1. Let us first draw the rough diagram of the rhombus.

2. Draw a line segment  $ \text{RU=3 cm} $ 

3. Draw the perpendicular bisector of  $ \text{AC} $  and let it bisect at  $ \text{O} $ 

4. Draw arcs of  $ \text{3}\text{.2 cm} $  on both sides of the perpendicular bisector. Let the arcs intersect the perpendicular bisector at  $ \text{B} $  and  $ \text{D} $ .

5. Join  $ \text{AB, AC, AD} $  and  $ \text{CD} $ .

Thus  $ \text{ABCD} $ is the required rhombus.

3. Draw the following:

A rectangle with adjacent sides of length  $ \text{5 cm} $  and  $ \text{4 cm} $ 

Ans: The opposite sides of the triangle are always equal. And all angles of the rectangle are of measure  $ 90{}^\circ  $ . The rectangle  $ \text{ABCD} $  can be drawn as follows:

1. Let us first draw the rough diagram of the rectangle.

2. Draw a line segment  $ \text{AB=5 cm} $ 

3. Draw an angle  $ \angle A={{90}^{{}^\circ }} $  and  $ \angle B=90{}^\circ  $  and draw a ray from both the points.

4. Draw an arc from  $ \text{A} $  and  $ \text{B} $  with the radius of  $ \text{4 cm} $ 

5. Join  $ \text{CD} $ 

Thus  $ \text{ABCD} $  is the required rectangle.

4. A parallelogram  $ \text{OKAY} $  where  $ \text{OK=5}\text{.5 cm} $  and  $ \text{KA=4}\text{.2 cm} $ .

Ans: Opposite sides of the parallelogram are equal and parallel to each other. The given parallelogram can be drawn as follows:

1. The rough sketch of the parallelogram  $ \text{OKAY} $  is drawn as follows:

2. Draw a line segment  $ \text{OK=5}\text{.5 cm} $  and a ray at a point in a convenient angle.

3. Draw a ray at a point  $ \text{O} $  parallel to the ray at  $ K $ . As the vertices  $ \text{A} $  and  $ \text{Y} $  are  $ \text{4}\text{.2 cm} $  away from the vertices  $ \text{K} $  and  $ \text{O} $  respectively, cut the line segment  $ \text{KA, OY} $  each of  $ \text{4}\text{.2 cm} $  from these rays.

4. Join  $ \text{AY} $ .

Topics Included in the Class 8 Maths Chapter 4 Practical Geometry

Given below is the list of topics discussed in Chapter 4 Practical Geometry.

NCERT Maths Solutions for Class 8 Chapter 4 - Summary

NCERT Solutions for Class 8 Maths Chapter 4 comprises of 5 exercises, and an overview of each of the sets is provided below. Take a look!

Exercise 4.1

In the first section of Maths Chapter 4 Class 8 solutions guides, students need to construct quadrilaterals with the given measurements. For instance, AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and AC = 7 cm and these are the measurements of a quadrilateral named ABCD. Similar measurements for rhombus, parallelograms, etc. are also given.

Exercise 4.2

The second section in Maths Class 8 Chapter 4 Solution set also contains question-related to the sketching of quadrilaterals. Here too, all the measurements are provided.

Exercise 4.3

Here again, you are asked to construct quadrilaterals, but along with distance measurements, angles are also laid out. So, these questions are a bit complex than the previous ones, and you must follow Maths NCERT Solutions Class 8 Chapter 5 to understand the steps well.

Exercise 4.4

Again in exercise 4.4 of NCERT Solutions Class 8 Maths Chapter 4, you are supposed to create diagrams or shapes using measurements and angles.

Exercise 4.5

The last exercise of NCERT solutions for class 8 Maths Chapter 4 consists of four questions, and you need to sketch four diagrams - a square, rectangle, rhombus and parallelogram. All the necessary information required to construct the shapes are given. However, if you face any difficulty in solving any of the sums, please make sure to go through the solutions. All the answers are written step by step, making it comprehendible for all students.

Key Takeaways from Class 8 Maths NCERT Solutions for Chapter 4 Practical Geometry

Class 8 students have already studied the topic of Construction in Practical Geometry in their previous classes, but here they will be introduced to the topic of construction of quadrilaterals. They will be using their previous knowledge in addition to the new concepts introduced in Chapter 4 Practical Geometry of NCERT Class 8 Maths.

Here are some key takeaways from Vedantu’s NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry:

• Students learn to construct quadrilaterals in the following four cases:

1. When four sides and one diagonal are given.

2. When three sides and two diagonals are given.

3. When two adjacent sides and three angles are given.

4. When three sides and two included angles are given.

5. In some special cases.

• Class 8 students learn to correctly draw quadrilaterals using a ruler and a compass.

• Children get to know the different properties of quadrilaterals that make them unique. 

• Class 8 Maths Chapter 4 NCERT Solutions with a step-wise, precise, well-written, and detailed explanation of all exercise questions helps enable students to draw the figures correctly without any error. 

So, what are you waiting for? Download Vedantu’s NCERT Solutions for Class 8 Maths Ch 4 and get a clear idea about what type of questions will come from this chapter, and also how to solve them appropriately. As a result, it will benefit you with commendable scores in exams.

Conclusion 

Vedantu's provision of NCERT Solutions for Class 8 Maths Chapter 4 - "Practical Geometry" is a valuable aid for students. These solutions offer clear and concise explanations, making the intricate subject of geometry more accessible and comprehensible. Vedantu's commitment to providing these solutions for free promotes inclusive access to high-quality educational resources, benefiting students from diverse backgrounds. These NCERT Solutions not only assist in understanding geometric concepts but also facilitate problem-solving skills crucial for mathematics. Vedantu's dedication to education is evident through its support, enabling students to excel in their maths studies and lay a solid foundation for future mathematical learning.