NCERT Solutions For Class 8 Maths Chapter 4

NCERT Solutions Class 8 Maths Practical Geometry

Ncert Solutions For Class 8 Maths Chapter 4 PDF Download

NCERT solutions class 8 maths chapter 4 applied practical geometry is one of the most common chapters from where questions are asked in class 8th mathematics. The NCERT solutions class 8 maths chapter 4 applied practical geometry is provided here so that learners can understand the concepts properly. Applied practical geometry is one of the most crucial topics in higher standard like 10th and 12th class, therefore students must learn this chapter effectively. NCERT solutions class 8 maths chapter 4 is prepared according to the latest syllabus of CBSE. Check the NCERT solutions class 8 maths chapter 4 pdf given below.

The NCERT solutions is one of the best tool to prepare science for class 8. Practical Geometry is a crucial chapter in CBSE class 8. Students must prepare this chapter well to score well in their exam. NCERT Solutions Class 8 Maths Practical Geometry is given below so that students can understand the concepts of this chapter in depth.

NCERT Solutions Class 8 Maths Chapter 4 Exercises

Exercise 4.1

Question 1:

Construct the following quadrilaterals.

i) Quadrilateral of PQRS

PQ = 4.5 cm

QR = 5.5 cm

RS =4 cm

PS = 6 cm

PR =7 cm

ii) Quadrilateral of HUMP

HU = 3.5 cm

UM =4cm

MP = 5 cm

PH = 4.5 cm

PU = 6.5 cm

iii) Parallelogram of CORE

OR = 6 cm

RE = 4.5 cm

E0 = 7.5 cm

iv) Rhombus of WEST

WE=4.5cm

ET=6cm

 

Answer:

i) Firstly , a rough sketch of this quadrilateral can be drawn as follows:

1

1. \(\Delta PQR\) can be constructed by using the given measurements as follows:

2

2. Vertex S is 6cm away from vertex P. Therefore, while taking P as centre, drawn an arc of radius 6cm.

3

3. Taking R as centre, draw an arc of radius 4cm, cutting the previous arc at point S. Join S to and R.

4

PQRS is the required quadrilateral.

 

(ii) Firstly, a rough sketch of this quadrilateral can be drawn as follows.

5

(1) \(\Delta HUP\) can be constructed by using the given measurements as follows.

6

(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.

7

(3) Join M to P and U.

JUMP is the required quadrilateral.

8

 

(iii)We know that opposite sides of a parallelogram are equal in length and also these are parallel to each other.

Hence, CE = OR, CO = ER

A rough sketch of this parallelogram can be drawn as follows.

9

(1) \(\Delta EOR\) can be constructed by using the given measurements as follows.

10

(2) Vertex M is 4.5 cm away from vertex 0 and 6 cm away from vertex E. Therefore, while taking 0 and E as centres, draw arcs of 4.5 cm radius and 6cm radius respectively. These will intersect each other at point M.

11

 

(3) Join C to 0 and E.

12

CORE is the required parallelogram.

 

(iv)We know that all sides of a rhombus are of the same measure.

Hence, WE = ES = ST = TW

A rough sketch of this rhombus can be drawn as follows.

13

(1) \(\Delta WET\) can be constructed by using the given measurements as follows.

14

(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 be intersecting each other at point S.

15

( 3 )  Join S to E and T.

16

WEST is the required rhombus.

 

 

EXERCISE 4.2

Question 1: Construct the geometry of the following quadrilaterals by the given data.

i) Quadrilateral of the word called GIFT

  GI= 4 cm

IF= 3 cm

TG= 2.5 cm

GF= 4.5 cm

IT= 4 cm

ii) Quadrilateral BOLD

OL =7.5 cm

 BL=6 cm

BD=6cm

LD =5 cm

 OD=10 cm

iii) Rhombus MEND

MN=5.6 cm

 DE=6.5 cm

 

Answer:

The rough text of the quadrilateral can be drawn here:

17

\(\Delta GIT\)can be constructed by following measurements

18

2) Vertex F is 4.5 cm away from vertex L and 5 cm away from vertex I. Therefore, while taking L and I as centres, draw arcs of 4.5 cm radius and 3 cm radius respectively, which will be intersecting each other at point F.

23

3) Join F to T and F to I.

24

BEAT is the required to form the quadrilateral.

 

ii) A rough sketch of this the diagram quadrilateral can be as follows.

21

1) \(\Delta GDL\)can be constructed by following measurements of the quadrilateral

22

2) Vertex O is 10 cm to the vertex D and 7.5 Cm 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.

23

3) Join O to L and B.

24

4) BOLD is the required quadrilateral.

 

iii) We know that the diagonals of a rhombus always bisect each other at \(90^{\circ}\).Let us assume that these are intersecting each other at a point O in this rhombus.

Hence EO=OD=3.25

A rough sketch of the rhombus can be drawn as follows:

25

1) Draw a line segment MN of 5.6 cm and also draw its perpendicular bisector. Let it intersect the line segment MN at point O.

26

2) Taking O as centre, draw arcs of 3.25 cm radius to intersect the perpendicular.

27

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

28

MEND is the required quadrilateral.

 

 

Exercise 4.3

Question 1:

 Construct the quadrilaterals.

(i) Quadrilateral EROM

Side MO = 6cm

Side OR = 4.5cm

\(\angle M=60^{\circ}\)

\(\angle O=105^{\circ}\)

\(\angle R=105^{\circ}\)

(ii)Quadrilateral NALP

PL = 4cm

LA = 6.5cm

\(\angle P=90^{\circ}\)

\(\angle A=110^{\circ}\)

\(\angle N=85^{\circ}\)

(iii)Parallelogram RAEH

EH = 5cm

AE = 6cm

\(\angle R=85^{\circ}\)

(iv) Rectangle YAKO

KO = 7cm

AK = 5cm

 

Answer:

(i)

(1) Sketch the figure of the quadrilateral

29

(2) Using the geometry tools draw the line segment OM 6cm and an angle of \(105^{\circ}\) at point O. As vertex R is 4.5cm away from the vertex O, cut a line segment RO of 4.5cm from this ray.

30

(3) Again, draw an angle of \(105^{\circ}\) at point R.

32

(4) Draw an angle of \(60^{\circ}\) at point M. Extend this ray so that it meets the previously drawn from R at point E.

33

EROM is the desired quadrilateral.

 

(ii)

(1) Total sum of the interior angles of quadrilateral is \(360^{\circ}\)

In quadrilateral NALP, \(\angle P+\angle L+\angle A+\angle N=360^{\circ}\)

\(90^{\circ}+\angle L+110^{\circ}+85^{\circ}=360^{\circ}\)

\(285^{\circ}+\angle L=360^{\circ}\)

\(\angle L=360^{\circ}-285^{\circ}=75^{\circ}\)

(2)Draw a rough diagram of quadrilateral

34

(3) Construct a line PL of 4cm and draw an angle of \(75^{\circ}\) at point L. As vertex A is 6.5cm away from vertex L, cut a line segment LA of 6.5cm from this ray.

35

(4) Draw another angle of \(110^{\circ}\) at point A.

36

(5) Construct an angle of \(90^{\circ}\) at point P. This line will meet touch the previously drawn ray from A at point N.

0

NALP is the required quadrilateral

 

(iii)

(1)Sketch a rough diagram of this quadrilateral as follows

37

(2)Construct a line segment EH of 5cm and an angle \(85^{\circ}\) at point E. As vertex A is at a distance of 6cm and 5cm, draw arcs from point H and A respectively. These will be intersecting each other at point R.

38

(3) Vertex H and A are 6 and 5cm away from vertex R. By taking radius as 6cm and 5cm, draw arcs from point H and A respectively. These will be intersecting each other at point R.

39

(4) Join H and A with point R

40

RAEH is the desired quadrilateral.

 

(iv)

(1) Draw a rough diagram of this quadrilateral as follows.

41

(2) Construct a line segment KO of 7cm and an angle of \(90^{\circ}\) at point K. As vertex A is 5cm away from vertex K, cut a line segment AK of 5cm from this ray.

42

(3) Draw two arcs of radius 5 cm from points O and K respectively. These arcs intersect at Y and A.

43

(4) Join AY

44

It is the desired rectangle YAKO

 

 

Exercise 4.4

Question 1:

Construct the given quadrilaterals.

(a) Quadrilateral ABCD

AB = 4cm

BC = 5cm

CD = 4.5cm

\(\angle B\) = \(60^{\circ}\)

\(\angle C\) = \(90^{\circ}\)

 

(b) Quadrilateral PQRS

PQ = 3.5cm

QR = 3cm

RS = 4cm

\(\angle Q\) = \(75^{\circ}\)

\(\angle R\) = \(120^{\circ}\)

 

Answer:

(a) Quadrilateral ABCD

(i) Draw a rough sketch of the given quadrilateral as given below.

45

(ii) Draw a line segment AB of 4 cm and mark an angle of \(60^{\circ}\) at point B. Cut a line segment BC of 5 cm from its ray as vertex B is 6 cm away from vertex B.

46

(iii) Draw an angle of \(90^{\circ}\) at point C. Cut a line segment DC of 4.5 cm from its ray as vertex C is 4.5 cm away from vertex C.

47

(iv) Join A to D.

48

Thus we get the required quadrilateral ABCD.

 

(b) Quadrilateral PQRS

(i) Draw a rough sketch of the given quadrilateral as given below.

49

(ii) Draw a line segment QR of 4 cm and point R make an angle of \(120^{\circ}\). Cut a line segment RS of 5 cm from its ray as vertex S is 5 cm away from vertex R.

50

(iii) Draw an angle of \(75^{\circ}\) at point Q. Cut a line segment QP of 4.5 cm from its ray as vertex P is 4.5 cm away from vertex Q.

51

(iv) Join P to S.

52

Thus we get the required quadrilateral PQRS.

 

 

Exercise 4.5

Question 1:

Draw the quadrilateral according to the given information.

 Draw the square BEST with the length 6.1 cm.

 

Answer:

BE = 6.1 cm

Steps to construct the square

(i) Draw BE = 6.1 cm. Construct an angle of \(90^{\circ}\) at point E and at point B.

53

(ii) Vertex T and S are 6.1 cm away from point B and E respectively. Therefore cut the line segments SE and BT of 6.1 cm.

54

(iii) Join T to S.

55

Thus we get the required square BEST.

 

Question 2:

Draw the rhombus according to the given information.

Measurements of the diagonals are 4.2 cm and 5.4 cm long.

 

Answer:

In any rhombus, diagonals are perpendicular to each other.

(i) Draw a line segment PR of 4.2 cm and draw a perpendicular bisector of PR. Let the point of intersection be M.

56

(ii) Draw arcs of \(\frac{5.4}{2}\) = 2.7 cm on both the sides of the perpendicular bisector. Let the point of intersection of arcs and perpendicular bisector as S and Q.

57

(iii) Join points S and Q with points P and R.

58

Thus we get the required rhombus PQRS.

 

Question 3:

Draw the rectangle according to the given information.

Measurements of the adjacent sides are 6 cm and 3 cm long.

 

Answer:

In rectangle, opposite sides have same lengths and measure of all the interior angles is \(90^{\circ}\).

(i) Draw line – segment PQ of 6 cm and draw an angle of \(90^{\circ}\) at point P and Q.

59

(ii) Vertices R and S are 3 cm away from point Q and P respectively. Thus, arcs cut the line – segments PS and QR, 3 cm away.

60

(iii) Join S to R.

61

 

Question 4:

Draw the parallelogram NICE according to the given information.

Measure of NI = 7 cm and IC = 4.5 cm long.

 

Answer:

In parallelogram, opposite sides have same length and they are parallel to each other.

(i) Draw a line – segment NI of 7 cm and a ray at point I at any angle.

62

(ii) Draw a ray from point O which is parallel to the ray at I. The vertices E and C are 4.5 cm away from vertices N and I respectively. Cut the line – segments IC and NE of 4.5 cm.

63

(iii) Join E to C.

64

Thus this is the required parallelogram NICE.

The Central Board of Secondary Education is one of the popular educational board in India. CBSE follow the NCERT curriculum to conducts its examinations for class 10 and class 12 respectively. Class 8 is one of the crucial stages in students academic life. The NCERT Solutions Class 8 Maths Practical Geometry is given so that students can understand the concepts of this chapter in depth.

We require three measurements (of sides and angles) to draw a unique triangle. Since 3 measurements were enough to draw a triangle, a natural question arises whether four measurements would be sufficient to draw a unique four sided closed figure, namely, a quadrilateral.

Some interesting facts about quadrilaterals are given below.

  • Five measurements can determine a quadrilateral uniquely.
  • A quadrilateral can be constructed uniquely if its two diagonals and three sides are known.
  • A quadrilateral can be constructed uniquely if the lengths of its four sides and a diagonal is given.
  • A quadrilateral can be constructed uniquely if its three sides and two included angles are given
  • A quadrilateral can be constructed uniquely if its two adjacent sides and three angles are known.
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