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Chapter 4: Practical Geometry

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.

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Thus this is the required parallelogram NICE.



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2 Comments

  1. Puneet Puneet
    May 25, 2018    

    Good

  2. Ashutosh Ashutosh
    May 25, 2018    

    Good questions

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