# NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry

Contents

## Class 8 Maths Chapter 4 Practical Geometry

In this chapter, we will discuss practical geometry. You will learn about the different types of lines and angles and how to measure them. You will also be introduced to basic concepts of geometry such as points, lines, angles, etc.

## Constructions

There are many ways to construct a line or circle. The most common method is to use a straightedge and compass. However, there are other methods that can be used as well.

One method is to use a ruler and a piece of string. First, tie one end of the string around the pencil. Then, hold the ruler at one end of the line you want to draw. Use the string to draw a line by holding the string taut and moving the ruler along it.

Another method is to use two nails and a piece of string. First, drive one nail into the board at one end of the line you want to draw. Then, tie one end of the string around the second nail. Drive the second nail into the board at the other end of the line. Finally, holding the string taut, use a pencil to trace the line between the two nails.

## Perpendicular Bisector of a Line Segment

The perpendicular bisector of a line segment is a line that passes through the midpoint of the line segment and is perpendicular to the line segment.

To find the equation of the perpendicular bisector of a line segment, we first need to find the midpoint of the line segment. The midpoint of a line segment is the point that is halfway between the two endpoints of the line segment. To find the midpoint, we average the x-coordinates of the endpoints and average the y-coordinates of the endpoints.

Once we have the coordinates of the midpoint, we can use them to write the equation of a line that passes through the midpoint and is perpendicular to the line segment. A line is perpendicular to a line segment if it has a slope that is the negative reciprocal of the slope ofthe line segment.

For example, if a line segment has a slope of 3, then a line that is perpendicular to it would have a slope of -1/3.

We can use this information to write the equation of our perpendicular bisector. We will use the point-slope form of a linear equation, which is:

y – y1 = m(x)

## Angle Bisector of a Line Segment

An angle bisector is a line that divides an angle into two equal parts. The angle bisector of a line segment is the line that passes through the midpoint of the segment and bisects the angle formed by the segment and the line.

To find the angle bisector of a line segment, first find the midpoint of the segment. To do this, simply average the x-coordinates of the endpoint and the y-coordinates of the endpoint. This will give you the coordinates of the midpoint. Then, draw a line through the midpoint that bisects the angle formed by the segment and the line.

## Division of a line segment in a given ratio

1. Division of a line segment in a given ratio:

The division of a line segment in a given ratio can be done using the compass and straightedge. First, mark the point where the division is to occur. Then, use the compass to draw two arcs that intersect at this point. The compass should be set to the desired ratio (for example, if the ratio is 2:1, set the compass to twice the length of the line segment). Finally, use the straightedge to connect the intersection points of the arcs. This will divide the line segment into the desired ratio.

## Construction of a Triangle Given its Base and one other Side

There are various methods that can be used to construct a triangle given its base and one other side. One method is to use a straightedge and compass. First, draw the base of the triangle. Then, use the compass to create an arc that intersects the base at two points. The point where the arc intersects the base is the third vertex of the triangle.

## Conclusion

Knowledge Glow provides NCERT Solutions a comprehensive and detailed explanation of the various concepts taught in the class 8 maths chapter 4 practical geometry. The solutions are designed in such a way that they help students understand the concepts better and also develop problem-solving skills.

### Chapter 4 – Practical Geometry Exercise 4.1 Solution

Question 1: Construct the following quadrilaterals:
AB = 4.5 cm, BC= 5.5 cm, CD = 4 cm, AD = 6 cm, AC = 7 cm

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

( i ) Given: AB = 4.5 cm, BC= 5.5 cm, CD = 4 cm, AD = 6 cm, AC = 7 cm
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.
It is required quadrilateral ABCD. R

( ii ) Given: JU = 3.5 cm, UM = 4 cm, MP = 5 cm, PJ = 4.5 cm, PU = 6.5 cm
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.

( 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 centers 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.

( iv ) Given: R 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 centers, which intersect at S.
(d) Again draw two arcs of 4.5 cm taking E and T as centers, which arcs of 4.5 cm take intersects at B.
(e) Join TS, ES, BT and EB. It is the required rhombus BEST.

### Chapter 4 – Practical Geometry Exercise 4.2 Solution

Question 1: Construct the following quadrilaterals:

LI = 4 cm, IF = 3 cm, TL = 2.5 cm, LF = 4.5 cm, IT= 4 cm.

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

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

(a) Draw a line segment LI = 4 cm.
(b) Taking radius 4.5 cm, draw an arc taking L as center.
(c) Draw an arc of 3 cm taking I as center which intersects the first arc at F.
(d) Join FI and FL.
(e) Draw another arc of radius 2.5 cm taking L as center and 4 cm taking I as center which intersect at T.
(f) Join TF, Tl and TI.
It is the required quadrilateral LIFT.

(ii) Given: OL= 7.5 cm, GL = 6 cm, GD = 6 cm, LD = 5 cm, OD = 10 cm
Steps of construction:

(a) Draw a line segment OL = 7.5 cm
(b) Draw arc of radius 5cm taking L as center and another arc of radius 10cm take O as center 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 center, which intersects at G.
(e) Join GD and GO.
It is the required quadrilateral GOLD.

(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 0, 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.

### Chapter 4 – Practical Geometry Exercise 4.3 Solution

Question 1: Construct the following quadrilaterals:

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

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

(i) Given: MO = 6 cm, ∠R = 105°, OR = 4.5 cm, ∠M = 60°, ∠O = 105°
Steps of construction

(a) Draw OR = 4.5 cm
(b) Draw two angles of 105° each at O and R with the help of protactor.
(c) Cut OM = 6 cm
(d) Draw an angle of 60° at M to meet the angle line through R at E
It is the required quadrilateral More.

(ii) Given: PL = 4 cm, LA = 6.5 cm, ∠P = 90°, ∠A = 110°, ∠N = 85°
Steps of construction

(a) Draw a line segment PL = 4 cm.
(b) Construct angle of 90° at P and produce the side PN.
(c) Construct angle of 75° at L and with L as center, draw an arc of radius 6 cm, which intersects at A.
(d) Construct A = 110° at A and produce the side AN which intersects PN at N.
It is the required quadrilateral PLAN.

(iii) Given: HE 5 cm, EA = 6 cm, R= 85
To construct: Parallelogram HEAR.
To find: HE 5 cm, EA = 6 cm, R= 85 [Sum of adjacent angle of ||gm is 180″] H= 180-85″ =95°
Steps of construction

(a) Draw a line segment HE = 5 cm.
(b) Construct H = 95° and draw an arc of radius 6 cm with center H. It intersects AR at R.
(c) Join RH.
(d) Draw R= E = 85 and draw an arc of radius 6 cm with E as a center which intersects RA at A.
(e) Join RA It is the required parallelogram HEAR.

(iv) Given: OK = 7 cm, KA = 5 cm
To construct: A rectangle OKAY.
Steps of construction:

(a) Draw a line segment OK = 7 cm.
(b) Construct angle 90° at both points 0 and K and produce these sides.
(c) Draw two arcs of radius 5 cm from points 0 and K respectively. These arcs intersect at Y and A.
(d) Join YA.
It is the required rectangle OKAY.

### Chapter 4 – Practical Geometry Exercise 4.4 Solution

Question 1. Construct the following quadrilaterals:
DE = 4 cm, EA = 5 cm, AR = 4.5 cm, ∠E = 60°, ∠A = 90°
TR = 3.5 cm, RU = 3 cm, UE = 4.5 cm, ∠R = 75°, ∠U = 120°

(i) Given: DE = 4 cm, EA = 5 cm, AR = 4.5 cm, ∠E 60°, ∠A = 90°
Steps of construction:

(a) Draw a line segment DE = 4 cm.
(b) At point E, construct an angle of 60″.
(c) Taking radius 5cm, draw an arc from point E which intersect at A.
(d) Construct ∠A= 90, draw an arc of radius 4.5 cm with center A which intersect at R.
(e) Join RD.
It is the required quadrilateral DEAR.

(ii) Given: TR=3.5 cm, RU = 3 cm, UE = 4 cm, ∠R= 75, ∠U = 120°
Steps of construction:

(a) Draw a line segment TR = 3.5 cm.
(b) Construct an angle 75 at R and draw an arc of radius 3 cm with R as center, which intersects at U.
(c) Construct an angle of 120 at U and produce the side UE.
(d) Draw an arc of radius 4 cm with U as center.
(e) Join UE and TE.
It is the required quadrilateral TRUE.

### Chapter 4 – Practical Geometry Exercise 4.5 Solution

Question 1. Draw the following:
The square READ with RE = 5.1 cm.
Construction: square READ with RE = 5.1 cm.
Steps of construction:

(a) Draw RE = 5.1 cm.
(b) Draw an angle of 90° at E and cut EA = 5.1 cm.
(c) Draw two arcs from A and R with radius 5.1 cm to cut each other at D.
Thus, READ is the required square.

Question 2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.

Given: Diagonals are 5.2 cm and 6.4 cm
Steps of construction:

(a) Draw AC = 6.4 cm.
(b) Draw the right bisector of AC at E.
(c) Draw two arcs with center E and radius = 5.2/2 = 2.6 cm to cut the previous diagonal at B and D.
(d) Join AD, AB, BC and DC.
Thus ABCD is the required rhombus.

Question 3. A rectangle with adjacent sides of lengths 5 cm and 4 cm.

Construction: Let the two adjacent sides of a rectangle PQRS be PQ = 5 cm and QR = 4 cm.
Steps of construction:

(a) Draw PQ = 5 cm.
(b) Draw an angle of 90° at Q and cut QR = 4 cm.
(c) Draw an arc with centre R and radius 5 cm.
(d) Draw another arc with centre P and radius 4 cm to meet the previous arc at S.
(e) Join RS and PS.
Thus, PQRS is the required rectangle

Question 4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?

Given: OK = 5.5 cm and KA = 4.2 cm
Steps of construction:

(a) Draw OK = 5.5 cm.
(b) Draw angle of any measure (say 60°) at K and cut KA = 4.2cm.
(c): Draw an arc with centre A and radius of 5.5 cm.
(d) Draw another arc with centre O and radius 4.2 cm to cut the previous arc at Y.
(e) Join AY and OY.
Thus, OKAY is the required parallelogram.
No, it is not a unique parallelogram. The angle at K can be of measure other than 60°.