ncert solutions for class 8 maths chapter 4 discussed about Practical Geometry.

Mensuration.in provided solutions to Exercises in 8 maths chapter 4 Practical Geometry.

## ncert solutions for class 8 maths chapter 4 exercise 4.1

Mensuration.in provided solutions to Exercise 4.1 in class 8 maths chapter 4

**Construct the following quadrilaterals.**

(i) Quadrilateral ABCD.

AB = 4.5 cm

BC = 5.5 cm

CD = 4 cm

AD = 6 cm

AC = 7 cm

solution to Exercise 4.1 (i)

(i) Draw a rough sketch of this quadrilateral 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, By taking A as centre, draw an arc of radius 6 cm.

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

ABCD is the required quadrilateral.

**Construct the following quadrilaterals.**

(ii) Quadrilateral JUMP

JU = 3.5 cm UM = 4 cm MP = 5 cm PJ = 4.5 cm PU = 6.5 cm

solution to Exercise 4.1 (ii)

(ii)Firstly, a rough sketch of this quadrilateral 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.

Construct the following quadrilateral.

(iii) Parallelogram MORE OR = 6 cm RE= 4.5 cm EO = 7.5 cm

solution to Exercise 4.1 (iii)

Steps of constructions:

Draw a rough figure.

1) Draw OR= 6 cm as a base.

2) Drawn two arcs of radius 7.5 cm and radius 4.5 cm by taking O and R as centres respectively, Which intersects at E.

3) Join OE and RE.

4) Draw an arc of 6 cm radius taking E as center.

5) Draw another arc of 4.5 cm radius taking O as center, which intersects at M.

6) Join OM and EM.

MORE is the required parallelogram.

Construct the following quadrilateral.

(iv)Rhombus BEST BE = 4.5 cm, ET = 6 cm

solution to Exercise 4.1 (iii)

Steps of constructions:

1) Draw a rough figure.

2) Draw TE= 6 cm as a base. And bisect it into two equal parts by drawing perpendicular bisector.

3) Draw two arcs of 4.5 cm taking E and T as centres, which intersect at S.

4) Again draw two arcs 4.5 cm by taking E and T as centers, which intersects at B.

5) join TS, ES, BT and EB.

BEST is required rhombus.

### ncert solutions for class 8 maths chapter 4 exercise 4.2

Mensuration.in provided solutions to Exercise 4.2 in class 8 maths chapter 4

Exercise 4.2 (i)

**Construct the following quadrilaterals.**

(i) quadrilateral LIFT

LI = 4 cm

IF = 3 cm

TL = 2.5 cm

LF = 4.5 cm

IT = 4 cm

solution to Exercise 4.2 (i)

Steps of constructions:

1 Draw a rough figure.

2 Draw line segment LI=4 cm as a base.

3 Taking L as centre, 4.5 cm as radius draw an arc of circle and Taking I as centre 3 cm as radius draw another arc of circle, which intersects at F.

4 Join LF and IF.

5 Taking L as centre 2.5 cm as radius draw an arc of circle and taking I as centre 4.5 cm as radius draw another arc of circle, which intersects at T.

6 Join LT, IT and TF

7 LIFT is the required quadrilateral.

Exercise 4.2 (ii)

(ii) Quadrilateral GOLD

OL 7.5 cm

GL= 6 cm

GD =6 cm

LD= 5 cm

OD= 10 cm

solution to Exercise 4.2 (ii)

Steps of constructions:

1 Draw a rough figure.

2 Draw line segment OL= 7.5 cm as base.

3 Taking L as centre 5 cm as radius draw an arc of circle and O as centre 10 cm as radius draw another arc of circle, which intersects at D.

4 Join LD and OD.

5 Taking D as centre 6 cm as radius draw an arc of circle and L as centre 6 cm as radius draw another arc of circle, which intersects at G.

6 Join LG, DG and OG.

7 GOLD is the required quadrilateral.

Exercise 4.2 (iii)

(iii) Rhombus BEND

BN = 5.6 cm

DE = 6.5 cm

Solution to Exercise 4.2 (iii)

Steps of constructions:

1 Draw a rough figure.

2 Draw line segment DE= 6.5 cm as base.

3 Draw a line XY⊥ line segment DE, let O be the intersection point.

4 Draw two arcs of radius 2.8 cm from intersection point O which intersects the line at B and N.

5 Join DB,BE,DN and EN

6 BEND is the required quadrilateral.

ncert solutions for class 8 maths chapter 4 exercise 4.3

Mensuration.in provided solutions to Exercise 4.3 in class 8 maths chapter 4

**Construct the following quadrilaterals.**

Exercise 4.3 (i)

(i) Quadrilateral MORE

MO = 6 cm

OR = 4.5 cm

∠M= 60°

∠0= 105°

∠R = 105°

Solution to Exercise 4.3(i)

Steps of constructions:

1 Draw a rough figure.

2 Draw a line segment MO=6cm as a base.

3 Construct ∠MOX=105° and taking O as centre 4.5 cm as radius draw an arc of circle. Which intersects OX ray at R.

4 Construct ∠ORY=105° and ∠OMZ=60° Let ray MZ and ray RY intersects at E.

5 MORE is the required quadrilateral.

Exercise 4.3 (ii)

(ii) Quadrilateral PLAN

PL = 4 cm, LA = 6.5 cm, ∠P = 90°, ∠A = 110°, ∠N =85°.

Solution to Exercise 4.3(ii)

To construct PLAN Quadrilateral.

First we need to find ∠L

∠L=360°-(110°+75°+90°)=360°-285°=75° {∵ sum of angles in a Quadrilateral is 360°}

Steps of constructions:

1 Draw a Rough figure.

2 Draw a line segment PL=4 cm as base.

3 Construct ∠PLX=75° and taking L as centre 6.5 cm as radius draw an arc of circle. Let this arc intersect ray LX at A.

4 Construct ∠LAZ=110° and ∠LPY=90°, which intersects at N.

5 PLAN is required Quadrilateral.

Exercise 4.3 (iii)

(iii) Parallelogram HEAR

HE=5 cm, EA=6 cm ∠R=85°

Solution to Exercise 4.3(iii)

Given HE=5 cm, EA=6 cm,∠R=85°

Steps of constructions:

In a parallelogram sum of adjacent angles=180°. And lengths of opposite sides are equal.

∠H=180°- ∠E=180°-85°=95°

1 Draw a Rough figure.

2 Draw a line segment HE=5 cm as base.

3 Draw ∠HEX=85° and ∠EHY =95°.

4 Taking E as centre 6 cm as radius draw an arc of circle, whichb intersects ray EX at A.

5 Taking H as centre 6 cm as radius draw another arc of circle, which intersects ray HY at R.

6 Join R and A. HEAR is required Quadrilateral (Parallelogram).

Exercise 4.3 (iv)

(iv) Rectangle OKAY

OK=7cm, KA=5 cm

Solution to Exercise 4.3(iv)

Steps of constructions:

In a rectangle opposite sides are equal. And measure of each angle=90°.

OK=AY=7 cm , KA=OY=5 cm ∠K=∠O=∠A=∠Y=90°

1 Draw a Rough figure.

2 Draw a line segment OK=7 cm as base.

3 Draw ∠OKP=90° and ∠KOQ =90°.

4 Taking K as centre 5 cm as radius draw an arc of circle, which intersects ray KP at A.

5 Taking O as centre 5 cm as radius draw another arc of circle, which intersects ray OQ at Y.

6 Join A and Y. OKAY is required Quadrilateral (rectangle).

**Construct the following quadrilaterals.**

#### ncert solutions for class 8 maths chapter 4 exercise 4.4

Mensuration.in provided solutions to Exercise 4.4 in class 8 maths chapter 4

Exercise 4.4 (i)

(i) Quadrilateral DEAR,

DE=4 cm, EA=5 cm, AR=4.5 cm, ∠E=60°, ∠A=90°.

Solution to Exercise 4.4(i)

Steps of constructions:

1 Draw a rough figure.

2 Draw a line segment DE=4 cm as a base.

3 Construct ∠DEX=60° and taking E as centre 5 cm as radius draw an arc of circle. Which intersects ray OX at A.

4 Construct ∠EAY=90° and taking A as centre 4.5 cm as radius draw an arc of circle which intersects ray AY at R.

5 Join DR. DEAR is the required quadrilateral.

Exercise 4.4 (ii)

(ii) Quadrilateral TRUE,

TR=3.5 cm, RU=3 cm, UE= 4 cm,∠R=75°, ∠U=120°

Solution to Exercise 4.4(ii)

Steps of constructions:

1 Draw a rough figure.

2 Draw a line segment TR=3.5 cm as a base.

3 Construct ∠TRX=75° and taking R as centre 3 cm as radius draw an arc of circle. Which intersects ray RX at U.

4 Construct ∠RUE=120° and taking U as centre 4 cm as radius draw an arc of circle which intersects ray UY at E.

5 Join TE. TRUE is the required quadrilateral.

Construct the following quadrilaterals.

#### ncert solutions for class 8 maths chapter 4 exercise 4.5

Mensuration.in provided solutions to Exercise 4.5 in class 8 maths chapter 4

Construct the following quadrilaterals.

Exercise 4.5 (1)

Draw the following.

1 The square READ with RE=5.1 cm

Solution to Exercise 4.5(1)

In a square each angle=90° and lengths of all sides are equal.

So ∠R=∠E=∠A=∠D=90°, RE=EA=AD=DR=5.1 cm.

Steps of constructions.

1 Draw a Rough figure.

2 Draw RE=5.1 cm as a base.

3 Draw ∠REX=90° and takin E as a centre 5.1 cm as a radius draw an arc of circle which intersects ray EX at A.

4 Taking A as centre 5.1 cm as radius draw an arc of circle and R as centre 5.1 cm as radius draw another arc of circle which intersects previous arc at D.

5 Join RD and AD. READ is required Square.

Exercise 4.5(2)

Draw the following.

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

Solution to Exercise 4.5(2)

Steps of constructions:

Let ABCD be required rhombus and AC=6.4 cm, BD=5.2 cm. We know that in rhombus diagonals bisects each other perpendicularly.

1 Draw a rough figure.

2 Draw line segment AC= 6.4 cm as base.

3 Draw a line XY perpendicular to line segment AC, let O be the intersection point.

4 Draw two arcs of radius 2.6 cm from intersection point O which intersects the line XY at B and D respectively. {∵OD=OB=5.2/2}

5 Join AD,AB,BC and CD

6 ABCD is the Rhombus.

Exercise 4.5 (3)

Draw the following.

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

Solution to Exercise 4.5 (3)

Let ABCD be required rectangle and AB=5 cm,BC=4 cm.

Steps of constructions:

1 Draw a Rough figure.

2 Draw a line segment AB=5cm as base.

3 Draw ∠ABX=90° and ∠BAY =90°.

4 Taking B as centre 4 cm as radius draw an arc of circle, which intersects ray BX at C.

5 Taking A as centre 4 cm as radius draw another arc of circle, which intersects ray AY at D.

6 Join C and D. ABCD is required Quadrilateral(rectangle).

Exercise 4.5 (4)

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

Solution to Exercise 4.5 (4)

To construct parallelogram we required 3 measurements but given only two measurements. To draw the figure let an angel 90°

1 Draw a Rough figure.

2 Draw a line segment OK=5.5 cm as base.

3 Draw ∠OKX=90° and taking K as centre 4.2 cm as radius draw an arc of circle which intersects ray KX at A.

4 Taking A as centre 5.5 cm as radius draw an arc of circle and O as centre 4.2 cm as radius draw another arc of circle which intersects previous arc at Y.

{∵In parallelogram lengths of opposite sides are equal.OK=YA=5.5 cm and KA=OY=4.2 cm}

5 Join OY and YA. OKAY is the required parallelogram.

## what is Practical Geometry

pratical geometry is to constructing geometrical figuers.ex-squar,triangle etc.with the help of tools found in your geometry box