NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry

 

NCERT solutions for class 8 maths chapter 4 Practical Geometry- In class VII you have already learnt how to draw triangles. To draw a unique triangle you require three measurements of sides and angles. In this chapter, you will learn some advance level constructions like- quadrilaterals. This chapter has introduced in the curriculum to make students  imagine things and creating a shape according to the given data. CBSE NCERT solutions for class 8 maths chapter 4 Practical Geometry is covering all these quadrilaterals problems in a very detailed manner. In this, you are going to learn how to construct a unique quadrilateral when different conditions are given like when the lengths of four sides with a diagonal are given, when three sides and two diagonals are given, when two adjacent sides with three angles are given, etc. Geometry is a subset of Mathematics in which you learn about shape, size & relative position of geometric figures like- Angle, Triangle, Rectangle and etc. You know in order to draw a triangle, you need the measurement of three straight lines likewise you can say that the measurement of four lines will help us in creating a rectangle. CBSE NCERT solutions for class 8 maths chapter 4 Practical Geometry can be a handy tool to make your preparation level good. In this chapter, there are a total of 4 exercises and 33 questions. Solutions of NCERT for class 8 maths chapter 4 Practical Geometry is having answers to all the 33 questions. Along with all these, NCERT solutions for other classes and subjects can be downloaded using the link. 

NCERT solutions for class 8 maths chapter 4 Practical Geometry Chapter: Introduction

Q1 Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, \angle A = 50\degree , AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral? Give reasons for your answer.

Answer:

No, he cannot construct a unique quadrilateral. The given measurements consist of 2 adjacent sides, the angle between these sides and 2 diagonals. It is not one of the specific combinations to construct a quadrilateral.

Moreover, on trying to construct using these, we find that the vertex C is not fixed and can be varied. Hence, the quadrilateral is not unique.

Solutions of NCERT for class 8 maths chapter 4 Practical Geometry Excercise: Constructing A Quadrilateral

Q (i) We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do this?

Answer:

No, any five measurements of the quadrilateral cannot deteremine a quadrilateral uniquely.

For example, consider a quadrilateral ABCD where AB, BC, CD are known and \angleA and \angleC are known. But this cannot uniquely determine the quadrilateral. 

Q (ii) Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and
AS = 6.5 cm? Why?

Answer:

Yes, we can draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and
AS = 6.5 cm (diagonal).

We know, diagonals of a parallelogram are equal. Hence, BT = AS =6.5 cm 

And opposite sides are equal. Hence, BS =AT = 6 cm;

Therefore, we have 2 diagonals and 3 sides to construct the parallelogram uniquely.

Q (iii) Can you draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm? Why?

Answer:

Yes, we can draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm.

Because all sides of a rhombus are equal. 

Therefore we have four sides and a diagonal to construct the rhombus uniquely.

Q (iv)  A student attempted to draw a quadrilateral PLAY where PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY= 6 cm, but could not draw it. What is the reason?
[Hint: Discuss it using a rough sketch].

Answer:

Given, a quadrilateral PLAY where PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY= 6 cm,

Rough sketch: 

We know, that sum of two sides of a triangle must be greater than the third side. But PL + PY < YL . Hence this is not a triangle. Therefore, this quadrilateral cannot be made.

NCERT solutions for class 8 maths chapter 4 Practical Geometry Excercise: 4.1

Q1 (i) Construct the following quadrilaterals.

Quadrilateral ABCD.    
AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AD = 6 cm
AC = 7 cm

Answer:

Given,

AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AD = 6 cm
AC = 7 cm

Step 1.  Using SSS condition ,draw \Delta ABC

Step 2. Now, CD= 4 cm. Using C as centre, draw an arc of radius = 4 cm

Step 3. Also, AD = 6 cm. Using A as centre, draw an arc of radius = 6 cm.

Step 4. Point D will be the point of intersection of the two drawn arcs. Join AD and CD to complete the quadrilateral.

 

ABCD is the required quadrilateral.

Q1 (ii) Construct the following quadrilaterals.

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

Answer:

Given,

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

Step 1.  Using SSS condition ,draw \Delta JUP

 Step 2. Now, PM = MP = 5 cm. Using P as centre, draw an arc of radius = 5 cm

.Step 3. Also, UM = 4 cm. Using U as centre, draw an arc of radius = 4 cm.

Step 4. Point M will be the point of intersection of the two drawn arcs. Join PM and UM to complete the quadrilateral.

JUMP is the required quadrilateral.

Q1 (iii) Construct the following quadrilaterals.

Parallelogram MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm    

Answer:

Given,

Parallelogram MORE

OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm 

We know, Opposite sides of a parallelogram are equal in length and also parallel to each other

Therefore,

EM = OR = 6 cm,

OM = RE = 4.5 cm

Step 1.  Using SSS condition ,draw \Delta ORE

 Step 2. Now, EM = OR = 6cm . Using E as centre, draw an arc of radius = 6 cm

Step 3. Also, OM = RE = 4.5 cm . Using O as centre, draw an arc of radius = 4.5 cm.

Step 4. Point M will be the point of intersection of the two drawn arcs. Join OM and EM to complete the parallelogram.

MORE is the required parallelogram.

Q1 (iv) Construct the following quadrilaterals.

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

Answer:

Given,

BEST is a rhombus where:

BE = 4.5 cm
ET = 6 cm (diagonal)

   We know that the sides of a rhombus are equal in length and opposite sides are parallel to each other. 

Therefore, the sides of Rohmbus BE = ES = ST = TB = 4.5 cm

 

Step 1.  Using SSS condition ,draw \Delta BET

 Step 2. Now, ES = 4.5 cm . Using E as centre, draw an arc of radius = 4.5 cm

Step 3. Also, TS = 4.5 cm . Using T as centre, draw an arc of radius = 4.5 cm.

Step 4. Point S will be the point of intersection of the two drawn arcs. Join ES and TS to complete the rhombus.

BEST is the required rhombus.

Solutions for NCERT class 8 maths chapter 4 Practical Geometry Topic: When Two Diagonals And Three Side Are Given

Q1 In the above example, can we draw the quadrilateral by drawing? ABD first and then find the fourth point C?

Answer:

No, it is not possible. We cannot draw  \Delta ABD first, because neither SSS nor SAS condition is there.

Q2 Can you construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm? Justify your answer.

Answer:

Yes, we can construct a quadrilateral with the given conditions. Here, two diagonals and three sides are given.

(Hint: Construct  \Delta PRS using SSS condition. Next, find point Q by drawing arcs from P and R. PQRS is the required quadrilateral.)

NCERT solutions for class 8 maths chapter 4 Practical Geometry Excercise: 4.2

Q1 (i) Construct the following quadrilaterals.

quadrilateral LIFT

LI = 4 cm

IF = 3 cm

TL = 2.5 cm

LF = 4.5 cm

IT = 4 cm

Answer:

Given,

LI = 4 cm

IF = 3 cm

TL = 2.5 cm

LF = 4.5 cm (Diagonal)

IT = 4 cm (Diagonal)

Steps of construction:

Step 1.  Using SSS condition ,draw \Delta LIF

Step 2. Now, LT = TL = 2.5 cm . Using L as centre, draw an arc of radius = 2.5 cm

Step 3. Also, IT  = 4 cm . Using I as centre, draw an arc of radius = 4 cm.

Step 4. Vertex T will be the point of intersection of the two drawn arcs. Join LT and TF to complete the quadrilateral.

LIFT is the required quadrilateral.

Q1 (ii) Construct the following quadrilaterals.

Quadrilateral GOLD

OL = 7.5 cm

GL = 6 cm

GD = 6 cm

LD = 5 cm

OD = 10 cm

Answer:

Given,

OL = 7.5 cm

GL = 6 cm

GD = 6 cm

LD = 5 cm

OD = 10 cm

Steps of construction: 

Step 1.  Using SSS condition ,draw \Delta OLD

 Step 2. Now, DG = GD = 6 cm . Using D as centre, draw an arc of radius = 6 cm

Step 3. Also, LG = GL = 6 cm . Using L as centre, draw an arc of radius = 6 cm.

Step 4. Vertex G will be the point of intersection of the two drawn arcs. Join G to D and G to O to complete the quadrilateral.

GOLD is the required quadrilateral.

Q1 (iii) Construct the following quadrilaterals.

Rhombus BEND
BN = 5.6 cm
DE = 6.5 cm

Answer:

Given, BEND is a rhombus.

BN = 5.6 cm (Diagonal)
DE = 6.5 cm (Diagonal)

We know that the diagonals of a rhombus bisect (cut in halves) each other at 90 degrees.

Steps of construction: 

Step 1.  Draw a line segment BN = 6.5 cm. With radius greater than half of BN, draw arcs on both sides of BN with B and N as the center. The line joining these two intersections is the perpendicular bisector of BN. Let it intersect BN at O (Therefore, O is the midpoint of BN. It will also be the midpoint of DE!)

 Step 2.  With O as the center, draw two arcs on the perpendicular bisector with radius = \frac{1}{2} DE = \frac{6.5}{2} = 3.25 cm  (Since, O is the midpoint of DE.)

Step 3. The intersecting points are vertices D and E.

Step 4. Join D to B and N. Also join E to B and N.

BEND is the required rhombus.


Q1.Construct the following quadrilaterals. (iii) Rhombus BEND BN = 5.6 cm DE = 6.5 cm
Edit Q

 

Solutions of NCERT for class 8 maths chapter 4 Practical Geometry Topic: When Two Adjacent Sides And Three Angles Are Known

Q1 Can you construct the above quadrilateral MIST if we have 100° at M instead of 75°?

Answer:

Yes, we can draw the quadrilateral MIST if we have 100° at M instead of 75°. (It will be a different quadrilateral!)

Moreover, The vertex T will now change. And since the sum of all interior angles of a quadrilateral is 360°, hence angle T will also change.

Q2 Can you construct the quadrilateral PLAN if PL = 6 cm, LA = 9.5 cm, ∠P = 75°, ∠L =150° and ∠A = 140°? (Hint: Recall angle-sum property).

Answer:

Given, Two adjacent sides and three angles. 

We know that sum of all the interior angles of a quadrilateral is 360°.

\therefore ∠P + ∠L +∠A + ∠N= 140° + 75° + 150° + ∠N = 360°

\implies ∠N = 360° - (140° + 75° + 150°) = - 5°

But this is not possible, because an angle of a qadrilateral cannot be negative.

Hence, We cannot construct a quadrilateral PLAN.

Q3 In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct as in the example above?

Answer:

Yes, we still need at least the measure of the angle between the two adjacent sides to create a unique parallelogram.

NCERT solutions for class 8 maths chapter 4 Practical Geometry Excercise: 4.3

Q1 (i) Construct the following quadrilaterals.

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

Answer:

Given, MORE is a quadrilateral.

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

Rough: 

Steps of construction:

Step 1. Draw a line segment MO = 6 cm. Construct an angle of 105° at O. As OR =4.5 cm, draw an arc with O as centre and radius = 4.5 cm. We get point R.

Step 2. Construct an angle of 105° at R.

Step 3. Construct an angle of  60° at M. This ray intersects the previous one at E. 

MORE is the required quadrilateral.

Q1 (ii) Construct the following quadrilaterals.

Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°

Answer:

Given, PLAN is a quadrilateral.

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

We know, Sum of all angles of a quadrilateral = 360°

\therefore ∠P + ∠L + ∠A+ ∠N = 360° \implies  90° + ∠L + 110° + 85° = 360°

\implies ∠L = 360° - 285° = 75°

Steps of Construction

Step 1. Draw a line segment PL = 4 cm. Construct an angle of 75° at L. As LA =6.5 cm, draw an arc with L as centre and radius = 6.5 cm. We get point A.

Step 2. Construct an angle of 110° at A.

Step 3. Construct an angle of  90° at P. This ray intersects the previous one at N. 

PLAN is the required quadrilateral.

Q1 (iii) Construct the following quadrilaterals.

Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°

Answer:

Given, HEAR is a parallelogram.
HE = 5 cm
EA = 6 cm
∠R = 85°

We know, Diagonally opposite angles of a parallelogram are equal.

Therefore, ∠E = ∠R = 85°

And, Opposite sides of a parallelogram are equal and parallel to each other.

Therefore, AR = HE = 5 cm
                  RH = EA = 6 cm

Steps of construction:

Step 1. Draw a line segment HE = 5 cm. Construct an angle of 85° at E. As EA = 6 cm, draw an arc with O as center and radius = 6 cm. We get to point A.

Step 2. With A as centre, draw an arc of radius = AR = HE = 5 cm.

Step 3. With H as centre , draw an arc of radius = HR = EA = 6 cm. The intersection of this arc with the previous arc is R.

Step 4. Join R to H and R to A. 

HEAR is the required parallelogram.

Alternatively,

Steps of construction

Step 1. Draw a line segment HE = 5 cm. Construct an angle of 85° at E. As EA = 6 cm, draw an arc with O as center and radius = 6 cm. We get to point A.

Step 2. Construct an angle of (180°-85° = 95°) at A.    (Adjacent angles in a parallelogram are supplementary)

Step 3. Construct an angle of (180°-85° = 95°) at H. This ray intersects the previous one at R. 

HEAR is the required parallelogram.

Q1 (iv) Construct the following quadrilaterals.

Rectangle OKAY
OK = 7 cm
KA = 5 cm

Answer:

Given, OKAY is a rectangle.
OK = 7 cm
KA = 5 cm

We know that all four angles of a rectangle are right angles (90^{\circ}) and opposite sides are equal

Therefore, OK = AY = 7 cm. KA = YO = 5 cm 

Steps of construction:

Step 1. Draw a line segment OK = 7 cm. Construct a right angle at K. With K a center, draw an arc of radius = 5 cm. We get to point A.

Step 2. Construct another right angle at A. With A as the center, draw an arc of radius = AY =OK = 7cm. We get the point Y.

Step 3. Join Y to O and Y to A 

OKAY is the required rectangle.

Solutions of NCERT for class 8 maths chapter 4 Practical Geometry Excercise: When Three Sides And Two Included Angles Are Given

Q1 In the above example, we first drew BC. Instead, what could have been be the other starting points?

Answer:

Other than BC, we could have drawn AB or CD first.

Q2 We used some five measurements to draw quadrilaterals so far. Can there be different sets of five measurements (other than seen so far) to draw a quadrilateral? The following problems may help you in answering the question.

(i) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and \angle B = 80\degree.

(ii) Quadrilateral PQRS with PQ = 4.5 cm, \angle P = 70\degree, \angle Q = 100\degree, \angle R = 80\degree
and \angle S = 110\degree.

Construct a few more examples of your own to find sufficiency/insufficiency of the
data for the construction of a quadrilateral.

Answer:

Yes, there can be other sets of five measurements to draw a quadrilateral.

Examples:

(i)  Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and \angle B = 80\degree.  (GIVEN)

            (All 4 sides and 1 angle)

(ii) Quadrilateral PQRS having PQ = 4.5 cm, \angle P = 70\degree, \angle Q = 100\degree, \angle R = 80\degree
and \angle S = 110\degree.

            (1 side, all 4 angles)

(iii) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, BD = 8 cm  (Included angle diagonal)  and \angle B = 80\degree.

            (3 sides, 1 included angle, and included angle diagonal)

(iv) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, BD = 8 cm ,\angle B = 80\degree and \angle C = 95\degree

            (2 sides, 2 angles (1 included), and included angle diagonal).

There can be plenty of other examples.

NCERT solutions for class 8 maths chapter 4 Practical Geometry Excercise: 4.4

Q1 (i) Construct the following quadrilaterals.

Quadrilateral DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°

Answer:

DEAR is a quadrilateral.
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°

Steps of construction:

Step 1. Draw a line segment DE = 4 cm. Construct an angle of 60° at E.

Step 2. With E as centre, draw an arc of radius = EA = 5 cm. We get point A.

Step 3. Construct an angle of 90° at A. With A as centre, draw an arc of radius = AR = 4.5 cm. We get the point R.

Step 4. Join R to D.

DEAR is the required quadrilateral.

Q1 (ii) Construct the following quadrilaterals.

Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
\angle R = 75\degree
\angle U = 120\degree

Answer:

TRUE is a quadrilateral.
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
\angle R = 75\degree
\angle U = 120\degree

Steps of construction:

Step 1. Draw a line segment TR = 3.5 cm. Construct an angle of 75° at R.

Step 2. With R as a center, draw an arc of radius = OR = 3 cm. We get point U.

Step 3. Construct an angle of 120° at U. With U as a center, draw an arc of radius = UE = 4 cm. We get the point E.

Step 4. Join E to T

TRUE is the required quadrilateral.

Solutions for NCERT class 8 maths chapter 4 Practical Geometry Excercise: Some Special Cases

Q1 How will you construct a rectangle PQRS if you know only the lengths PQ and QR?

Answer:

A rectangle is a special case of a parallelogram whose opposite sides are equal and all the angles are 90^{\circ}.

Therefore, we know all 4 sides and all 4 angles.

(Steps:

1. Draw PQ. Draw right angle at R and Q. Draw an arc of radius = QR = PS to get the points S and R.

3. Join S to R. PQRS is the required rectangle.)

Q2 Construct the kite EASY if AY = 8 cm, EY = 4 cm and SY = 6 cm (Fig 4.26). Which properties of the kite did you use in the process?

Answer:

We know, in a Kite one diagonal perpendicularly bisects the other diagonal.

Steps of construction:

Step 1. Draw a  line segment AY= 8 cm. Draw the perpendicular bisector of AY. 

Step 2. With Y as a center and radius = EY = 4 cm,  draw an arc on one side of the perpendicular bisector. The intersection is point E.

Step 3. Again, With Y as a center and radius = SY = 6 cm,  draw an arc on the other side of the perpendicular bisector. The intersection is point S.

Step 4. Join E to A and Y. And also join S to A and Y.

EASY is the required kite.

NCERT solutions for class 8 maths chapter 4 Practical Geometry Excercise: 4.5


Q1 The square READ with RE = 5.1 cm.

Answer:

Given, READ is a square side RE = 5.1 cm.

We know, All the sides of a square are equal and all angles are right angles.

Therefore, RE = EA = AD = DR = 5.1 cm

Steps of construction:

Step 1. Draw a line segment RE = 5.1 cm. Construct a right angle at R and E.

Step 2. With R and E as centre, draw an arc of radius = EA = RE = 5.1 cm. We get the point D and A respectively.

Step 3. Join D to A.

READ is the required square.

Q2 Draw a rhombus whose diagonals are 5.2 cm and 6.4 cm long.

Answer:

Let ABCD be the rhombus such that:

AC = 6.4 cm  (longer diagonal) and BD = 5.2 cm (shorter diagonal)

We know that the diagonals of a rhombus bisect (cut in half) each other perpendicularly, i.e at 60°

In other words, the midpoint of the diagonals coincide.

Steps of construction:

Step 1. Draw a line segment AC =6.4 cm. Now, construct the perpendicular bisector of AC. Let it intersect AC at O.

    (Therefore, O is the midpoint of both the diagonals AC and BD)

Step 2. With O as center and radius half of BD = \frac{5.2}{2} = 2.6\ cm , draw two arcs on both sides of AC intersecting the perpendicular bisector at B and D.

Step 3. Join B to A and C. Also join D to A and C.

ABCD is the required rhombus.

Q3 Draw a rectangle with adjacent sides of lengths 5 cm and 4 cm.

Answer:

Let ABCD be the rectangle such that :

Where AB = 5 cm and BC = 4 cm.

We know that the opposite sides of a rectangle are equal and all the angles are 90°.

Steps of construction:

Step 1. Draw a line segment AB = 5 cm. Construct an angle of 90° at A and B. 

Step 2. Construct an arc of radius =AD =  BC = 4 cm with A and B as the centre . The intersection points are C and D.

Step 3. Join C to D.

ABCD is the required rectangle.

Q4 Draw a parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?

Answer:

Given, OKAY is a parallelogram where OK = 5.5 cm and KA = 4.2 cm.

Steps of construction:

Step 1. Draw a line segment OK = 5.5 cm. Draw a ray KX from point K (With any suitable angle). Extend OK in the direction of KO.

Now we make a ray parallel to KX from O.

Step 2. With K as the center and a suitable radius, draw an arc cutting both OK and KX at P and Q respectively.

Step 3. With the same radius and O as the center, draw an arc cutting extended OK at P'.

Step 4. With P as the center, measure PQ using the compass. Using this as radius, cut the previous arc with P' as the center and mark it Q'. Draw a ray OZ passing through Q'.

Step 5. With radius = KA = 4.2 cm, cut two arcs on OZ and KX with O and K as centers respectively. These intersections points are Y and A respectively. Join Y to A.

OKAY is the required parallelogram. It is not unique as the angle can be varied keeping opposite sides parallel to each other.

NCERT solutions for class 8 maths: Chapter-wise

Chapter -1

NCERT solutions for class 8 maths chapter 1 Rational Numbers

Chapter -2 

Solutions of NCERT for class 8 maths chapter 2 Linear Equations in One Variable

Chapter-3

CBSE NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals

Chapter-4

NCERT solutions for class 8 maths chapter 4 Practical Geometry

Chapter-5

Solutions of NCERT for class 8 maths chapter 5 Data Handling

Chapter-6

CBSE NCERT solutions for class 8 maths chapter 6 Squares and Square Roots

Chapter-7

NCERT solutions for class 8 maths chapter 7 Cubes and Cube Roots

Chapter-8

Solutions of NCERT for class 8 maths chapter 8 Comparing Quantities

Chapter-9

NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities

Chapter-10

CBSE NCERT solutions for class 8 maths chapter 10 Visualizing Solid Shapes

Chapter-11

NCERT solutions for class 8 maths chapter 11 Mensuration

Chapter-12

NCERT Solutions for class 8 maths chapter 12 Exponents and Powers

Chapter-13

CBSE NCERT solutions for class 8 maths chapter 13 Direct and Inverse Proportions

Chapter-14

NCERT solutions for class 8 maths chapter 14 Factorization

Chapter-15

Solutions of NCERT for class 8 maths chapter 15 Introduction to Graphs

Chapter-16

CBSE NCERT solutions for class 8 maths chapter 16 Playing with Numbers

NCERT solutions for class 8: Subject-wise

How to use NCERT solutions for class 8 maths chapter 4 Practical Geometry?

  • You must have knowledge about the classification of geometric figures.
  • You should go through some examples to understand the pattern of constructions.
  • Once you have a clear idea, you can draw the shapes for unsolved problems given in the practice exercises.
  • NCERT solutions for class 8 maths chapter 4 can be used to help while practicing.
  • Keep working hard and happy learning!
 

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