NCERT Solutions Class 8 Maths Chapter 4 Practical Geometry – Here are all the NCERT solutions for Class 8 Maths Chapter 4. This solution contains questions, answers, images, explanations of the complete chapter 4 titled Practical Geometry of Maths taught in class 8. If you are a student of class 8 who is using NCERT Textbook to study Maths, then you must come across chapter 4 Practical Geometry. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry in one place.
NCERT Solutions Class 8 Maths Chapter 4 Practical Geometry
Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 8 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.
For a better understanding of this chapter, you should also see summary of Chapter 4 Practical Geometry , Maths, Class 8.
| Class | 8 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 4 |
| Chapter Name |
Practical Geometry |
NCERT Solutions Class 8 Maths chapter 4 Practical Geometry
Class 8, Maths chapter 4, Practical Geometry solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm,
∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique
quadrilateral? Give reasons for your answer.
Ans : Missing
Q.2: (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?
(ii) Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and
AS = 6.5 cm? Why?
(iii) Can you draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm? Why?
(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].
Ans : Missing
Q.3: Construct the following quadrilaterals.
(i) Quadrilateral ABCD.
AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AD = 6 cm
AC = 7 cm
(ii) Quadrilateral JUMP
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
Ans : (i) Firstly, a rough sketch of this quadrilateral can be drawn as follows.
(1) \( \triangle \mathrm{ABC} \) can be constructed by using the given measurements as follows.
(2) Vertex D is 6 cm away from vertex A. Therefore, while taking Aas 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.
(ii) Firstly, a rough sketch of this quadrilateral can be drawn as follows.
(1) \( \Delta \mathrm{JUP} \) 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.
(iii) We know that opposite sides of a parallelogram are equal in length and also these are parallel to each other.
Hence, ME = OR, MO = ER
A rough sketch of this parallelogram can be drawn as follows.
(1) \( \triangle \mathrm{EOR} \) can be constructed by using the given measurements as follows.
(2) Vertex M is 4.5 cm away from vertex O and 6 cm away from vertex E. Therefore, while taking O and E as centres, draw arcs of 4.5 cm radius and 6 cm radius respectively. These will intersect each other at point M.
(3) Join M to O and E.
MORE is the required parallelogram.
(iv) We know that all sides of a rhombus are of the same measure.
Hence, BE = ES = ST = TB
A rough sketch of this rhombus can be drawn as follows.
(1) \( \Delta \mathrm{BET} \) can be obstructed by using the given measurements as follows.
(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.
(3) Join S to E and T.
BEST is the required rhombus.
Q.4: . In the above example, can we draw the quadrilateral by drawing ∆ABD first and then find the fourth point C?
Ans : Missing
Q.5: 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.
Ans : Missing
NCERT / CBSE Book for Class 8 Maths
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