**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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### NCERT Solutions Class 8 Maths chapter 4 Practical Geometry- Video

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### Download NCERT Solutions Class 8 Maths chapter 4 Practical Geometry In PDF Format

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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

You can download the NCERT Book for Class 8 Maths in PDF format for free. Otherwise you can also buy it easily online.

- Click here for NCERT Book for Class 8 Maths
- Click here to buy NCERT Book for Class 8 Maths

### All NCERT Solutions Class 8

- NCERT Solutions for Class 8 English
- NCERT Solutions for Class 8 Hindi
- NCERT Solutions for Class 8 Maths
- NCERT Solutions for Class 8 Science
- NCERT Solutions for Class 8 Social Science
- NCERT Solutions for Class 8 Sanskrit

### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Class 1 | Class 2 | Class 3 |

Class 4 | Class 5 | Class 6 |

Class 7 | Class 8 | Class 9 |

Class 10 | Class 11 | Class 12 |

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