RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

Class 8 : RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev

The document RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8

PAGE NO 18.8:

Question 1:

Construct a quadrilateral ABCD in which AB  = 3.8 cm, BC  = 3.4 cm, CD  = 4.5 cm, AD  = 5 cm and ∠B  = 80°.

ANSWER:

Steps of construction:

Step I: Draw AB = 3.8 cm.

Step II: Construct ∠ABC = 80°.

Step III : With B as the centre and radius 3.4 cm, cut off BC = 3.4 cm.

Step IV: With C as the centre and radius 4.5 cm, draw an arc.

Step V :With A as the centre and radius 5.3 cm, draw an arc to intersect the arc drawn in Step IV at D.

Step VI: Join AD, BC and CD to obtained the required quadrilateral.

 

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev


Question 2:

Construct a quadrilateral ABCD, given that AB  = 8 cm, BC  = 8 cm, CD  = 10 cm, AD  = 10 cm and ∠A  = 45°.

ANSWER:

Steps of Construction:

Step I: Draw AB = 8 cm.

Step II: Construct ∠BAD = 45°.

Step III : With A as the centre and radius 10 cm, cut off AD = 10 cm.

Step IV: With D as the centre and radius 10 cm, draw an arc.

Step V :With B as the centre and radius 8 cm, draw an arc to intersect the arc drawn in Step IV at C.

Step VI: Join  BC and CD to obtained the required quadrilateral.

 

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev


Question 3:

Construct a quadrilateral ABCD in which AB  = 7.7 cm, BC  = 6.8 cm, CD  = 5.1 cm, AD  = 3.6 cm and ∠C  = 120°.

ANSWER:

Steps of construction:

Step I: Draw DC = 5.1 cm.

Step II: Construct ∠DCB = 120°.

Step III : With C as the centre and radius 6.8 cm, cut off BC = 6.8 cm.

Step IV: With B as the centre and radius 7.7 cm, draw an arc.

Step V :With D as the centre and radius 3.6 cm, draw an arc to intersect the arc drawn in Step IV at A.

Step VI: Join AB and AD to obtained the required quadrilateral.

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev

 


Question 4:

Construct a quadrilateral ABCD in which AB  =  BC  = 3 cm, AD  =  CD  = 5 cm, and ∠B  = 120°.

ANSWER:

Steps of construction:

Step I: Draw AB = 3 cm.

Step II: Construct ∠ABC = 120°.

Step III : With B as the centre and radius 3 cm, cut off BC = 3 cm.

Step IV: With C as the centre and radius 5 cm, draw an arc.

Step V :With A as the centre and radius 5 cm, draw an arc to intersect the arc drawn in Step IV at D.

Step VI: Join AD and CD to obtained the required quadrilateral.

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev

 

Question 5:

Construct a quadrilateral ABCD in which AB  = 2.8 cm, BC  = 3.1 cm, CD  = 2.6 cm, and DA  = 3.3 cm and ∠A  = 60°.

ANSWER:

Steps of construction:

Step I: Draw AB = 2.8 cm.

Step II: Construct ∠BAD = 60°.

Step III : With A as the centre and radius 3.3 cm, cut off AD = 3.3 cm.

Step IV: With D as the centre and radius 2.6 cm, draw an arc.

Step V :With B as the centre and radius 3.1 cm, draw an arc to intersect the arc drawn in Step IV at C.

Step VI: Join BC and CD to obtained the required quadrilateral.

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev


Question 6:

Construct a quadrilateral ABCD in which AB  =  BC  = 6 cm, AD  =  DC  = 4.5 cm and ∠B  = 120°.

ANSWER:

Steps of construction:

Step I: Draw AB = 6 cm.

Step II: Construct ∠ABC = 120°.

Step III : With B as the centre and radius 6 cm, cut off BC = 6 cm.

Now, we can see that AC is about 10.3 cm 

which is greater than AD+CD = 4.5+4.5 = 9 cm.

We know that sum of the lengths of two sides of triangle is always greater than the third side but here, the sum of AD and CD is less than AC.

So, construction of the given quadrilateral is not possible.

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5), Class 8, Maths Class 8 Notes | EduRev

Complete Syllabus of Class 8

Dynamic Test

Content Category

Related Searches

MCQs

,

mock tests for examination

,

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5)

,

Maths Class 8 Notes | EduRev

,

Semester Notes

,

past year papers

,

Maths Class 8 Notes | EduRev

,

Important questions

,

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5)

,

shortcuts and tricks

,

Exam

,

pdf

,

study material

,

Class 8

,

Class 8

,

video lectures

,

RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 5)

,

Class 8

,

Summary

,

ppt

,

practice quizzes

,

Viva Questions

,

Extra Questions

,

Objective type Questions

,

Maths Class 8 Notes | EduRev

,

Previous Year Questions with Solutions

,

Sample Paper

,

Free

;