Class 8 Exam  >  Class 8 Notes  >  RD Sharma Solutions for Class 8 Mathematics  >  RD Sharma Solutions - Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths

Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

PAGE NO 18.13:

Question 1:

Construct a quadrilateral ABCD given that AB  = 4 cm, BC  = 3 cm, ∠A  = 75°, ∠B  = 80° and ∠C  = 120°.

ANSWER:

Steps of construction:

Step I: Draw AB = 4 cm.

Step II: Construct ∠XAB = 75° at A and ∠ABY = 80° at B.

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

Step IV: At C, draw ∠BCZ = 120° such that it meets AX at D. 

The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 2:

Construct a quadrilateral ABCD, where AB  = 5.5 cm, BC  = 3.7 cm, ∠A  = 60°, ∠B  = 105° and ∠D  = 90°.

ANSWER:

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

i.e., ∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠C = 105°

Steps of construction:

Step I: Draw AB = 5.5 cm.

Step II: Construct ∠XAB = 60° at A and ∠ABY = 105°.

Step III: With B as the centre and radius 3.7 cm, cut off BC = 3.7 cm.

Step IV: At C, draw ∠BCZ = 105° such that it meets AX at D.

The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 3:

Construct a quadrilateral PQRS, where PQ  = 3.5 cm, QR  = 6.5 cm, ∠P  = ∠R  = 105° and ∠S  = 75°.

ANSWER:

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

i.e., ∠P + ∠Q + ∠R + ∠S = 360°⇒ ∠Q = 75°

Steps of construction:

Step I: Draw PQ = 3.5 cm.

Step II: Construct ∠XPQ = 105° at P and ∠PQY = 75° at Q.

Step III: With Q as the centre and radius 6.5 cm, cut off QR = 6.5

Step IV: At R, draw ∠QRZ = 105° such that it meets PX at S.

The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 4:

Construct a quadrilateral ABCD when BC  = 5.5 cm, CD  = 4.1 cm, ∠A  = 70°, ∠B  = 110° and ∠D  = 85°.

ANSWER:

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

i.e., ∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠C = 95°

Steps of construction:

Step I: Draw BC = 5.5 cm.

Step II: Construct ∠XBC = 110° at A and ∠BCY = 95°.

Step III: With C as the centre and radius 4.1 cm, cut off CD = 4.1 cm.

Step IV: At D, draw ∠CDZ = 85° such that it meets BY at A.

The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 5:

Construct a quadrilateral ABCD, where ∠A  = 65°, ∠B  = 105°, ∠C  = 75° BC  = 5.7 cm and CD  = 6.8 cm.

ANSWER:

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

i.e., ∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠D = 115°

Steps of Construction:

Step I: Draw BC = 5.7 cm.

Step II: Construct ∠XBC = 105° at B and ∠BCY = 105° at C.

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

Step IV: At D, draw ∠CDZ = 115° such that it meets BY at A. 

The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics


Question 6:

Construct a quadrilateral PQRS, in which PQ  = 4 cm, QR  = 5 cm, ∠P  = 50°, ∠Q  = 110° and ∠R  = 70°.

ANSWER:

Steps of construction:

Step I: Draw PQ = 4 cm.

Step II: Construct ∠XPQ = 50° at P and ∠PQY = 110° at Q.

Step III : With Q as the centre and radius 5 cm, cut off QR = 5 cm.

Step IV: At R, draw ∠QRZ = 70° such that it meets PX at S. 

The quadrilateral so obtained is the required quadrilateral.

Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

The document Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics 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
88 docs

Top Courses for Class 8

FAQs on Chapter 18 - Practical Geometry (Constructions) (Part - 2), Class 8, Maths RD Sharma Solutions - RD Sharma Solutions for Class 8 Mathematics

1. What are the different types of constructions covered in Chapter 18 of RD Sharma Solutions?
Ans. Chapter 18 of RD Sharma Solutions covers various types of constructions such as constructions of tangents to a circle, constructions of triangles, constructions of quadrilaterals, and constructions of regular polygons.
2. How can I construct a tangent to a circle using RD Sharma Solutions?
Ans. To construct a tangent to a circle using RD Sharma Solutions, follow these steps: 1. Draw the circle. 2. Take a point outside the circle and join it to the center of the circle. 3. Bisect the line segment joining the center of the circle to the external point. 4. Draw a perpendicular to the bisector from the external point. 5. The perpendicular drawn is the required tangent to the circle.
3. How can I construct an equilateral triangle using RD Sharma Solutions?
Ans. To construct an equilateral triangle using RD Sharma Solutions, follow these steps: 1. Draw a line segment AB of any length. 2. With A as the center, draw an arc of the same length as AB. 3. With B as the center, draw another arc of the same length as AB, intersecting the previous arc. 4. Join the points of intersection of the two arcs with line segments. 5. The triangle formed by these line segments is an equilateral triangle.
4. How can I construct a square using RD Sharma Solutions?
Ans. To construct a square using RD Sharma Solutions, follow these steps: 1. Draw a line segment AB of any length. 2. With A as the center, draw an arc of the same length as AB. 3. With B as the center, draw another arc of the same length as AB, intersecting the previous arc. 4. Join the points of intersection of the two arcs with line segments. 5. Extend the line segments on both sides to form a square.
5. How can I construct a regular hexagon using RD Sharma Solutions?
Ans. To construct a regular hexagon using RD Sharma Solutions, follow these steps: 1. Draw a line segment AB of any length. 2. With A as the center, draw an arc of the same length as AB. 3. With B as the center, draw another arc of the same length as AB, intersecting the previous arc. 4. Join the points of intersection of the two arcs with line segments. 5. Extend the line segments on both sides to form a regular hexagon.
88 docs
Download as PDF
Explore Courses for Class 8 exam

Top Courses for Class 8

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Objective type Questions

,

Summary

,

study material

,

Exam

,

Extra Questions

,

Class 8

,

Chapter 18 - Practical Geometry (Constructions) (Part - 2)

,

Chapter 18 - Practical Geometry (Constructions) (Part - 2)

,

pdf

,

ppt

,

Sample Paper

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

,

Viva Questions

,

Previous Year Questions with Solutions

,

practice quizzes

,

Free

,

shortcuts and tricks

,

mock tests for examination

,

Important questions

,

Class 8

,

past year papers

,

video lectures

,

Chapter 18 - Practical Geometry (Constructions) (Part - 2)

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics

,

Semester Notes

,

Class 8

,

MCQs

;