# Ex 11.2, 3 - Chapter 11 Class 9 Constructions (Important Question)

Last updated at Oct. 5, 2018 by Teachoo

Last updated at Oct. 5, 2018 by Teachoo

Transcript

Ex 11.2, 3 Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR − PQ = 2 cm. Steps of Construction: Draw base QR of length 6 cm 2. Now, let’s draw ∠ Q = 60° Let the ray be QX Open the compass to length PR – PQ = 2 cm. Note: Since PR – PQ = 2 cm, (PQ – PR) is negative So, QD will be below line QR From point Q as center, cut an arc on ray QX. (opposite side of QR). Let the arc intersect QX at D 4. Join RD Now, we will draw perpendicular bisector of RD 6. Mark point P where perpendicular bisector intersects RD Join PR ∴ Δ PQR is the required triangle

Class 9

Important Questions for Exam - Class 9

- Chapter 1 Class 9 Number Systems
- Chapter 2 Class 9 Polynomials
- Chapter 3 Class 9 Coordinate Geometry
- Chapter 4 Class 9 Linear Equations in Two Variables
- Chapter 5 Class 9 Introduction to Euclid's Geometry
- Chapter 6 Class 9 Lines and Angles
- Chapter 7 Class 9 Triangles
- Chapter 8 Class 9 Quadrilaterals
- Chapter 9 Class 9 Areas of parallelograms and Triangles
- Chapter 10 Class 9 Circles
- Chapter 11 Class 9 Constructions
- Chapter 12 Class 9 Herons Formula
- Chapter 13 Class 9 Surface Areas and Volumes
- Chapter 14 Class 9 Statistics
- Chapter 15 Class 9 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.