# Ex 11.1, 2 - 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 11.1, 2 Construct an angle of 45° at the initial point of a given ray and justify the construction . Steps of construction Draw a ray OA. Taking O as center and any radius, draw an arc cutting OA at B. 3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C. 4. With C as center and the same radius, draw an arc cutting the arc at D 5. With C and D as centers and radius more than 1/2 CD, draw two arcs intersecting at P. 6. Join OP. Thus, ∠ AOP = 90° Now we draw bisector of ∠ AOP 7. Let OP intersect the original arc at point Q 8. Now, taking B and Q as centers, and radius greater than 1/2 BQ, draw two arcs intersecting at R. 9. Join OR. Thus, ∠ AOR = 45° Justification We need to prove ∠ AOR = 45° Join OC & OB Thus, OB = BC = OC ∴ Δ OCB is an equilateral triangle ∴ ∠ BOC = 60° Join OD, OC and CD Thus, OD = OC = DC ∴ Δ DOC is an equilateral triangle ∴ ∠ DOC = 60° Join PD and PC Now, In Δ ODP and Δ OCP OD = OC DP = CP OP = OP ∴ Δ ODP ≅ Δ OCP ∴ ∠ DOP = ∠ COP So, we can say that ∠ DOP = ∠ COP = 1/2 ∠ DOC ∠ DOP = ∠ COP = 1/2 × 60° = 30° Now, ∠ AOP = ∠ BOC + ∠ COP ∠ AOP = 60° + 30° = 90° Now, Join QR and BR In Δ OQR and Δ OBR OQ = OB QR = BR OR = OR ∴ Δ OQR ≅ Δ OBR ∴ ∠ QOR = ∠ BOR ∠ QOR = ∠ BOR = 1/2 ∠ AOP ∠ DOP = ∠ COP = 1/2 × 90° = 45° Thus, ∠ AOR = 45° Hence justified

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

About the Author

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.