Last updated at Feb. 25, 2017 by Teachoo

Transcript

Ex 10.2,13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Given : Let ABCD be the quadrilateral circumscribing the circle with centre O. ABCD touches the circle at points P,Q,R and S To prove: Opposite sides subtend supplementary angles at centre i.e. ∠ AOB + ∠ COD = 180° & ∠ AOD + ∠ BOC = 180° Construction: Join OP, OQ, OR & OS Proof: Let us rename the angles In Δ AOP and Δ AOS AP = AS AO = AO OP = OS ∴ Δ AOP ≅∆ AOS ∠ AOP = ∠ AOS i.e. ∠ 1 = ∠ 8 Similarly, we can prove ∠2 = ∠3 ∠5 = ∠4 ∠6 = ∠7 Now ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360° ∠ 1 + ∠ 2 + ∠ 2 + ∠ 5 + ∠ 5 + ∠ 6 + ∠ 6 + ∠ 1 = 360° 2 (∠ 1 + ∠ 2 + ∠ 5 + ∠ 6) = 360° ∠ 1 + ∠ 2 + ∠ 5 + ∠ 6 = (360°)/2 (∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180° ∠ AOB + ∠ COD =180° Hence both angle are supplementary Similarly, we can prove ∠ BOC + ∠ AOD =180° Hence proved

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 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.