Ex 5.5, 3

Transcript

Ex 5.5, 3 In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD=2AB? Give reasons for your answer. We draw a diagram Let Radius = r And, let distance of chord CD from center be d ∴ ON = d Given that distance of chord AB from the centre is twice the distance of another chord CD from the centre ∴ OM = 2 × ON = 2d We need to find how CD and AB are related Now, we know that Perpendicular from the center to the chord, bisects the chord Now, ∆ AOM & ∆CON are right angled triangles Applying Baudhāyana–Pythagoras theorem Since OM ⊥ AB We can write AM = BM = (𝑨𝑩 )/𝟐 Since ON ⊥ CD We can write CN = DN = 𝑪𝑫/𝟐 In ∆ AOM By Pythagoras theorem 〖𝑶𝑨〗^𝟐=〖𝑶𝑴〗^𝟐+〖𝑨𝑴〗^𝟐 Putting OA = r, OM = 2d 𝒓^𝟐=〖(𝟐𝒅) 〗^𝟐 + 〖𝑨𝑴〗^𝟐 𝑟^2=4𝑑^2+ 〖𝐴𝑀〗^2 𝑟^2−4𝑑^2=〖𝐴𝑀〗^2 〖𝐴𝑀〗^2=𝑟^2−4𝑑^2 𝑨𝑴=√(𝒓^𝟐−𝟒𝒅^𝟐 " " ) Since AB = 2 × AM AB = 𝟐√(𝒓^𝟐−𝟒𝒅^𝟐 " " ) In ∆ CON By Pythagoras theorem 〖𝑶𝑪〗^𝟐=〖𝑶𝑵〗^𝟐+〖𝑪𝑵〗^𝟐 Putting OC = r, ON = d 𝒓^𝟐=〖𝒅 〗^𝟐 + 〖𝑪𝑵〗^𝟐 𝑟^2−𝑑^2=〖𝐶𝑁〗^2 〖𝐶𝑁〗^2=𝑟^2−𝑑^2 𝑪𝑵=√(𝒓^𝟐−𝒅^𝟐 " " ) Since CD = 2 × CN CD = 𝟐√(𝒓^𝟐−𝒅^𝟐 " " ) The question asks if CD = 2 × AB But CD = 2√(𝑟^2−𝑑^2 " " ) AB = 2√(𝑟^2−4𝑑^2 " " ) Thus, CD ≠ 2 × AB Hence, our answer is no