Question 18 - End-of-Chapter Exercises - Chapter 5 Class 9 - I’m Up and Down, and Round and Round (Ganita Manja

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Question 18 Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm . Find the radius of the circle. Let’s draw the figure Let our two chords be AB = 10 cm & CD = 24 cm with mid-points M & N respectively We are given they are on same side of center of circle And, distance between them is MN = 7 cm We need to find radius of circle i.e. r Let distance between CD & center be x i.e. let ON = x And, distance between them is MN = 7 cm We need to find radius of circle i.e. r Let distance between CD & center be x i.e. let ON = x Since MN is distance between the two chords ∴ MN ⊥ AB & MN ⊥ CD We know that Perpendicular from the center bisects the chord Thus, M is mid-point of AB AM = MB = 10/2 = 5 cm And, N is mid-point of CD CN = ND = 24/2 = 12 cm We use Pythagoras Theorem to find radius In ∆ OMA AM = 5 cm OM = MN + ON = 7 + x OA = Radius = r Since ∆ OMA is a right angled triangle By Pythagoras Theorem OA^2=OM^2+AM^2 𝒓^𝟐=〖(𝟕+𝒙)〗^𝟐 + 𝟓^𝟐 𝑟^2=7^2+𝑥^2+2 × 7 × 𝑥+25 𝑟^2=49+𝑥^2+14𝑥+25 𝒓^𝟐=𝒙^𝟐+𝟏𝟒𝒙+𝟕𝟒 In ∆ OCN CN = 12 cm ON = x OC = Radius = r Since ∆ ONC is a right angled triangle By Pythagoras Theorem O𝐶^2=𝑂𝑁^2+𝐶𝑁^2 𝑟^2=𝑥^2+12^2 𝒓^𝟐=𝒙^𝟐+𝟏𝟒𝟒 Comparing (1) & (2) Since their Left side is same (𝑟^2) So, their Right side would be the same too 𝒙^𝟐+𝟏𝟒𝒙+𝟕𝟒=𝒙^𝟐+𝟏𝟒𝟒 𝑥^2−𝑥^2+14𝑥=144−74 14𝑥=70 𝑥=70/14 𝒙=𝟓 But, we need to find Radius r Putting x = 5 in (2) 𝑟^2=𝑥^2+144 𝑟^2=5^2+144 𝑟^2=25+144 𝑟^2=169 𝑟^2=13^2 𝑟=√(13^2 ) 𝒓=𝟏𝟑 cm Thus, radius is 13 cm