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Ex 10.4, 6 - A circular park of radius 20 m is situated - Ex 10.4

Ex 10.4, 6 - Chapter 10 Class 9 Circles - Part 2
Ex 10.4, 6 - Chapter 10 Class 9 Circles - Part 3
Ex 10.4, 6 - Chapter 10 Class 9 Circles - Part 4

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Ex 10.4, 6 A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. Given: Circular park with radius 20m Let Ankur, Syed and David be donated by points A,S & D resp. Given all three are sitting at equal distances, i.e., AS = SD = AD To find: Length of AS = SD = AD Explanation: Let AS = SD = AD = 2x In ASD, all sides are equal, ∴ ASD is an equilateral triangle We draw OP ⊥ SD So, SP = DP = 1/2 SD ⇒ SP = DP = 2𝑥/2 = x Join OS & AO In Δ OPS By Pythagoras theorem OS2 = OP2 + PS2 (20)2 = OP2 + x2 400 = OP2 + x2 400 – x2 = OP2 OP2 = 400 – x2 OP = √("400 – x2" ) Now, AP = AO + OP √("3" )x = 20 + √("400 – x2" ) √("3" )x – 20 = √("400 – x2" ) √("400 – x2" ) = √("3" )x – 20 Squaring both sides (√("400 – x2" ))^2= (√("3" )x – 20)2 400 – x2 = (√3x)2 + (20)2 – 2 ×(√3x) × (20) "400 – x2 =" 3x2 + 400 – 40√3x "400 – 400 + 40" √3 "x =" 3x2 + x2 "40" √3 "x =" 4x2 4x2 = 40√3x "x2" /"x" = 40/4 √3 x = 10√3 m AS = SD = AD = 2x = 2 × 10√3 = 20√3 m ∴ Length of string of each phone is 20√3 m

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.