Two parallel chords of lengths 30 cm and 16 cm are drawn on opposite sides of the centre of a circle
of radius 17 cm , find the distance between the chords .
Answers
Step-by-step explanation:
Consider AB and CD as the chords of circle with centre O It is given that AB = 30cm and CD = 16cm Join the lines OA and OC We know that AO = 17cm and CO = 17cm Construct OM ⊥ CD and OL ⊥ AB Perpendicular from the centre of a circle to a chord bisects the chord We know that AL = ½ × AB By substituting the values AL = ½ × 30 So we get AL = 15cm We know that CM = ½ × CD By substituting the values CM = ½ × 16 So we get CM = 8cm Consider △ ALO Using the Pythagoras theorem it can be written as AO2 = OL2 + AL2 By substituting the values 172 = OL2 + 152 So we get OL2 = 172 – 152 On further calculation OL2 = 289 – 225 By subtraction OL2 = 64 By taking the square root OL = √64 OL = 8cm Consider △ CMO Using the Pythagoras theorem it can be written as CO2 = CM2 + OM2 By substituting the values 172 = 82 + OM2 So we get OM2 = 172 – 82 On further calculation OM2 = 289 – 64 By subtraction OM2 = 225 By taking the square root OM = √225 OM = 15cm So the distance between the chords = OM + OL By substituting the values Distance between the chords = 8 + 15 = 23cm Therefore, the distance between the chords is 23 cm.Read more on Sarthaks.com - https://www.sarthaks.com/727461/two-parallel-chords-lengths-30cm-16cm-drawn-opposite-sides-the-centre-circle-radius-17cm
Answer:
The yhydj yeh rhe geen dudh rhe dudh when yeh eggs yeh hu eb heb dHdb hbddhb hdbh dudh hd d ke hebben yeh heb heb hd hd. hd d. hd.