A chord of length 42 cm is drawn in a circle having diameter 58 cm. What is the minimum distance of other parallel chord of length 40 cm in the same circle from 42cm long chord?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 1 cm.
We have the following figure,
Diametre = 58 cm, radius = 29 cm
MN is the minimum distance between the chords
OM is the perpendicular on chord AB which divides the chord into two equal halves
Hence, BM = AM = 29 cm
Using Pythagoras theorem in triangle OMA, we get
⇒ OA2 = OM2 + AM2
⇒ 292 = OM2 + 212
⇒ OM2 = 841 - 441
⇒ OM2 = 400
⇒ OM = 20 cm
Similarly using Pythagoras theorem in triangle ONC, we get
⇒ ON2 = 292 - 202
⇒ ON = 21
Hence, the distance between parallel chord = OM - ON = MN = 21 - 20 = 1 cm