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 42 cm long chord?
Answer & explanation
Correct answer: option 2
We know that,
H2 = B2 + P2
We have,
The diameter of the circle = 58 cm
The length of one chord = 42 cm
The length of another chord = 40 cm
AB and PQ are two chords, and O is the center of the circle.
M is the midpoint of AB and N is the midpoint of PQ
OB = OQ = 29 cm [radius of the circle]
AB = 42 cm and OB = 29
Then,
AM = MB = \(\frac{42}{2}\) = 21
In ΔMOB
(OB)2 = (OM)2 + (MB)2
= (29)2 = (OM)2 + (21)2
= 841 = (OM)2 + 441
= OM = 20 cm
Now,
PQ = 40 cm and OB = 29
NQ = PN = \(\frac{40}{2}\) = 20
In ΔONQ
(OQ)2 = (ON)2 + (NQ)2 = (29)2 = (ON)2 + (20)2
= 841 = (ON)2 + 400
= ON = 21 cm
So according to the question. = 21 – 20 = 1