Question
Download Solution PDFA 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 (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
The diameter of the circle = 58 cm
The length of one chord = 42 cm
The length of another chord = 40 cm
Formula used:
H2 = B2 + P2
Calculation:
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 = 40 cm and OB = 29
Then,
AM = MB = 40/2 = 20 cm
In ΔMOB
(OB)2 = (OM)2 + (MB)2
⇒ (29)2 = (OM)2 + (20)2
⇒ 841 = (OM)2 + 400
⇒ (OM)2 = (841 – 400)
⇒ (OM)2 = 441
⇒ OM = 21 cm
Now,
PQ = 42 cm and OB = 29
NQ = PN = 42/2 = 21 cm
In ΔONQ
(OQ)2 = (ON)2 + (NQ)2
⇒ (29)2 = (ON)2 + (21)2
⇒ 841 = (ON)2 + 441
⇒ (ON)2 = (841 – 441)
⇒ (ON)2 = 400
⇒ ON = 20 cm
So according to the question.
⇒ 21 - 20 = 1
∴ The minimum distance between the two chords is 1 cm.
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