radius of a circle is 5cm.if length of a chord is 6cm.Find the distance of chord from the centre of circle.
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Step-by-step explanation:
Hello Mate!
In ∆ABM,
AB² = BM² + AM²
AB² - AM² = BM²
6² - AM² = BM² _(i)
In ∆BMO
BO² = BM² + OM²
5² - OM² = BM² _(ii)
From (i) and (ii) we get,
5² - OM² = 6² - AM²
AM² = 36 - 25 + OM²
Since OM = AO - AM
AM² = 9 + ( AO - AM )²
AM² = 9 + ( 5 - AM )²
AM² = 9 + 25 + AM² - 10AM
10AM = 36
AM = 3.6 cm
In ∆AMC
AC² = AM² + CM²
6² = 3.6² + CM²
36 - 12.96 = CM²
√23.04 = CM
4.8 = CM.
Since AO is perpendicular bisector of chord BC.
BM = CM
BM + CM = BC
2CM = BC
2(4.8) = BC
9.6 cm = BC
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