AD is a diameter of a circle and AB is a chord If AB = 30 cm and AB= 24cm then the distance of AB from the centre of the circle is
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U write the value of AB twice
U write the value of AB twice
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AD is the diameter of the circle of length is AD = 34 cm
AB is the chord of the circle of length is AB = 30 cm.
Distance of the chord from the centre is OM.
Since the line through the centre to the chord of the circle is the perpendicular bisector, we have
∠OMA = 90° and AM = BM.
∴ ΔAMC is a right triangle.
Apply Pythagorean Theorem
OA2 = AM2 + OM2 --------(1)
Since the diameter AD = 34 cm., radius of the circle is 17 cm.
Thus,
OA = 17 cm
Since AM = BM and AB = 30 cm, we have AM = BM = 15 cm.
Substitute the values in equation (1), we get
OA2 = AM2 + OM2
172 = 152 + OM2
OM2 = 289 – 225
OM2 = 64
OM = 8.
Distance of the chord from the centre is 8 cm.
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