Question

Ad is the diameter of a circle and ab is a chord if ad=34cm and ab=30cm then find the distance of ab from the center

Answer 1

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.

Answer 2

Hey mete here is ur answer...

Answer:

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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