Math, asked by makkarjai1512, 12 days ago

# AD is a diameter of a cricle and ab is a chord if AB=34cm ,AB =30cm find the distance of AB from center​

0

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.

Apply Pythagorean Theorem

∴ ΔAMC is a right triangle.

OA2 = AM2 + OM2 --------(1)

Since the diameter AD = 34 cm., radius of the circle is 17 cm.

OA = 17 cm

Thus,

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

OM = 8.Distance of the chord from the centre is 8 cm.

OM2 = 64

0

AD is a diameter of a circle and AB is chord. If AD= 34 cm and AB =30 cm, the distance of AB from the centre of the circle is

AD is a diameter of a circle and AB is chord. If AD= 34 cm and AB =30 cm, the distance of AB from the centre of the circle is: (A) 17cm (B) 15cm (C) 4cm (D)8cm

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