Question

AD is a diameter of a circle and AB is a chord. If AD = 10 cm, AB = 8 cm, the distance of AB from the centre of the circle is​

Answer 1

Answer:

the answer is 3cm

Step-by-step explanation:

since diameter is 10 cm

radius is 5cm

also length of chord is 8cm

it is a property that the perpendicular drawn from the centre to the chord bisects the chord ie 1:1 ratio

now we get a right triangle with hypotenuse is 5cm and base is 4cm

applying pytagorus theorem we get height of perpendicular as 3cm which is the distance from the centre to chord.

Answer 2

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.