Question

in the given figure ​

Answer 1

Answer:

8cm

Step-by-step explanation:

we know that a perpendicular line drawn from a chord to the centre of circle , divides the chord into two equal parts

therefore AD is half AC

AD=1/2×12

AD=6

RADIUS IS HALF THE DIAMETER

THEREFORE AO WILL BE 10cm

Taking triangle AOD

AO is the hypotenuse

AD is the base

DO is the perpendicular

we know that

(hyp)²= (perp)²+(base)²

10²=6²+(perp)²

100-36=(perp)²

64=(perp)²

8= perp

therefore the distance between the chord AC and the centre of the circle is 8cm

Answer 2

Answer:

8cm

Step-by-step explanation:

OA = radius and AB = diameter = 20cm, AC = 12cm.

2OA = AB.

Therefore, OA = 10cm

Using similar triangles,

$\frac{AB}{AC} = \frac{OA}{AD}$

$\frac{20}{12} =\frac{10}{AD}$

Therefore, on solving we get AD = 6cm.

Using Pythagoras Theorem on Triangle AOD,

$AD^{2} + OD^{2}= OA^{2}$

$6^{2} + OD^{2} = 10^{2}$

$OD^{2} = 64$

$OD = 8 cm$

Therefore, distance of chord AC from O id 8cm.

Hope it helps!

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