Question

c is the centre of circle whose radius is 10 cm find the distance of a chord if length of chord is 12cm​

Answer 1

$\: \: \: \: \: \: \: \: {AC}^{2} = {AB}^{2} + {BC}^{2} \\ = > AC = \sqrt{ {(12)}^{2} + {(10)}^{2} } \\ = > AC = \sqrt{144 +100} \\ = > AB = \sqrt{244 } \\ = > AC =2 \sqrt{61} \: cm$

Answer 2

Answer:

8 cm

Step-by-step explanation:

The radius divides the chord into two equal parts.

Let the length of the chord be AB.

Let point P be the point where the radius cuts the chord.

The lengths:

AP = BP

We divide this by 2 to get :

12/2 = 6 cm

The radius to the point where the chord cuts the circle gives the hypotenuse of the right angled formed.

Since the radius is 10 cm we have a right angled triangle whose sides are as follows :

a = h

b = 6

c = 10 cm

By Pythagoras theorem we can get the length a = h

We apply this as follows :

10^2 - 6^2 = 64

h = Square root of 64 = 8

So the distance of the chord from the center is 8 cm.