Two chords |AB| = 10cm and|CD| = 8cm are of a circle of radius 12cm find the distance of each of the chords from the center​

Let O be the center of the circle, and let E and F be the midpoints of chords AB and CD, respectively.

Since the perpendicular bisectors of the chords pass through the center of the circle, we can find the distance of each chord from the center by finding the distance between O and the midpoint of the chord.

First, we need to find the length of the perpendicular bisectors of the chords. To do this, we can use the Pythagorean theorem:

For chord AB:

Let x be the distance from O to the midpoint E of AB. Then, we have a right triangle OEB with hypotenuse OE = 12 cm and one leg EB = 5 cm (half of AB). Using the Pythagorean theorem, we can find the other leg:

x^2 + EB^2 = OE^2

x^2 + 5^2 = 12^2

x^2 = 144 - 25

x^2 = 119

x = sqrt(119) ≈ 10.91 cm

Therefore, the distance of chord AB from the center is approximately 10.91 cm.

For chord CD:

Let y be the distance from O to the midpoint F of CD. Then, we have a right triangle OFD with hypotenuse OF = 12 cm and one leg FD = 4 cm (half of CD). Using the Pythagorean theorem, we can find the other leg:

y^2 + FD^2 = OF^2

y^2 + 4^2 = 12^2

y^2 = 144 - 16

y^2 = 128

y = sqrt(128) ≈ 11.31 cm

Therefore, the distance of chord CD from the center is approximately 11.31 cm.