Two circles are having radii 9 cm and 12 cm. The distance between their centres is 15 cm. What is the length (in cm) of their common chord?
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6.8
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13.6
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7.2
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14.4
The radii are 9 and 12, and the distance between centers is 15. Since 9^2 + 12^2 = 81 + 144 = 225 = 15^2, the triangle formed by the centers and the intersection point is a right triangle. The common chord is twice the altitude to the hypotenuse. Altitude = (9 * 12) / 15 = 7.2. Chord = 2 * 7.2 = 14.4.
Let the centers of the two circles be O1 and O2, and let the common chord intersect the line segment O1O2 at point P. The line joining the centers is the perpendicular bisector of the common chord, so we can use the Pythagorean theorem on the right triangles formed at P. Setting the half-length of the common chord, let's call it x, equal in both right triangles gives 9 squared minus x1 squared equals 12 squared minus x2 squared, where x1 plus x2 equals 15; this yields the distances from the centers to the chord as 9 cm and 6 cm. Finding the half-chord length using the radius 9 gives x equals the square root of (9 squared minus 9 squared) which is not valid, so correcting the distance x1 to 9 squared minus 81 is zero means using x1 equals 9 for the first circle gives a half-length of zero, which indicates a recalculation where x1 is actually 81/15 or 5.4; thus, using the larger circle gives the half-chord length as the square root of (12 squared minus 9 squared) which is 7.2. The full length of the common chord is twice this half-length, resulting in 14.4 cm.