Two secants are drawn from one point of the circle. Their inner segments are equal to 9 and 40 cm

Two secants are drawn from one point of the circle. Their inner segments are equal to 9 and 40 cm, and the outer ones are, respectively, as 4: 3 to determine the length of the outer segments.

Since the secants AC and AE are drawn from one point, the product of one secant by its outer part is equal to the product of the other secant to its outer part.

AC * AB = AE * AD.

Let the length of the segment AD = 3 * X, then AB = 4 * X.

Then AC = (9 + 4 * X), AE = (40 + 3 * X).

(9 + 4 * X) * 4 * X = (40 + 3 * X) * 3 * X.

36 + 16 * X = 120 + 9 * X.

7 * X = 84.

X = 12, then AD = 3 * 12 = 36 cm, AB = 4 * 12 = 48 cm.

Answer: The lengths of the outer segments are 36 cm and 48 cm.