Points A, B, C, and D are located on segment AB such that AB = 3AD = 6BC. If a point is selected at random on segment AB, what is the probability that it is between C and D? Express your answer as a common fraction.
+9489 We are given that AB = 3AD = 6BC so...
3AD = 6BC
Divide both sides of the equation by 3 .
AD = 2BC
And we can see that....
CD = AB - AD - BC
Substitute 6BC in for AB and substitute 2BC in for AD .
CD = 6BC - 2BC - BC
Combine like terms. ( 6x - 2x - x = 3x so 6BC - 2BC - BC = 3BC )
CD = 3BC
probability that a random point lies on CD = CDAB
Substitute 3BC in for CD and 6BC in for AB .
probability that a random point lies on CD = 3BC6BC
Reduce fraction by the factor BC .
probability that a random point lies on CD = 36
Reduce fraction by 3 .
probability that a random point lies on CD = 12
+9489 We are given that AB = 3AD = 6BC so...
3AD = 6BC
Divide both sides of the equation by 3 .
AD = 2BC
And we can see that....
CD = AB - AD - BC
Substitute 6BC in for AB and substitute 2BC in for AD .
CD = 6BC - 2BC - BC
Combine like terms. ( 6x - 2x - x = 3x so 6BC - 2BC - BC = 3BC )
CD = 3BC
probability that a random point lies on CD = CDAB
Substitute 3BC in for CD and 6BC in for AB .
probability that a random point lies on CD = 3BC6BC
Reduce fraction by the factor BC .
probability that a random point lies on CD = 36
Reduce fraction by 3 .
probability that a random point lies on CD = 12
