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.

Guest Jul 6, 2018

#1**+1 **

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 = \(\frac{\text{CD}}{\text{AB}}\)

Substitute 3BC in for CD and 6BC in for AB .

probability that a random point lies on CD = \(\frac{3\text{BC}}{6\text{BC}}\)

Reduce fraction by the factor BC .

probability that a random point lies on CD = \(\frac36\)

Reduce fraction by 3 .

probability that a random point lies on CD = \(\frac12\)

.hectictar Jul 7, 2018

#1**+1 **

Best Answer

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 = \(\frac{\text{CD}}{\text{AB}}\)

Substitute 3BC in for CD and 6BC in for AB .

probability that a random point lies on CD = \(\frac{3\text{BC}}{6\text{BC}}\)

Reduce fraction by the factor BC .

probability that a random point lies on CD = \(\frac36\)

Reduce fraction by 3 .

probability that a random point lies on CD = \(\frac12\)

hectictar Jul 7, 2018