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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.

Jul 6, 2018

#1
+8720
+1

We are given that   AB  =  3AD  =  6BC   so...

Divide both sides of the equation by  3 .

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  =  $$\frac12$$

.
Jul 7, 2018

#1
+8720
+1

We are given that   AB  =  3AD  =  6BC   so...

Divide both sides of the equation by  3 .

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