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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.

 

Guest Jul 6, 2018

Best Answer 

 #1
avatar+7155 
+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
avatar+7155 
+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

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