In rectangle ABCD, if point P (not shown) is to be randomly selected from line segment AB, what is the probability that the length of line segment PC will be less than √45?Adding point \(P\) and line segment \(PC\) to the the figure, we get the following.
We see that we have a right triangle with sides \(BP\) and \(BC\) and hypotenuse \(PC\).
So, the length of \(PC\) will be \(√45\) when \(BP^2 + 6^2 = (√45)^2\).
In other words, the length of \(PC\) will be \(√45\) when \(BP^2 + 36 = 45\)
So, when \(PC = √45\), \(BP^2 = 9\), and \(BP = 3\).
So, for PC to be less than √45, BP must be less than 3.
Since the length of side \(AB\) is \(8\), \(3/8\) of the points on \(AB\) are less than \(3\) from \(B\).
(Note: While it's true that the point such that \(BP = 3\) cannot be included in the points on \(AB\) that are less than \(3\) from \(B\), since a point is infinitely small, the length of \(BP\) can effectively be up to \(3\), and thus \(3/8\) of the points on \(AB\) could be \(P\).)
Thus, the probability that \(PC\) is less than \(√45\) is \(3/8\).
A. 3/8
B. 3/5
C. 4/7
D. 5/8
E. 3/4The correct answer is (A).