Exercise 36 Page 394

We have been given two tasks. First, we should find the length of all the segments. Then, we can calculate the probability that one of them is longer than $3.$

Find the Lengths

To find all of the lengths, we can use the Segment Addition Postulate. Let's start with segments $AC$ and $CD.$ Together, they form the segment $AD.$ With the Segment Addition Postulate, we can write the following equation. From the exercise, we know that $AC=CD.$ Since $AC$ and $CD$ are equal, they must be half as long as $AD.$ We are told that the length of $AD$ is $12.$ Therefore, the lengths of $AC$ and $CD$ are both $6.$ Now that we know the length of $AC,$ we can calculate the lengths of $AB$ and $BC.$ We will use the Segment Addition Postulate and the fact that $AC$ is $6.$ Next, let's use the fact that $AB=BC.$ Therefore, we can replace $BC$ with $AB$ and solve the equation for $AB.$ We have one length left, $BD.$ It will be the sum of $BC$ and $CD.$ We have now calculated the lengths of all of the segments.

Segment	Length
$AB$	$3$
$BC$	$3$
$AC$	$6$
$CD$	$6$
$BD$	$9$

Find the Probability

Previously, we calculated the lengths of the segments and, from the exercise, we know that $AD=12.$ To find the probability that a segment is greater than $3,$ we need to identify the lengths that are greater than $3.$

Segment	Length
$AB$	$3$
$BC$	$3$
$AC$	$6$
$CD$	$6$
$BD$	$9$
$AD$	$12$

There are four segments with a length greater than $3.$ To calculate the probability, we divide the favorable outcomes, $4,$ by the possible outcomes, $6.$ The probability of choosing a segment with a length greater than $3$ is $\frac{2}{3}.$