Exercise 36 Page 18

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. AC+CD=AD 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. 12/2=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. BC+CD=BD ⇒ 6+3=9 We have now calculated the lengths of all of the segments.

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.

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. 4/6=4÷2/6÷2=2/3 The probability of choosing a segment with a length greater than 3 is 23.