Question

Can someone please help me ??? :)

Answer 1

Explanation:

AD=12

so as C mid point of AD , then AC=CD=12/2=6

as B is mid point of AC, then AB=BC=6/2=3

BD=BC+CD=3+6=9

AD=12 .

The probability of getting a random segment greater than 3 is :

Total number of segments greater than 3 is 4 which are AC,CD,AD and BD.

the total number of segments are 6.

Probablity=number of segments greater than 3 / total number of segments =4/6=2/3.

3 out of 5 segments are greater than 3 unit.

Answer 2

Answers:

• AB = 3
• BC = 3
• BD = 9
• AC = 6
• CD = 6
• AD = 12
• The probability is 2/3

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Explanation

Since AC and CD combine to form AD, and because AC = CD, we know that C is the midpoint of AD.

This would mean AD = 12 splits in half to get AC = 6 and CD = 6.

Put another way:

AC + CD = AD

6 + 6 = 12

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Since AC = 6 and AB = BC, we can then split things further in half to get AB = 3 and BC = 3 as well. Use the same idea as the previous section. Point B is the midpoint of AC.

From there we have the following steps

BD = BC + CD

BD = 3 + 6

BD = 9

For more information, search out "midpoint" and also "segment addition postulate".

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The last part of the question asks if we selected a segment at random, what is the probability it is larger than 3 units.

Here are all the segment lengths

• AB = 3
• BC = 3
• BD = 9
• AC = 6
• CD = 6
• AD = 12

There are 6 segments listed above. Of those 6, there are 4 of them that are greater than 3 units long (BD, AC, CD and AD)

The probability we're after is therefore 4/6 = 2/3

As extra info, 2/3 = 0.667 = 66.7% approximately