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# Problem: Need Help plz give a good explanation

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Problem:

In quadrilateral ABCD, we have AB=3,BC=6,CD=4 and DA=4.

If the length of diagonal AC is an integer, what are all the possible values for AC? Explain your answer in complete sentences.

Hint(s):

Draw a picture!

Apply the triangle inequality. There is more than one triangle to think about.

Guest Aug 8, 2017
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#1
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how did you post that priority question

GirlsRbetteR  Aug 9, 2017
#2
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Uh, do you mean trapezoid? Because given your measurements, the figure can't be a parallelogram. Either only AB and CD or BC and DA can be parallel.

Mathhemathh  Aug 9, 2017
#3
+214
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I did stuff, and here's what I got:

First of all, apparently you can't even make a trapezoid with your measurements; only two sides are congruent, and none can be parallel. Secondly, you can draw a figure like this using the 3-4-5 right triangle:

C

4  /

/  D

|\

4 |  \   5

|__\

A   3  B

Pretend there is a line connecting B and C that measures 6

Now we know the length of DB, 5. This means that DB cannot be the longest diagonal. The longest diagonal has to be longer than any of the sides. So we have to find a number that is greater than 6 BUT less than 8, because of the triangle inequality theorem (thanks) which states that the sum of any 2 sides must be greater than the 3rd side. The sum of the sides of the triangle formed is 4+4, or 8, which means it can't be greater than 8. And since its length is an integer, AC has to be 7. That's all I could think of.

Mathhemathh  Aug 9, 2017
#4
+89752
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AC must be greater than 3 and less then 8 so AC can be 4,5,6 or 7 units long

Melody  Aug 9, 2017
#5
+808
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Another way to solve this problem is to use what the hint suggests-- the triangle inequality theorem.

The triangle inequality theorem states that the sum of the lengths of two different sides of a triangle is greater than the third remaining side. Here is a picture that demonstrates this theorem. It might be easier to understand that way:

Let's apply this to the current problem. Here is a picture to reference as I solve:

Knowing the triangle inequality theorem, let's just worry about 1 triangle at a time. Let's worry about $$\triangle ABC$$. Let's use the theorem to determine the possible lengths that $$\overline{AC}$$ can be:

 $$AB+BC>AC$$ $$AB+AC>BC$$ $$AC+BC>AB$$ Replace the known values into these inequalities and solve. $$3+6>AC$$ $$3+AC>6$$ $$AC+6>3$$ $$9>AC$$ $$AC>3$$ $$AC>-3$$

With so many solutions, sometimes it easier me to sketch a graph of all the solutions and figure out where the solutions overlap. Here is a picture of the number line graph:

Where is the area of overlap? The area of overlap is from 3 to 9. Therefore, we can express the possible lengths of AC as the following compound inequality: $$3 . We aren't done yet! We now have to consider the other triangle, \(\triangle ACD$$. Again, we will have to consider and use the triangle inequality theorem again:

 $$AD+DC>AC$$ $$DC+AC>AD$$ $$AC+AD>DC$$ Yet again, plug in the known values. $$4+4>AC$$ $$4+AC>4$$ $$AC+4>4$$ Solve. $$8>AC$$ $$AC>0$$ $$AC>0$$

For this one, you could use a number line graph again, but I will just use logic to figure out the set of possible lengths AC can be in this case. In two of our inequalities, it was determined that $$AC>0$$. If $$8>AC$$, or, as I like to think of it, $$AC<8$$, this means that the compound inequality for AC in this case is \(0 . Now, let's look at the restriction we placed on AC before.

\(3

\(0

AC has to comply to both restrictions because it is in both triangles, which means that the final compound inequality is

\(3 . Therefore, the only possible integers are 4, 5, 6, and 7.

Bam! Done!

TheXSquaredFactor  Aug 9, 2017
#6
+808
+1

Uh oh! My syntax appears to be funky, so the compound inequalities did not display properly.

The \(3 should be 3 < x < 9

The \(0 should be 0 < x < 8

The final \(3 should be 3 < x < 9

TheXSquaredFactor  Aug 9, 2017
edited by TheXSquaredFactor  Aug 9, 2017
edited by TheXSquaredFactor  Aug 9, 2017
#6
+75302
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Very nice, X2   !!!!

CPhill  Aug 9, 2017

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