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7)                    The distance from Capital City to Little Village is 660 miles. From Capital City to Mytown is 310 miles, from Mytown to Yourtown is 200 miles, and from Yourtown to Little Village is 150 miles. How far is it from Mytown to Little Village?

 

 

9)        Part a.                           A triangle has integer length sides. If two sides of the triangle are 16 and 21, how many possible lengths are there for the third side?

 

Part b.                                    An obtuse triangle has integer length sides. If two sides of the triangle are 16 and 21, how many possible lengths are there for the third side?

Guest Feb 13, 2018
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 #1
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7 - See the answer to this one here:

https://web2.0calc.com/questions/the-distance-from-capital-city-to-little-village-is-660-miles-from-capital-city-to-mytown-is-310-miles-from-mytown-to-yourtown-is-200-miles-and-from

Guest Feb 13, 2018
 #2
avatar+82937 
+1

 

9)        Part a.   A triangle has integer length sides. If two sides of the triangle are 16 and 21, how many possible lengths are there for the third side?

 

Either  

 

Unknown Side +  16 > 21

Unknown Side >  5

 

or

 

16 + 21  > Unknown Side

37 > Unknown Side   ⇒    Unknown Side  <  37

 

So.....the possible integer lengths range from     6 -  36  =   36 - 6 + 1   =  31 possible lengths

 

 

cool cool cool

CPhill  Feb 13, 2018
 #3
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What about part b?
 

Guest Feb 13, 2018
 #4
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I'm waiting for the WolframAlpha website to come back up....it makes the solving of these types of problems way easier with the computations......hopefully....it will be back up soon !!!

 

 

cool cool cool

CPhill  Feb 13, 2018
 #5
avatar+1720 
+1

For the second question, it is possible to utilize something known as the Pythagorean Inequality Theorem. The theorem actually has two parts, but the relevant portion for this particular problem is that if the square of a triangle's longest side length is greater than the the sum of the squares of the shorter side lengths, then the triangle is obtuse. This wording is quite verbose, I know, so I decided to create a visual aide as a reference for you! 

 

 

Let's first try and identify what the longest side length is of this particular triangle is. We know that 16 and 21 are already defined lengths, but we also have the unknown side length, which is labeled as s. I will follow the visual aide above, and I will consider the longest side length to be of length "c." In the visual aide above, the shorter side lengths are labeled "a" and "b."

 

It is easy to exclude the side with length 16 as the longest side length since we know that there exists a side with length 21, which is longer. However, beyond this, there are two possibilities. The possibility is that:

 

  • The side with length s is the longest side.
  • The side with length 21 is the longest side.

 

Of course, we do not know which one is the longest side, so we will just have to assume both cases. Let's begin with assuming that "s" is the longest side, or our "c."

 

\(c^2>a^2+b^2\)"s" will be the longest side, so plug it into "c." Plug the other two side lengths in any order.
\(s^2>21^2+16^2\)Now, solve for "s" to see what the possibilities are.
\(s^2>441+256\) 
\(s^2>697\)Take the square root of both sides. This results in an absolute value inequality. 
\(|s|>\sqrt{697}\)Of course, the absolute value splits the equation into two parts. 
\(s>\sqrt{697}\)\(s<-\sqrt{697}\)

 

Of course, let's remember that "s" is an integer side length of a triangle, so "s" must be positive. We can reject the second inequality because those values include ones less tha zero.
\(s>\sqrt{697}\) 

 

The above algebra only solved the first case where we assumed that "s" is the longest side. Now, let's assume that the side with length 21 is the indeed the longest. 

 

\(21^2>16^2+s^2\)Let's solve for "s" here by simplifying first and foremost. 
\(441>256+s^2\)Subtract 256 from both sides.
\(185>s^2\\ s^2<185\)For me, I can interpret the inequality better when the variable is placed on the left hand side of the equation. Take the square root of both sides.
\(|s|<\sqrt{185}\)Now, solve the absolute value inequality. 
\(s<\sqrt{185}\)\(s>-\sqrt{185}\)

 

Unlike the previous inequality, we can combine this into one compound inequality.
-√185 < s < √185Of course, yet again, "s" is a side length, so it should be greater than zero.
0 < s < √185 

 

Of course, let's take clean this up. We know, by the given information, that we only care about integer solutions, so let's calculate the radicals.We now have set restrictions for "s."  \(\)

 

√697 ≈ 26.40

√185 ≈ 13.60

 

We can then make the restrictions the following      

 

We are not done yet, though! We have to take into account the side length restriction that Cpill figured out in part a! He found that 5 < s < 37

 

Therefore, integer sides with lengths 6 to 13 or sides with lengths 27 to 36 are possible. We can find the total number of integer solutions by figuring out how many integers are in this range. 

 

From [6,13], there are 8 integers

From [27,36], there are 10 integers

 

Altogether, there are 18 integer solutions. 

 

*Unfortunately, LaTeX appears to be funky when compound inequalities are introduced, so I have resorted to a different method to portray them.

TheXSquaredFactor  Feb 13, 2018
edited by TheXSquaredFactor  Feb 13, 2018
edited by TheXSquaredFactor  Feb 13, 2018
 #6
avatar+82937 
0

Very nice, X2  !!!!

 

Yours  (and hectictar's )  visual presentations are always well done   (as opposed to being medium-rare......!!!  )

 

 

cool cool cool

CPhill  Feb 14, 2018

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