document.write( "Question 1203040: Please help me with this
\n" ); document.write( "The sides of a triangle have lengths 11, 15, and k where k is an integer. For how many values of k, is
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Algebra.Com's Answer #838279 by math_helper(2461) $\"\"$ $\"About$

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document.write( "This amounts to 13 values of k. \r
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document.write( "Go to this site:
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document.write( "https://www.triangle-calculator.com/?what=sss&a=15&b=11&c=19&submit=Solve\r
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document.write( "Plug in 11,15 as two sides and then 'experiment' with other values for the 3rd side.   You will find integer values 5-10 and 19-25 all form obtuse triangles.  The site will draw the triangles so you can see where the obtuse angle is.  
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document.write( "Best wishes. 
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document.write( "Another approach is to pick a k such that you have a valid triangle.  This means the triangle inequalities must each hold for the value of k.  For this problem:  11+15>k, k+11>15, and k+15>11.   Then, for each triple (11,15,k) pick the longest side and compute the square of each side-length (121,225,k^2).  Call the longest side 'c'.  If  c^2 is greater than the sum of the other two squared values, you have an obtuse triangle.   
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document.write( "Example:
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document.write( " Say k=19  ... the triangle inequalities hold, check. 
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document.write( " The values of the squares of each side-lengths are:  (121,225,361)
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document.write( "    Since 361 > 225+121, the triangle is obtuse.
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document.write( "Another example:
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document.write( " Say k=12...  the triangle inequalities hold, check.
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document.write( " Squared side-lengths are (121,225,144)
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document.write( "      However, since 225 < 121+144, we do NOT have an obtuse triangle.
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