The lengths of two sides of a triangle are 10 and 17. The length of the third side as well as the area of the triangle are integers. What is the sum of all possible lengths of the third side?

Answer 1

the 3rd side of a triangle is always less than the sum of the other two sides: c < a+b, c<10+17, c< 27

the 3rd side is also always greater than the difference of the other 2 sides: c>a-b or c>b-a, c>17-10, c>7

7<c<27

c = 8,9,10,11, 12, 13, 14, 15, 16, 17, 18, 19,20,21,22,23. 24,25,25,26. twenty possible integers. narrows it down a bit.

exclude 7 or 27 as that collapses the "triangle" into a straight line

but then you add the area has to be an integer

that eliminates more than just the 3rd side must be an integer.

Your instructor or textbook author may be a little sadistic. Maybe this is "just" an extra credit problem? Only way to justify it.

third side = 9

Using Heron's formula, semi-perimeter = 36/2 = 18

Area = square root of (18(18-9)(18-10)(18-17) = sqr(18x9x8x1)= sqr(18x72) = sqr1296= 36 which is an integer, as 9 is also.

9 is the 3rd side

or 3rd side could be 21 leading to an integer area

semi- perimeter= (21+10+17)/2= 48/2=24

Area = sqr[(24)(24-21)(24-10)(24-17)]=sqr(24)(3)(14)(7)=84