Question 1104581
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For one of them, the side of length 20 is the longest side.  In that case, x+10 has to be greater than 20, so the shortest possible integer value for the third side is 11.<br>
For the other, the 10 and 20 are the shortest sides; then the third side must be less than 30, so the largest possible integer value for the third side is 29.<br>
Since two of the sides in the two triangles are the same length, the difference in perimeters is the difference in the lengths of the third sides.  So the greatest possible difference in the perimeters of the two triangles is 29-11 = 18.