To solve this question, we have to:
Doing this, we get that: 6 different such triangles can be made.
Isosceles triangle:
In an isosceles triangle, two sides have the same measure. An example is given by the picture at the end of this answer.
Perimeter:
The perimeter of a polygon, in this case a triangle, is the sum of all it's sides.
Constraint:
The sum of the lengths of the two smaller sides of a triangle has to be greater than the length of the greater side.
This triangle:
Thus, the perimeter is:
[tex]P = 2x + y = 28[/tex]
Now, for the values of x, we find if the values of y generate a triangle, respecting the constraint.
x = 1
[tex]2 + y = 28 \rigtharrow y = 26[/tex]
1 + 1 < 26, thus, the constraint is not respected.
x = 2
[tex]4 + y = 28 \rightarrow y = 24[/tex]
2 + 2 < 24, thus, the constraint is not respected.
x = 3
[tex]6 + y = 28 \rightarrow y = 22[/tex]
3 + 3 < 22, thus, the constraint is not respected.
x = 4
[tex]8 + y = 28 \rightarrow y = 20[/tex]
4 + 4 < 20, thus, the constraint is not respected.
x = 5
[tex]10 + y = 28 \rightarrow y = 18[/tex]
5 + 5 < 18, thus, the constraint is not respected.
x = 6
[tex]12 + y = 28 \rightarrow y = 16[/tex]
6 + 6 < 16, thus, the constraint is not respected.
x = 7
[tex]14 + y = 28 \rightarrow y = 14[/tex]
7 + 7 = 14, thus, the constraint is not respected.
x = 8
[tex]16 + y = 28 \rightarrow y = 12[/tex]
8 + 8 > 12, thus, the constraint is respected, and this is the first possible triangle.
x = 9
[tex]18 + y = 28 \rightarrow y = 10[/tex]
9 + 9 > 10, thus, the constraint is respected, and this is the second possible triangle.
x = 10
[tex]20 + y = 28 \rightarrow y = 8[/tex]
8 + 10 > 10, the constraint is respected, and this is the third possible triangle.
x = 11
[tex]22 + y = 28 \rightarrow y = 6[/tex]
11 + 6 > 11, the constraint is respected, and this is the fourth possible triangle.
x = 12
[tex]24 + y = 28 \rightarrow y = 4[/tex]
4 + 12 > 12, the constraint is respected, and this is the fifth possible triangle.
x = 13
[tex]26 + y = 28 \rightarrow y = 2[/tex]
13 + 2 > 13, the constraint is respected, and this is the sixth possible triangle.
x = 14
[tex]28 + y = 28 \rightarrow y = 0[/tex]
Cannot have a triangle with a side of 0, so this is not possible.
After the tests, we have that 6 different such triangles can be made.
A similar question is given at https://brainly.com/question/6139098
