Answer:
The length of the shortest side of the triangle is 10 units.
Step-by-step explanation:
Let a be the shortest side of the isosceles triangle and b be the two congruent sides.
The congruent sides b are each one unit longer than the shortest side. Hence:
[tex]b=a+1[/tex]
The perimeter of the isosceles triangle is given by:
[tex]\displaystyle P_{\Delta}=b+b+a=2b+a[/tex]
This is equivalent to the perimeter of a square whose side lengths are two units shorter than the shortest side of the triangle. Let the side length of the square be s. Hence:
[tex]s=a-2[/tex]
The perimeter of the square is:
[tex]\displaystyle P_{\text{square}}=4s=4(a-2)[/tex]
Since the two perimeters are equivalent:
[tex]2b+a=4(a-2)[/tex]
Substitute for b:
[tex]2(a+1)+a=4(a-2)[/tex]
Solve for a. Distribute:
[tex]2a+2+a=4a-8[/tex]
Simplify:
[tex]3a+2=4a-8[/tex]
Hence:
[tex]a=10[/tex]
The length of the shortest side of the triangle is 10 units.
