The longest side of an acute isosceles triangle is 8 centimeters . rounded to the nearest tenth, what is the smallest possible length of one of the two congruent sides?

The smallest possible length of congruent sides of an acute isosceles triangle must be  <  than the length of congruent sides of the right isosceles triangle

so 
The sides of a right isosceles triangle are in proportion: √2/2:√2/2:1

if  the longest side of an acute isosceles triangle is > 8 centimeters 
then
the smallest possible length of one of the two congruent sides
is 8*√2/2- >4√2=5.66=5.6
5.6<5.66 is ok

we know that the two smallest sides of a triangle have to add up to be longer than the longest side.

2x>8 > x>4
therefore the solution is the interval (4,5.6)
the smallest possible length rounded to the nearest tenth is 4.1

the answer is 4.1