The longest side of an acute isosceles triangle is 12 centimeters, Rounded to the nearest tenth, what is the smallest possible length of one of the two congruent sides?
O 6.0 cm
O 6.1cm
O 8,4 cm
O 8,5 cm

38+105= 143, 39+104=143, and you will find then adding the numbers on opposite sides, moving inwards yields the answer 138 every time. because of this, you’re going to multiply 138 times the number of pairs, which is 34. therefore, 34*138= 4692(i’m sorry if it’s the wrong numbers, but it’s the right concept)