Question

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?

6.0 cm
6.1 cm
8.4 cm
8.5 cm

Answer 1

Let the congruent sides have length x and the base (the longest side) be 12. For an acute triangle, the square of the longest side must be less than the sum of the squares of the other two sides:
144 < x^2 + x^2 = 2x^2
x^2 > 72 => x > √72 = 6√2 ≈ 8.4853

Rounded to the nearest tenth this is 8.5 cm.