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?

Answer 1

12/sqrt (2)=6/sqrt (2), 8.484cm. Round it up to 8.5cm

Answer 2

The smallest possible length would be 8.5 cm.

Explanation:

Since, an acute isosceles triangle having two congruent sides with three acute interior angles,

Given,

The longest side of an acute isosceles triangle is 12 centimeters,

Let x be the side length of the each of the congruent sides,

So, by the property of acute isosceles triangle,

`x^2+x^2\geq (12)^2`

`2x^2\geq 144`

`x^2\geq 72`

`\implies x\geq √(72)=8.48528137424\approx 8.5`

Hence, the smallest possible length of one of the two congruent sides is 8.5 cm