Question

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? 4.1 cm 4.0 cm 5.6 cm 5.7 cm

Explanation:

Step1: Define variables

Let the congruent sides = $x$, longest side = $8$ cm.

Step2: Apply acute triangle rule

For acute triangles, the square of the longest side must be less than the sum of the squares of the other two sides:
$$8^2 < x^2 + x^2$$

Step3: Simplify the inequality

$$64 < 2x^2$$
$$x^2 > 32$$

Step4: Solve for $x$

$$x > \sqrt{32} = 4\sqrt{2} \approx 5.6568$$

Step5: Round to nearest tenth

$5.6568$ rounded to the nearest tenth is $5.7$

Answer:

5.7 cm