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

Explanation:

Step1: Define variables and condition

Let the congruent sides be $x$. For an acute triangle, the square of the longest side must be less than the sum of the squares of the other two sides:
$12^2 < x^2 + x^2$

Step2: Simplify the inequality

Calculate $12^2$ and combine terms:
$144 < 2x^2$

Step3: Isolate $x^2$

Divide both sides by 2:
$x^2 > \frac{144}{2} = 72$

Step4: Solve for $x$

Take the square root of both sides:
$x > \sqrt{72} \approx 8.485$

Step5: Round to nearest tenth

Round $8.485$ to the nearest tenth:
$x \approx 8.5$

Answer:

D. 8.5 cm