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?

Respuesta :

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

Answer:

The smallest possible length would be 8.5 cm.

Step-by-step 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,

[tex]x^2+x^2\geq (12)^2[/tex]

[tex]2x^2\geq 144[/tex]

[tex]x^2\geq 72[/tex]

[tex]\implies x\geq \sqrt{72}=8.48528137424\approx 8.5[/tex]

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