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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?

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12/sqrt (2)=6/sqrt (2), 8.484cm. Round it up to 8.5cm
User Snnlankrdsm
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7.8k points
4 votes

Answer:

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

User Meeech
by
8.3k points

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