SOLUTION: 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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Question 987380: 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?
Found 2 solutions by Boreal, solver91311:
Answer by Boreal(15235) About Me  (Show Source):
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If the triangle were a right triangle, the sides would be 12/sqrt (2)= 6 sqrt(2), 8.484 cm. If the angle were obtuse, even a little, the two sides would be shorter, so the sides have to be longer, and we will round up to 8.5 cm.

Answer by solver91311(24713) About Me  (Show Source):
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The limiting shape for the shortest congruent sides on an acute isosceles triangle is a right isosceles triangle. So calculate the measure of the congruent sides of a right isosceles triangle with a base that measures 12 cm. Hint: The sides of a right isosceles triangle are in proportion: .

The smallest possible length of congruent sides of an acute isosceles triangle must be strictly less than the length of congruent sides of the right isosceles triangle, so once you have calculated a decimal approximation for the side of the right isosceles triangle, round your answer DOWN (because if you simply round off, you might have an answer that is just a little larger than the side of the right isosceles triangle, and therefore would be the length of the side of an obtuse isosceles triangle)

Boreal has it backwards. The congruent sides of any obtuse isosceles right triangle are shorter than the measure of the right isosceles triangle with the same base. Consider an obtuse isosceles triangle with an apex angle of 179 degrees. Such a triangle would have congruent sides that measured only slightly more than one-half the base. The congruent sides of a right isosceles triangle are (very roughly) about 3/4 the measure of the base. And an acute isosceles triangle with an apex angle that measured 1 degree would have congruent side lengths that were very nearly half of the entire perimeter, leaving a very tiny proportion for the base.

John

My calculator said it, I believe it, that settles it