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2^(1/2)a

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FBEDAC

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F, E, A, B, C, D (probably wrong lol)

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The maximum area occurs when there is a 90 degree angle.
According to phyatogrean theorem, the length of a triangle is a^2+b^2=c^2
since a and b are equal it becomes 2x^2= c^2
therefore the length is sqrt 2 times x.

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F, E, B, C , A, D

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F, E, A, D,B, C

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0.5a^2

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sqrt(2) times a

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Since the area of an isosceles triangle is maximized when the triangle is a right triangle, we can use the Pythagorean Theorem of a^2+b^2=c^2 to solve this problem. In this case, we will define the two equal sides as having length b, and the unknown side as side a.

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F,E,A,B,C,D

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F, E, A, B, C ,D

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FEABCD

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We want a 90 degree angle, so we can first thinkl of our triangle as half of an equilateral, using height and two sides with length A we can deduce the length of the other side.
We can compute the derivative to max.
We will have a critical point -0 and eventually arrive at x = a/root2

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A = ½[√(a² − b² ⁄4) × b]

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F E C

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F E A B D C

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the length of the third side that maximizes the area of the triangle is a

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THE TRIANGLE MUST BE RIGHT ISOSCELES TO GET THE AREA

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feabcd

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f b d a e c

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the third side length should be the square root of 2*a^2.

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FABEDC