Question 1206464
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A triangle has two {{{highlight(cross(edges))}}} <U>sides</U> of length 5. What length should be chosen for the third side 
of the triangle so as to maximize the area within the triangle?
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I will give two geometric solutions here.



<pre>
It is clear that your triangle is isosceles.


Take an other copy (instance) of this triangle and attach these two triangles "base-to-base".


You will get a rhombus with the side length of 5 units.


Now ask yourself - which rhombus with the side length of 5 units has maximum area ?


Place such rhombus on its side to a horizontal line .


The rhombus is a deformable quadrilateral, so deform it to get maximum area.


It is clear that the maximum area is achieved when the rhombus is a square.


So, the maximum possible area of a rhombus is 5 x 5 = 25 square units,
and the diagonal of the rhombus is  {{{5*sqrt(2)}}} units then,

giving the optimum size to the third side of the triangle of  {{{5*sqrt(2)}}} units.    <U>ANSWER</U>
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Another possible solution is even simpler. &nbsp;&nbsp;See below.



<pre>
The area of this triangle is half the product of the two given sides by the sine of the concluded angle

    area of the triangle = {{{(1/2)*5*5*sin(alpha)}}}.


{{{sin(alpha)}}} is maximal when {{{alpha}}} is the right angle.  Then the area of the triangle is maximal

    area of the triangle = {{{(1/2)*25}}} = 12.5  square units,


and the opposite side  "c"  is, obviously,  c = {{{sqrt(5^2+5^2)}}} = {{{5*sqrt(2)}}}.
</pre>

Solved in two different ways, &nbsp;for your better understanding.