I found a problem on my Google+ page recently (posted by Julia Soh), which I first misread. If a triangle has side lengths 5,6 and X and area X, what is the largest possible value of X? It is unusual to ask for a length to be equal to an area, of course.

We can do it in Euler Math Toolbox as follows. Have a look at the following sketch (done with my program C.a.R.)

Then it is clear what the equations of the problems should be.

>X &= sqrt((5+a)^2+b^2)

                                2          2
                          sqrt(b  + (a + 5) )

>area &= 5*b/2

                                  5 b
                                  ---
                                   2

>sol &= solve([X^2=area^2,a^2+b^2=36],[a,b])[4]

                                        11/2
              4 sqrt(42) - 4      sqrt(2     sqrt(3) sqrt(7) + 212)
         [a = --------------, b = ---------------------------------]
                    5                             5

>&X with sol | expand, %()

                        3/2    11/2
                    4 42      2     sqrt(3) sqrt(7)
               sqrt(------- + --------------------- + 53)
                      25               25

 10.2394299444

You see that I selected solution number 4 from the four solutions of the problem. I tried the other solutions, and they were smaller.

After I did all this, I had a longer look at the sketch, and noticed that the height of this triangle must be 2. I feel a bit ashamed not to have seen this immediately. For it yields the obvious solution without much work.

\(X = \sqrt{6^2-2^2}+\sqrt{5^2-2^2}\)

As always, a bit of geometry helps for a geometric problem.