The key intermediate STATEMENT is that this triangle ABC is a right-angled triangle.


One can prove it by different ways.  I choose this one:


    The fact that the point M is equidistant from the points A, B and C means that the point M is the center of the circle 
    subscribed about the triangle ABC.

    AB is the diameter of this circle.

    Hence, the angle C is the right angle, since it leans on the diameter.


Thus we have the right angled triangle ABC.

Its hypotenuse AB has the length 5 + 5 = 10 units     (given !).

The sum of the legs is the perimeter minus hypotenuse = 24 - 10 = 14 units.


Thus we have this system of two equations for the legs x and y:

x + y = 14,          (1)
x^2 + y^2 = 10^2.    (2)


From (1) express y = 14-x and substitute it into (2).
You will get the quadratic equation for x:

x^2 + (14-x)^2 = 100,

x^2 + 196 - 28x + x^2 = 100,

2x^2 - 28x + 96 = 0,

x^2 - 14x + 48 = 0,

(x-6)*(x-8) = 0.


Thus we have this pair of solutions (x,y) = (6,8)  or  this pair  (x,y) = (8,6).


In any case, the area of the triangle ABC is  = 24 square units.