Question 408568
Either using Heron's formula or drawing an altitude onto the base of length 2 will work (since the triangle is isosceles and we can use Pythagorean theorem). I'll present the first solution since it works for any triangle of given side lengths.


Applying Heron's formula, we have semiperimeter = (5 + 5 + 2)/2 = 6. Thus, the area of the triangle is


{{{sqrt((6)(6-5)(6-5)(6-2)) = sqrt(24) = 4.899}}} sq meters.