One leg of a right isosceles triangle is 8 feet. what is the area of the triangle?

"One leg of a right isosceles triangle is 8 feet. What is the area of the triangle?"

If one leg is 8 ft, so is the corresponding leg.  That leaves the third leg unknown.

If you represent the length of the unknown third leg by x, and regard it as the base of your triangle, then the area of the triangle is

A = (1/2)(base)(height)      BUT, we don't have the height.  That "8" represents the length of each of the equal legs of this isosceles triangle.
We must find the height of the triangle.  The Pyth. Thm. applies here:
the hypotenuse is 8, and one leg is (x/2).  Thus, the following is true:

(height)^2 + (x/2)^2 = 8^2, or   (height)^2 = 64 - x^2/4

Thus, the height is the positive square root of 64 - x^2/4.

We can now write an expression for the area of this triangle:

A = (1/2)(base)(height)  =  (1/2) (x) (sqrt(64-x^2/4)).

Example:  suppose that x=4.  Then x/2 = 2.

The area of this particular triangle would be   A = (1/2)(4)(sqrt(64-4), or
                                                                            = 2 sqrt(60), or
                                                                        A = 2sqrt(4)sqrt(15), or
                                                                            = 4sqrt(15)

Important:  realize that this is an example, not the actual answer.

Can you provide the length of the base of this triangle, please?