16)

The large triangle is isosceles. The altitude to the base bisects the base. The base is $$y$$. The altitude is $$4$$ cm. The two equal sides of the isosceles triangle are the hypotenuses of the two right-angled triangles formed by the altitude. Let $$x$$ be half of the base, so $$y = 2x$$. Using Pythagorean theorem, $$x^2 + 4^2 = ( ext{side})^2$$. This problem is unsolvable with the given information.