The diagonal of a square measures $$6\sqrt{2}$$ cm. The measure of the diagonal of a square whose area is twice that of the first square is:

Let's assume the side of the first and second squares are $$a_1\ and\ a_2$$ respecively.

The diagonal of a square measures $$6\sqrt{2}$$ cm.

$$\sqrt{2}a_1 = 6\sqrt{2}$$

$$a_1=6$$    Eq.(i)

As given in the question, the area of the second square is twice the first square.

$$2\times\ a_1\times a_1=a_2\times a_2$$

Put  Eq.(i) in the above equation.

$$6\times6\times2\ =a_2\times a_2$$

$$a_2=6\sqrt{\ 2}$$

diagonal of the second square = $$\sqrt{\ 2}a_2 = \sqrt{\ 2}\times 6\sqrt{\ 2}$$

= 12 cm