Discussion

The ratio of the area of a square to that of the square drawn on its diagonal is?

Answer: D

Let the original square have side \( a \), so its diagonal is \( a\sqrt{2} \).

The second square is built on that diagonal as its side:

\[ A_{1} = a^{2}, \qquad A_{2} = \left(a\sqrt{2}\right)^{2} = 2a^{2} \]

\[ \frac{A_{1}}{A_{2}} = \frac{a^{2}}{2a^{2}} = \frac{1}{2} \quad\Longrightarrow\quad 1 : 2 \]

The square on the diagonal is always exactly double - the diagonal is longer by \( \sqrt{2} \), and areas scale by the square of that.

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