*ACD*. Since

*ABCD*is a square,

*AD*=

*CD*, so we're looking at an isosceles right triangle (a.k.a. a 45°-45°-90° triangle).

If you don't know it by heart, take a look at the beginning of the section for the Reference Information table, which tells you that a triangle like this with legs

*s*will have a hypotenuse of

*s*√2.

Here's the tricky bit: the hypotenuse of triangle

*ACD*is just plain old 4. There's no radical to be found. All that means though, is that the legs already had a √2 in them.

Think of it this way: what do we have to do to

*s*√2 to make it be just

*s*? We divide by √2! So we do the same thing with our hypotenuse of 4: we divide it by √2. So here's what we can get from that:

*which can be rewritten as:*

The area of the square is going to be (2√2)

^{2}. That's 2

^{2}(√2)

^{2}, or

**(4)(2) = 8. Answer choice (A).**

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