4/28/10

Test 3 Section 5 - #5 (page 526)

To start, just look at triangle 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 22(√2)2, or (4)(2) = 8. Answer choice (A).

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