<p>I got this problem from the Kaplan's 12 Practice Tests for the SAT:</p>
<p>The lengths of two sides of a triangle are (x-2) and (x+2), where x>2. Which of the following ranges includes all and only the possible values of the third side y?</p>
<p>A. 0<y<x
B. 0<y<2x
C. 2<y<x
D.4<y<x
E. 4<y<2x</p>
<p>The answer is supposed to be "B", but I do not understand how that is possible when x>2 meaning y has to be greater than 4.. (3+2= 5) and (3-2= 1), therefore y HAS to be greater than the difference (5-1=4) right?</p>
<p>Thanks for the help</p>
<p>Masamune929:
You're right. "B" would be a wrong answer.</p>
<p>One way to visualize it is to imagine the (x+2) side being horizontal, and the (x-2) being attached to it on the left, free to swing either way (no pun intended). If the (x-2) side is almost at a 3 o'clock position, the third side y would have a length nearly equal to the (x+2) - (x-2) = 4. If the (x-2) side is almost at a 9 o'clock position, y would be almost equal to the sum of (x+2) and (x-2) i.e. 2x . </p>
<p>Since x > 2, 2x must be > 4. The answer is "E" .</p>
<p>Thanks for the explanation..I'm going to check for any more errors I think were made in my SAT book -..-</p>