<p>This is from the 10 Real SATs, third edition. I couldn't find a link to this on the gathered threads so i was wondering if anyone could help me out.</p>
<p>pg 488, #24</p>
<p>Points P,Q,R lie in a plane. If distance bt P and Q is 5, and distance bt Q and R is 2, which of the following could be the distance bt P and R?
I.3
II.5
III.7</p>
<p>the answer is e), i, ii, iii</p>
<p>I can't seem how it could be choice II)5</p>
<p>your help is appreciated!</p>
<p>I think you could have an isoceles triangle with base of 2 (QR) and two legs of 5 (PQ) and (QR).</p>
<p>I think the key is to not assume (as the question does not state) that the points lie on a line - they lie in a plane, which means there can be any order between them. </p>
<p>I drew a line from Q to P with length 5 and then I drew a line skew to that line from Q to R of length 2. If you rotate this line visually (so that Q acts as the center of a circle with QR the radius) you can see how the distance between P and R could be any number between 3 and 7, inclusive.</p>
<p>Web Site Explanation:
Points P, Q, and R lie in the same plane, but not necessarily in the same line. Therefore, Q could either be between P and R, so that PR is 7, or R could be betweenP and Q, so that PR is 5. OR, P, Q, and R can be the vertices of a 3-4-5 right triangle, with PQ the hypotenuse being 5, which would make PR equal to 3. </p>
<p>I understand 3 and 7. But if R is between P and Q how is its distance 5? if P and Q's distance is 5.</p>
<p>Try drawing out what Gospy explained above. (An isoceles triangle with two sides of length 5 and one of length 2)</p>
<p>Remember, the third leg has to be equal or greater than the difference between the other two and equal to or lesser than the sum of the other two. </p>
<p>Therefore, x has to be equal to or greater than 3
and x has to to be equal to or lesser than 7</p>
<p>so it could be any of the 3...</p>