<p>The Cube shown above has edges of length 2, and A and B are midpoints of two of the edges. What is the length of AB.</p>
<p>I can't explain it properly but basically Point A is on the bottom edge of the front cube face and Point B is on the face that is perpendicular to the front face and is on the right edge. Anyways if you have the College Board SAT second edition 2009? 2010? it is Practice Test 7, Section 9, Question 15.</p>
<p>Thanks.</p>
<p><a href=“http://talk.collegeconfidential.com/sat-preparation/969566-sat-help-freakout.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/969566-sat-help-freakout.html</a></p>
<p>Quoted: To solve, you do the Pythagorean theorem twice. The first to solve for the hypotenuse of length from the top corner of point B to the base of A. 2^2 + 1^1 = 5. So the hypotenuse of the the first right triangle is root 5. Root 5 is now one of the legs of the new triangle. Plug it in again. 5 (the square and root cancel) + 1^1 = 6. Take the root of that so root 6.</p>
<p>quick questions=quick mistakes</p>
<p>Thank you! I knew you needed to do Pythagoras’ Theorem for that question but I did not where to draw the right angle. I was drawing it so from Point A to Point B was the hypotenuse but it actually is from the top of the line where Point B is situated to Point A or have I missed something??</p>
<p>And thanks for the link xiggi! I also struggle with functions. How was that answer 3/2??? I get you have to plug in a-2 for x.</p>
<p>Edit: Thanks for the explanation but I still do not understand.</p>