<p>(Remove spaces)</p>
<p><a href="http://i.imgur">http://i.imgur</a>. com/MNf739c.png
Answer: B</p>
<p><a href="http://i.imgur">http://i.imgur</a>. com/lcUhnC8.png
Answer: A
For future reference, how would you go about solving problems like these? It's very hard to imagine questions like that in my head, involving folding boxes along lines and whatnot.</p>
<p><a href="http://i.imgur">http://i.imgur</a>. com/NTT7iUw.png
Answer: D</p>
<p>problem no.1 and 3 share the same type of trianlges, the ones that have the side ratio of 1:2:sqrt(5) and the angles of 30,60 and 90. </p>
<h1>1 is because QP is twice as large as TP (here I’m assuming the cube is regular) so its the triangle with the angle measurements of 30 60 and 90 so angle QPT is 60 degrees.</h1>
<h1>2 is pretty obvious, as you can try folding those shapes in your head, and if you find it hard to do so, you could try playing around with paper pieces to construct a cube youself (it actually helps)</h1>
<p>and in #3, angle ABC is similar to angle DEF meaning both of them are 30 degrees, leaving angles BAC and EDF to be 60 degrees. As mentioned above, this kind of triangle has side ratio of 1:2:sqrt(5) so side DF is half the length of DE, which is 6, option D</p>
<p>1) Key words: midpoint, cube, square</p>
<p>From those three words we can know TP=2(QP)</p>
<p>Special triangle (memorize these): <a href=“http://upload.wikimedia.org/wikipedia/commons/thumb/6/68/30-60-90.svg/150px-30-60-90.svg.png”>http://upload.wikimedia.org/wikipedia/commons/thumb/6/68/30-60-90.svg/150px-30-60-90.svg.png</a></p>
<p>Let’s assign some variables…TP=A, QP=2(A), TQ= SQRT(3)(A)</p>
<p>Any triangle with that side ratio is a 30-60-90.</p>
<p>If one recognizes the side ratio it becomes pretty logical (even if you forget the corresponding degrees). If you forget the degrees, I believe? you can solve this with trigonometry…I’m not sure though because I slept through class. </p>
<p>If you don’t memorize the special triangles at all you should still be able to eliminate all but A/B based on knowledge of how angle degrees work. I’m pretty sure all figures are drawn to scale unless stated otherwise (you might want to look that up), so you could possibly even eliminate all but B just by looking at it and having knowledge of how angles work.</p>
<p>2) You could try redrawing the 3d structure on a piece of paper as you solve, eliminating answers as you go? sorry don’t know how to help you with this.</p>
<p>3) This is the same special triangle as the one in problem 1. 180-90=90. X+Y (the unlabeled angle)=90. X=30. 90-30=60. It’s a 30-60-90 triangle. Thus the hypotenuse=2a. Short leg=a. Long leg=SQRT(3)(a). DF=short leg. DF=A
Hypotenuse=DE=12=2a. 12/2=6</p>
<p>OOPS LOL. Haven’t practiced math section in a while. I forgot the 2 common special triangles are listed in reference materials. Definitely have them on insta-recall though to speed everything up.</p>
<p>First one, you know the sides of the square are the same as the sides of the cube. So line segment PT is half of QP. PT is the adjacent side to the angle theta (the one being asked about) and QP is the hypotenuse of triangle QPT. Cosine of theta is adj/hyp which in this case is (1/2)/1 = 1/2. The angle theta that makes cosine = 1/2 is 60 degrees.</p>
<p>Third one. You know that in a 30-60-90 triangle, the side opposite the 30 degree angle is half of the hypotenuse. Therefore if angle DEF is 30, its opposite side, DF, will be 12/2 = 6.</p>
<p>I can’t really offer much help on the second one besides telling you to try to visualize it. You don’t really don’t have to visualize the whole thing, though, since some of them can be eliminated after the first fold or so.</p>
<p>Thanks for the responses everyone - I realized my stupid mistake in the 3rd question (I looked at the 30-60-90 diagram wrong) </p>
<p>Thanks for the responses everyone - I realized my stupid mistake in the 3rd question (I looked at the 30-60-90 diagram wrong) </p>
