<p>If you've got the Collegeboard blue book, its Practice test 7 section 5 q20 and section 8 q15.....i'm going to attempt to draw the figure for question 20 section 5 but the one in section 8 is just too complicated. So if you've got the book and got the answer right or if you've subscribed to the online course and have access to the explanations, could you PLEASE help me out????!?!?</p>
<p>hi i have the answer and explanation for number 20 for practice test 7 section 5...</p>
<p>The problem states that the five line segments in the figure are all congruent, but it does not give you a value, so you can pick one. i made the lengths equal to 1. once you draw diagonal AC, you have made four equilateral triangles. the two diagonals make a 90 degree corner in the triangles. since line BD is equal to 1, and you halved it, the line made from the center of the shape to D is equal to .5 </p>
<p>since you have the length of the hypotenuse DC (1) and the length of one side (.5) you can find the length of the line made from the center of the shape to point C using the pythag. theorum..</p>
<p>.5^2 + x^2 = 1</p>
<p>so the other side = .86</p>
<p>double that to get 1.72, the length of AC</p>
<p>since they are asking for the ratio of the length of AC to BD, the answer is 1.72 to 1, or the square root of 3 to 1, letter choice B.</p>
<p>thanks very much!</p>
<p>Q. 796 / 20.</p>
<p>AC intersects BD in point O.
In rhombus diagonals bisect the angles and are perpendicular.
The triangle AOD is 30-60-90deg (ABD is 60-60-60),
AO=root(3) x OD, so</p>
<h1>AC/BD = AO/OD = root(3) = root(3)/1.</h1>
<h1>Look up Q. 807 / 15 in Updated Consolidated List of Blue Book Math Solutions.</h1>
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<p>SAT Preparation forum is a better place to ask these questions.</p>