<p>I'm not "seeing" the solution.</p>
<p>I can get every side except BC. By visual estimation it looks like it's 4, which yields the correct perimeter of 20, but I can't seem to prove why BC = 4.</p>
<p>Thanks.</p>
<p>I’m not really sure either.</p>
<p>Points BCE make a triangle. BE and CE have length 4 and assuming that the triangle is equilateral, then BC would also be length 4. </p>
<p>I’m not really sure how to confirm that it’s equilateral though…</p>
<p>It doesn’t say that it isn’t drawn to scale so you can just measure and compare. :)</p>
<p>To prove the middle triangle (BEC) is equilateral . . .</p>
<p>Draw three triangles inside ABCD. AB, AE, CD, DE, BE, and EC are all radii. That makes triangles AEB and CED equilateral, so angles AEB and CED are both 60. Then angle BEC has to be 60 to make the line AD, so the middle triangle has to be equilateral (two sides equal and included angle 60).</p>
<p>Whew!</p>
<p>I believe I got this. It took me longer than I’d have liked. A lot longer!</p>
<p>Segments AB and AE are radii of circle A, so they measure 4; segments DE and DC are radii of circle D, so they measure 4; segments EB and EC are radii or circle E, so they measure 4.</p>
<p>This means triangles ABE and DCE are equilateral, since all their sides measure 4.</p>
<p>This means that the measures of angles BAE and CDE are 60 degrees.</p>
<p>Segment AD is a diameter of circle E, so its measure is 8.</p>
<p>Now, consider the quadrilateral ABCE. Segments BA, AE and EC all measure 4. Angle A measures 60 degrees. Because angle ACE forms a linear pair with angle CED, and the measure of angle CED is 60, we conclude that the measure of angle ACE is 120 degrees. Because consecutive angles of this quadrilateral are supplementary, the quadrilateral is a parallelogram. (This one happens to be a rhombus, too, but a rhombus is a parallelogram.) Opposite sides of a parallelogram are congruent, so BC = AE = 4. (You could have done all this same reasoning with quadrilateral EDCB.)</p>
<p>Now you have lengths for all four sides of the trapezoid: AD + DC + CB + BA = 8 + 4 + 4 + 4 = 20.</p>
<p>Oh, blast! Daisie’s way is a lot shorter.</p>
<p>Everybody’s answer is fine, but the simple way, DRAW A GIANT TRIANGLE! Extend the line from AB and DC. Through geometry, you’ll know that you created a 60-60-60 triangle. Now as an equilateral triangle, you’ll be able to prove that BC segment is the midpoint of the “midway” point of the triangle and therefore 1/2 the base, the base being 8.</p>
<p>Oh. I thought the point of the SAT was to find the most time-consuming answer possible. Did I miss something?</p>
<p>This really isn’t as difficult as it seems at first glance.</p>
<p>AD is obviously 8, since the radius of each circle is 4.</p>
<p>AB and CD are each 4, since they are also radii of the circles.</p>
<p>BC is the kind of tricky part, but when you think about it, not really. Since all three circles are congruent and the radii are in a perfectly straight line, B has to be exactly in the middle of AE horizontally, so from B to a line drawn straight up from point E would be 2 units. Same goes for C, so BC is 4.</p>
<p>Thus the answer is B) 20.</p>
<p>Sikorsky: lol
Andevan: the way I looked at it in the beginning, but had no idea how I actually thought it… Good job putting it in to words!</p>
<p>Yeah, I had trouble putting it into words too. The concept seemed so simple to me as soon as it “clicked” for me, but then when I started typing up how I got the solution I had to go back and look at the drawing again to figure out how to put it into words. :P</p>
<p>Hopefully the OP can make some sense of all our answers.</p>