<p>your explanation makes sense, but what is the fault with mine.</p>
<p>When you draw the perpendicular bisector, you don't split the 60 degree angle into 30-30.</p>
<p>yeah you dont split it into 30-30, the 60 stays and a complement of 30 is formed and the right angle on the bottom is formed too. but the answer is twenty</p>
<p>can we agree you form a right triangle w/ the perpendicular bisector.</p>
<p>can we agree that the radii of circle one a 4 and 4.</p>
<p>draw three triangles like you said except do this within the trapezoid. ok. </p>
<p>So not only do triangles form but parallelograms. Theorems from parallelgrams tell us opposite sides are congruent. Therefore each triangle is equilateral. (reflexive property x=x, x=x)</p>
<p>Okay so an equilateral triangle has 60 degree. Perpendicular angles bisect angles. A 30-60-90 results. </p>
<p>You receive 2√3.</p>
<p>Use the 45-45-90 from the rectangle. The top is clearly 2√3. 16+2√3. I hold by answer. It seems it uses the same methodology you do, but I employ only the two circles.</p>
<p>So basically:</p>
<p>Divide into three equilateral triangles in trapezoid.
Use a perpendicular bisector to divide the 60 degree angle in the equilateral triangle into a 30-60-90</p>
<p>apply 45-45-90</p>
<p>top is 2√3</p>
<p>I used the trapezoid and the two circles it was in.</p>
<p>Well you guys are probably right, but this was the only one that was bugging me. Everything else I got.</p>
<p>Uh.. yea. You don't get a 45-45-90 from drawing a diagnol of that rectangle. I know what you're doing but you don't even need to do that to find the top. If you see your picture, and draw one perpendicular bisector, you can clearly see that it is 2root3, sure. But that measn that the hypoteneuse for that triangle is 4, which is also a part of the equilateral triangle (you even agree that it is equilateral). So, if it is 4, then the top is 4.</p>
<p>but their are two answers, isn't the question kind of ambigous</p>
<p>The third circle is above the trapezoid in the diagram.</p>
<p>People said it's 20 but yeah i got it wrong too
i put 16 + 2√3</p>
<p>I put 16 + 2√3 too. When I think about it now, it didn't say "not drawn to scale." Yeah, it was THAT simple. :\ Read too into it...man..</p>
<p>I did the EXACT same thing carbon... we should have realized that since the question was in the middle of the question block, that it would not have been that obvious. shucks.</p>
<p><a href="http://i36.tinypic.com/10nfexg.jpg%5B/url%5D">http://i36.tinypic.com/10nfexg.jpg</a></p>
<p>I'm pretty sure the angle in the bottom right was labeled as 60 degrees.</p>
<p>^I'm pretty sure it wasn't,</p>
<p>oh well - then I don't remember how I figured out you could make it 3 equilateral triangles.</p>
<p>I was conflicted between these two answers as well...At the end I just put 16+2 root 3....But After I did that, I thought it might actually be 20.</p>
<p>it is 20</p>
<p>through geometry.</p>
<p>the intersection of the two equal circles lies on the perpendicular bisector of the line joining their radii</p>
<p>I don't think you can dispute spidey, he's like our math junkie :D</p>
<p>Carbon2, here's a little flaw I think that's in your solution. How can you assume that the rectangle was a square (only squares for 45,45,90 triangle when a diagnol is employed)?
I highly doubt the rectangle was a square. In fact, if I follow your solution, I find that the base of the rectangle to be 4 and the height to be 2root3.
Of course, I might be self-defending myself since I chose 20 myself. Please correct me if i'm wrong.</p>