<p>Does anyone remember what the choices were for the math question with the three circle and the trapezoid. (finding the perimeter) I think I chose 20 but what were the other choices. (Anyone remember 16+2 root 3.</p>
<p>Anybody think the answer was 20.</p>
<p>This was the only question I had trouble with on the math section of the SAT I 10/4.</p>
<p>The trapezoid with three circles. I understand the methodology of how the answer is 20 but if you add up the base and the two sides you get 16 and then you can divide the trapezoid into a 30-60-90 triangle, which makes the base 2 and the side 2root3. A square is formed having side 2 root 3. Therefore the perimeter is 16+2 root 3 (~19.46)</p>
<p>answer was 20</p>
<p>But i dont think you can assume it is a 30 60 90 triangle since it wasnt specified</p>
<p>What was the answer to the trapezoid problem with the three circles?</p>
<p>What letter choice was 20? Sorry, you probably don't remember but try to.</p>
<p>The answer was 20. I think it was choice a or b.</p>
<p>I get where you are coming by. I just can't recall where the third circle was nbafan135. 20 makes sense but it seems to be too obvious. </p>
<p>I drew a perpendicular bisector from the top of the trapezoid to the base of the trapezoid. This formed a right triangle. Next the two radii willl form an equilateral triangle, 60, 60, 60. Therefore the 30-60-90 forms due to the perpindicular bisector. 30-60-90 would make a 2, 2root3 and 4 triangle (leg, leg, and hypotenuse respectively). Ok with that said, the leg 2 root 3 is part of the rectangle (in between the two triangles). Rectangles form a 45-45-90 triangle when a diagnol is employed. Therefore the top portion would be 2 root 3 and the final answer would be 4+4+4+4+2 root 3 which equal 16+2root 3 (~19.46)</p>
<p>I drew a perpendicular bisector from the top of the trapezoid to the base of the trapezoid. This formed a right triangle. Next the two radii willl form an equilateral triangle, 60, 60, 60. Therefore the 30-60-90 forms due to the perpindicular bisector. 30-60-90 would make a 2, 2root3 and 4 triangle (leg, leg, and hypotenuse respectively). Ok with that said, the leg 2 root 3 is part of the rectangle (in between the two triangles). Rectangles form a 45-45-90 triangle when a diagnol is employed. Therefore the top portion would be 2 root 3 and the final answer would be 4+4+4+4+2 root 3 which equal 16+2root 3 (~19.46)</p>
<p>^was that D or E</p>
<p>the choices went something like this:
A. 16
B. 20
C. 24
D. 16 + somethingroot
E. 16 + anothersomethingroot</p>
<p>I get where you are coming by. I just can't recall where the third circle was nbafan135. 20 makes sense but it seems to be too obvious. </p>
<p>I drew a perpendicular bisector from the top of the trapezoid to the base of the trapezoid. This formed a right triangle. Next the two radii willl form an equilateral triangle, 60, 60, 60. Therefore the 30-60-90 forms due to the perpindicular bisector. 30-60-90 would make a 2, 2root3 and 4 triangle (leg, leg, and hypotenuse respectively). Ok with that said, the leg 2 root 3 is part of the rectangle (in between the two triangles). Rectangles form a 45-45-90 triangle when a diagnol is employed. Therefore the top portion would be 2 root 3 and the final answer would be 4+4+4+4+2 root 3 which equal 16+2root 3 (~19.46)</p>
<p>what letter did you get? was it D or E?</p>
<p>Carborn2, where on the trapezoid did you draw the perpendicular bisector from? One of the top 2 vertices?</p>
<p>16+ 2√3 I got for the answer. Does my answer make sense.</p>
<p>was that answer letter D or E? Because I put E down.</p>
<p>nah, u're wrong, just admit it
i looked at the problem a bunch of times and the answer was crystal clear 20</p>
<p>^I'm not saying 20 is wrong or anything, but could you explain the problem? How did you get 5 as the length of the chord?</p>
<p>We can all agree that the base of the trapezoid and the two sides of the trapezoid add up to 16. However, I drew a perpendicular bisector from the top left point of the trapezoid to its base. The two radii form a equilateral triangle. The bisector divides the top left angle into 30-30. So a 30-60-90 triangle can be formed on both sides of the trapezoid. Apply the properties of the 30-60-90. Voila you get 2√3. Next in the middle is a rectangle (remember this subtle fact: rectangles are squares). So a 45-45-90 can be formed by drawing a diagnol. This results in the top of the trapezoid being 2√3. Add it to the rest and you get a perimeter for the trapezoid of 16+ 2√3. Tell me how you got 20. Its seems that 20 is the consensus. Maybe I overthought the question.</p>
<p>Ok i'll explain the 20 answer, to settle all disputes.</p>
<p>Draw the three circles and the trapezoid. Now, from the top two vertices of the trapezoid, draw an angled line that extends to the middle of the bottom part, splitting the bottom base into 4-4 (it is 8 fully). So, if you have those two lines drawn you will clearly see 3 triangles. The triangles on the outsides are 2 sides of length 4, and one "unknown". however, that last side is also 4, since it is also a radius. So now you hvae the middle triangle which has 2 side lengths of 4 also. To find the top, you have to notice that if the two triangles ont he sides are 60-60-60 in angle, the middle triangle's middle angle (one at the bottom) will be 60 since 60+60+x=180, x=60. So if that last angle is 60, and you have two side lengths of 4 meaning those two other angles are the same, 60+2x=180, x=60. Equilateral again. So, the last side length is 4.</p>
<p>If you don't understand something just ask. You can do this w/o drawing the cirlces, just draw the trapezoid if that makes things easier</p>