<p>I have two math questions from the SAT 2007-2008 Practice Exam.
Here are the questions:</p>
<p>ImageShack</a> - Image Hosting :: mathqshm7.png</p>
<p>According to the answer key, choice D is correct for #14. How is this possible? Isn't the perimeter of the entire cirlce 12pi? So how can half the circle be 12pi?</p>
<p>According to the answer key, choice E is correct for #16. I thought that drawing line CP would create two 30-60-90 triangles.</p>
<p>The first one is easier than you think. It was stated that the radii of the smaller circles were three each. Automatically, their diameters are 6, which is the same number that is equal to the radius of the large circle. The question asks for the diameter of the shaded area, but that involves several things. Firstly, the circumferences of the smaller circles are each 6pi, and the circumference of the large circle 12pi. Half the circumference of the large circle is 6pi, which would cover a part of the shaded area. Then, half of the circumferences of the smaller circles are 3pi each. The two circumference terms added together would cover the rest of the shaded area. 6pi+3pi+3pi=12pi. If I had an image, I would show it better, but I may have to leave soon.</p>
<p>As for the second question, I thought the answer was two. Apparently, I was wrong. Also, can you mentioned where you obtained the answers?</p>
<p>The answer is (A). The leg cannot be greater than either hypotenuse, so 2 is the only one that works. You must've misread the answer key because I'm 100% sure that's correct. You can't assume it'll make two 30-60-90 triangles, but you can assume you've made right triangles. Be careful with your assumptions.</p>
<p>yea i just took this exam yesterday and 2 is the correct answer. Arachnotron gave the correct explanation. </p>
<p>(I got #14 wrong too haha)</p>
<p>Well, for the second question there, one doesn't even need to know that AC=6. That particular length is never used. However, one does need to know that BC is a hypotenuse, which is always the numerically highest side on a right triangle, so the only good answer should be 2. The link below is a better image for solving the problem.</p>
<p><a href="http://i42.tinypic.com/2eckdhu.jpg%5B/url%5D">http://i42.tinypic.com/2eckdhu.jpg</a></p>
<p>Whoops, choice A is correct for #16. Thanks for the explanations. But for #14, the perimeter of half the big circle is 6pi. Isn't the shaded region just half the big circle (it's missing half a small circle on the left and adds half a small circle on the right... 6pi -3pi + 3pi)?</p>
<p>Yes but you are looking for perimeter, so you can't just add and subtract as if they were areas. You have to do 6pi + 3pi (for one small semicircle) and + 3pi (for the other small semicircle).</p>
<h1>14</h1>
<p>The perimeter of the shaded region is just half the circumference of the smaller circle + half of the circumference of the other smaller circle + half the circumference of the large circle. Or, 3pi + 3pi + 6pi, = 12pi.</p>
<h1>16</h1>
<p>Point P forms right triangle BPC. As everyone has already said, the legs of a right triangle are always less than the hypotenuse (in this case, BC = 3). Only one choice is less than 3, which is A) 2.</p>
<p>thanks everyone, got it.</p>
<p>OMG-- It's a double arbelos</p>
<p>wait for #14 you do NOT? need to add the perimeter of the dotted line (of the shaded smaller circle) ? why nott</p>
<p>The dotted line part of the shaded smaller circle is not part of the perimeter. It is inside the shape, perimeter is only the outside.</p>
<p>You need to redraw the second problem to see it better to get the answer.</p>
