<p>This is a scan of a question from the 2010 TPR book:</p>
<p><a href=“http://i584.■■■■■■■■■■■■■■■/albums/ss284/sear_squid/SAT/Picture2-1.png[/url]”>http://i584.■■■■■■■■■■■■■■■/albums/ss284/sear_squid/SAT/Picture2-1.png</a></p>
<p>I’m not sure how to go about solving this problem. The shaded region includes the four “pea pod” looking shapes. Do circle rules still apply here?</p>
<p>Apologies if this is super confusing. It’d be easier to draw this out, but I can’t, so…</p>
<p>The radius of the semicircles is 10. Split the semicircle containing points A and D down the middle (so from the center of the diagram, let’s call it point E, to the midpt of AD). Now connect points D and E to form a 45-45-90 right triangle. So, DE, the base of the triangle, equals 10sqrt2, and the altitude is 5sqrt2. So the area of the triangle is 50. </p>
<p>Next, the sector containing the triangle is just (1/4)((10^2)pi), or just 25pi. Subtracting the area of the triangle from the area of the sector will give you half of a “pea pod”. 25pi - 50.</p>
<p>So in order to get the area of all four pea pods, multiply this number by 8. So: 8(25pi - 50). Factor out the 25 and get 8*25(pi - 2), or 200(pi - 2), which is choice C.</p>
<p>Yes you can use whatever circle rules you wish.</p>
<p>One way to do it is let the area of the “pea pod” region be p and the area of the other four unshaded regions be q, so that the area of the square is 4p + 4q. We have</p>
<p>4p + 4q = 400 → p + q = 100</p>
<p>The area of one of the semi-circles is 2p + q = 50pi. We have a system of two equations:</p>
<p>p + q = 100
2p + q = 50pi</p>
<p>Subtracting the 1st equation from the second we get p = 50pi - 100 = 50(pi - 2). Multiply by 4 to find the area of the entire shaded region, 200(pi-2), C.</p>
<p>You could try integrals, dividing this into 4 sections and integrating, then using trigonometric substitution.</p>
<p>…or you could do it an easier way.
Remove 2 opposing unshaded regions and you see you have half of a circle on each side which is basically equal to one circle. That circle’s area is 100pi since its radius is half the square length. The area of the giant square is 400. So 2 of the unshaded regions is 400-100pi, meaning 4 of the unshaded regions is 800-200pi. So the shaded region is</p>
<p>(big square - the 4 unshaded regions)
= 400-(800-200pi) = 200pi-400 = 200(pi-2) = C</p>
<p>approximate pi by 3. Area of rectangle is 400. The area of the clover is somewhere around half that of the rectangle ~200. </p>
<p>e) this is larger than area of rectangle → can’t be correct
d) this is ~3/4 times the of the area of rectangle → too large, so incorrect
a) this is ~ 3/20 times the area of the rectangle → too small, so incorrect
b) this is ~ 1/10 times the area of the rectangle → too small, so incorrect</p>
<p>This leaves c, which is about 1/2 of the area, so its the right answer. The only answer besides c which is not obviously wrong is d…</p>
<p>^ Eh, I’m not sure if I’d use approximations like that. It might work on the SAT, but not elsewhere (or where two of the answer choices are really close together).</p>
<p>Here is a really simple way, I broke the square into 4 equal squares, each with 1 clover leaf. I knew that since the area of the whole square is 400, 1/4 of that is 100. I knew the area of the whole circle is 100 pi so 1/4 of that is 25 pi. I took 100 and got the equation 100-25pi. This is the missing part of the side of the clover on each of the 4 mini squares. Since there are 8 of those missing parts. I multiplied it by 8. 8(100-25pi) = 800-200pi. I took the area of the whole square which is 400 and subtracted 800-200pi getting 400-800+200pi. This equals -400+200pi. If you distribute 200 out you will get C. It is much easier to explain this on paper.</p>
<p>^^Kyrix1 explained it really well^^</p>
<p>@rspence, well, this is a SAT board/problem. FWIW, this type of method is suggested in many prep books (especially when you don’t know how to proceed directly).</p>
<p>This sort of thinking is actually quite useful in the real world too. For example, the best engineers not only know how to solve problems rigorously, but also how to perform sanity checks and back of the envelop design. Common sense is valuable…</p>
<p>True, it definitely helps to know how to approximate, especially for problems where you just want a rough estimate or as a check. There are other problems where an exact answer is needed, for example, if you are programming a missile to destroy a target and you want exact coordinates. Or if it’s a fill-in-the-blank/grid in on an exam.</p>
<p>Every problem on the SAT can be solved for an exact answer (i.e. without using “approximations” or plugging in numbers) but I do agree, for certain problems (not all) using an approximation works as well. I personally have never used those techniques.</p>
<p>It’s funny you should say that, because The Princeton Review actually suggests that you approximate. Even though it’s fine to do this on the SAT, I much rather prefer knowing how to thoroughly solve the problem.</p>
<p>Yeah, I basically ignored all of Princeton Review’s math strategies and still got an 800.</p>
<p>Another problem occurs when it’s a grid-in and you need an exact answer (or an answer contained within a range of possible answers).</p>
![http://i584.■■■■■■■■■■■■■■■/albums/ss284/sear_squid/SAT/Picture2-1.png[/url]](http://i584.photobucket.com/albums/ss284/sear_squid/SAT/Picture2-1.png%5B/url%5D){kind=link}
