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The article also demonstrates a “hard way” using the Pythagorean Theorem and the Quadratic Formula to solve for a and b, deriving e and f and confirming the same perimeter result.",{"@graph":14,"@context":73},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/that-annoying-sat-circle-question-explanation-and-solution-strategy/267092/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/that-annoying-sat-circle-question-explanation-and-solution-strategy/267092.png","ImageObject",300,407,{"name":42,"@type":43},"Berry Peter","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-30","2026-09-14",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",11,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"How is g determined from the circle information?","Question",{"text":63,"@type":64},"g is identified as a fraction of the circle’s circumference based on the geometry shown. Using the radius 6, it evaluates to 3√3.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"Why does the “easy way” focus on e + f instead of e and f individually?",{"text":68,"@type":64},"The perimeter needs e + f + c + g, and g and c are already known. Computing the sum e + f avoids unnecessary work when e and f are not required separately.",{"name":70,"@type":61,"acceptedAnswer":71},"What changes in the “hard way” approach?",{"text":72,"@type":64},"The hard way computes e and f by first solving for a and b. It uses the Pythagorean Theorem and then the Quadratic Formula to obtain b, followed by deriving a, then e and f from radius relations.","https://schema.org",{"og:url":32,"og:type":75,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":77,"canonical":32},"index,follow",{"doc_id":79,"site_id":7},267092,1789414750,{"code":4,"msg":82,"data":83},"success",[84,88,92,95,100,105,110,115,120,123,127],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":89,"show_sort_weight":90,"slug":91},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":93,"slug":94},70,"exam",{"id":96,"doc_module":4,"doc_module_name":25,"category_name":97,"show_sort_weight":98,"slug":99},5,"Comic",60,"comic",{"id":101,"doc_module":4,"doc_module_name":25,"category_name":102,"show_sort_weight":103,"slug":104},6,"Technology",50,"technology",{"id":106,"doc_module":4,"doc_module_name":25,"category_name":107,"show_sort_weight":108,"slug":109},7,"Healthcare",40,"healthcare",{"id":111,"doc_module":4,"doc_module_name":25,"category_name":112,"show_sort_weight":113,"slug":114},8,"Research & Report",30,"research-report",{"id":116,"doc_module":4,"doc_module_name":25,"category_name":117,"show_sort_weight":118,"slug":119},9,"Religion & Spirituality",20,"religion-spirituality",{"id":118,"doc_module":4,"doc_module_name":25,"category_name":121,"show_sort_weight":118,"slug":122},"World Cup","world-cup",{"id":124,"doc_module":4,"doc_module_name":25,"category_name":125,"show_sort_weight":124,"slug":126},10,"Lifestyle","lifestyle",{"id":128,"doc_module":4,"doc_module_name":25,"category_name":129,"show_sort_weight":96,"slug":130},19,"General","general",{"code":4,"msg":82,"data":132},{"doc_id":79,"user_id":133,"nickname":42,"user_avatar":134,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":135,"file_id":136,"file_url":137,"file_type":138,"file_size":139,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":33,"language":140,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":12,"update_tm":80,"read_time":124},1374402524268,"https://ap-avatar.wpscdn.com/davatar_3d24733baf745e90a7e4bdd5f77d97b2","That Annoying SAT Circle Question  \nby Jon Dreyer  \nWeb: [www.passionatelycurious.com](www.passionatelycurious.com)  \nAn SAT question that once stumped me (embarrassingly, since I was working with a student!) starts with something like this picture:  \nIt asks a simple question:  \nGiven that a + b = 8 and that the radius of the circle is 6, what is the perimeter of the area bounded by e, c, f and g?  \nAt this point, the best thing to do before proceeding is to try to solve the problem on your own. After you make a serious attempt, whether or not you succeed, read on. (This is true any time you are about to read an explanation of how to solve a math problem!)  \nThis looks like a 􀀐divide and conquer􀀑 kind of problem. Just 􀀜nd e, c, f, and g. Then add them up. This approach seems to work for awhile. Let’s start with g.  \nTo 􀀜nd g, just notice that g is just 14 ( ~~ ~~39600~~ ~~ ) of a circle whose radius is 6, so  \ng = 14 􀀁 2􀀙 􀀁 r 1 = 􀀙 􀀁 6 2 = 3􀀙  \nFinding c appears hard until you realize it is one of the two diagonals of rectangle ABCR. The other diagonal is a radius of the circle, and we already know that the radius is 6, so  \nc= 6  \nBut 􀀜nding e and f is harder.  \nIf we can’t 􀀜nd e and f, it’s worth asking whether you really need them. After all we are really trying to 􀀜nd g + c + e + f. Since we already know g and c, might it be easier to 􀀜nd e + f rather than to 􀀜nd each individually? That turns out to be true.  \nAn important problem-solving (and test-taking!) principle here is not to divide-and-conquer more than is necessary. Here’s a completely di􀀛erent, much simpler problem that shows this principle:  \nIf m + n = 4 and p + q + r = 6, what is the average of m, n, p, q and r? Normally to 􀀜nd an average we would try to 􀀜nd all those values, then add, and then divide by 5 (the number of values being averaged) . But in this case, we can’t 􀀜nd the values. We notice instead that  \naverage = m+ n+ p+ q + r 5  \n(m+ n) + (p+ q + r)  \n=  \n5  \n4 + 6  \n=  \n5  \n= 2  \nNotice that we never had to 􀀜nd the individual values!  \n[In what follows, lower case letters represent segment lengths and upper case letters represent points.]  \nThe Easy Way  \nIt turns out that we can more easily 􀀜nd the sum e+ f than we can 􀀜nd either e or f individually. That’s what I call the 􀀐easy way􀀑:  \nThe radius is still 6 so  \nRS = RT = 6 (1)  \nThey also told us that  \na+ b= 8 (2)  \nFrom that and the fact that ABCR is a rectangle,  \nAR + CR = 8 (3)  \nSince RS and RT are radii,  \nRS + RT= 6 + 6 = 12 (4)  \nSince the picture tells us that AS + AR = RSand CR + CT = RT ,  \nRS + RT = 12  \nAS + AR + CR + CT = 12  \nAR + CR + AS + CT = 12 (commutative law)  \n8 + AS + CT = 12 (see eqn3) AS + CT = 4  \ne+ f = 4 (5)  \nSo the whole perimiter is  \ne+ f + c+g = e+ f + 6 + 3p = 4 + 6 + 3p  \n= 10+ 3p (6)  \nThe hard way  \nJust for fun, let’s see what happens if we do it the 􀀐hard way􀀑 (􀀜nd e and f separately) . The Pythagorean Theorem says that  \na2 + b2 = c2 = 62 = 36  \nSubstituting 8 􀀀 b for a from equation (2):  \n(8 􀀀b)2 + b2 = 36  \nSimplifying:  \n(8 􀀀 b)2 + b2 = 36 64􀀀 16b + b2 + b2 = 36 64􀀀 16b + 2b2 = 36 2b2 􀀀 16b + 28 = 0 b2 􀀀 8b + 14 = 0  \nWe cannot factor the left side of that equation. So it’s time to pull out the big gun for quadratic equations: the Quadratic Formula. I’m going to rewrite that equation using the variable x so we don’t confuse the variable in the equation (now called x) with the b in the quadratic formula (coe􀀞cient of x term):  \nx2 􀀀 8x + 14 = 0  \n8 􀀆 p64 􀀀4(1)(14)  \nx =  \n2(1)  \n8 􀀆√64 􀀀 56  \n=  \n2  \n8 􀀆√8  \n=  \n2  \n= 4 􀀆√2  \nSo the two possible values for b, (which we were just calling x), are  \nb = 4 􀀆√2 (7)  \nTo keep things simple, let’s make an arbitrary decision:  \nb = 4 +√2  \nSince a+ b= 8, then  \na = 8 􀀀 b  \n= 8 􀀀 􀀐4 +√2 􀀑  \n= 4 􀀀√2  \n(If it were the other way around, it would work similarly.) So then because e and b make up a radius, we have  \ne+ b = 6  \nSubstituting, we get  \ne = 6 􀀀 b  \n= 6 􀀀 􀀐4 +√2 􀀑  \n= 2 􀀀√2  \nSimilarly, because f and a a","cbCaic1RqpoClpn6","https://ap.wps.com/l/cbCaic1RqpoClpn6","pdf",65710,"English","# Solution Overview\n## Compute g and c\n## Easy way: find e + f\n## Hard way: Pythagorean theorem and quadratic formula\n## Final result confirmation","[{\"question\":\"How is g determined from the circle information?\",\"answer\":\"g is identified as a fraction of the circle’s circumference based on the geometry shown. Using the radius 6, it evaluates to 3√3.\"},{\"question\":\"Why does the “easy way” focus on e + f instead of e and f individually?\",\"answer\":\"The perimeter needs e + f + c + g, and g and c are already known. Computing the sum e + f avoids unnecessary work when e and f are not required separately.\"},{\"question\":\"What changes in the “hard way” approach?\",\"answer\":\"The hard way computes e and f by first solving for a and b. It uses the Pythagorean Theorem and then the Quadratic Formula to obtain b, followed by deriving a, then e and f from radius relations.\"}]","That Annoying SAT Circle Question - Explanation and Solution Strategy | PDF"]