tag:blogger.com,1999:blog-7691251033406320222.post7560411869184694348..comments2024-03-26T04:19:38.862-07:00Comments on kitchen table math, the sequel: help desk - why square root of 5?Catherine Johnsonhttp://www.blogger.com/profile/03347093496361370174noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-7691251033406320222.post-7227692918810127562011-03-25T10:57:09.766-07:002011-03-25T10:57:09.766-07:00Remember to use your test taking skill of glancing...Remember to use your test taking skill of glancing at the answers in the process of deciding your approach. Since these answers involve fractions and square roots, no need to waste precious time in converting to decimal.<br /><br />E is recognizable to those who have an enrichment/honors course or math club that included the Golden Ratio topic. It is introduced in Dolciani PreAlgebra: An Accelerated Course.lgmnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-22162704247893224692011-03-25T09:20:51.077-07:002011-03-25T09:20:51.077-07:00Richard I, you saw that .618 and knew it was Phi, ...Richard I, you saw that .618 and knew it was Phi, didn't you? :)rockynoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-42839643534488048332011-03-24T22:43:12.750-07:002011-03-24T22:43:12.750-07:00Can I just add a recommendation to do Michael'...Can I just add a recommendation to do Michael's extra credit writing assignment?<br /><br />As soon as I saw 0.618, I knew what the answer was and all that was left was how to get to the quadratic equation that leads to that number. A bit like a mathematical version of Jeopardy!!<br /><br />For the record,<br /><br />AD=1<br />BC=1<br />AE=1/2<br />BE=1/2<br />FB=x<br />BH=x<br />HC=1-x (not needed, but it looks incomplete without it)<br />FE = x +1/2 = CE<br /><br />Pythagoras on BEC to give<br /><br />BC^2+BE^2=CE^2<br />1^2+(1/2)^2=(x+1/2)^2<br /><br />5/4 = x^2 + x +1/4<br /><br />and from that you get<br /><br />x^2 + x = 1<br /><br />The rest is left as "an exercise for the reader" (how I love that phrase!!)Richard Inoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-78676527097407166672011-03-24T20:16:33.930-07:002011-03-24T20:16:33.930-07:00Students familiar with problems like this expect t...Students familiar with problems like this expect to see sqrt(2)/2, or sqrt(5)/2, or sqrt(3)/2 in problems involving right triangles or equilateral triangles. One reason they know to leave them in that form because they expect to use the Pythagorean theorem again to solve more steps, and they'll be squaring them up again in the future. But after a while, they know to leave them in that form because that's just how the problems are usually written.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-61385323119027043082011-03-24T19:11:06.759-07:002011-03-24T19:11:06.759-07:00It is a rare person who can see a decimal and figu...It is a rare person who can see a decimal and figure out what the radical would be. You really want to get in the habit of working in whole numbers and leaving square roots once you get there, especially for SAT math where the answer is rarely the decimal.<br /><br />Having said that, it is worth memorizing the square root of 2 (1.41) and ln 2 (.693) so you can see them and know what the original value was.ChemProfnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-43546062544930379042011-03-24T16:43:56.738-07:002011-03-24T16:43:56.738-07:00The embarassing part isn't that I used decimal...The embarassing part isn't that I used decimal calculations, it's that I didn't perceive that's what I came up with .618 as the answer.<br /><br />I couldn't see how you would end up with a radicand in your answer (radicand?? that's the term, right?)<br /><br />It didn't jump out at me that 1/2 x 1/2 would produce a _/5 <br /><br />(squre -rt 5 written out as sqr rt 5)<br /><br />inflexible thinking<br /><br />major, major inflexible thinking<br /><br />I DON'T LIKE IT!Catherine Johnsonhttps://www.blogger.com/profile/03347093496361370174noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-50162280685005779002011-03-24T16:08:25.576-07:002011-03-24T16:08:25.576-07:00No reason to be embarrassed. The urge to "ca...No reason to be embarrassed. The urge to "calculate" (express your answer as a decimal) can be hard to resist. But it should almost <i>always</i> be saved for the <i>end</i> of a problem. Otherwise how else would you be able to recognize that the answer to a certain problem is (to take just one made-up example) pi / sqrt(3), when all you see is 1.8148?Michael Weissnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-87313562004143055892011-03-24T15:42:21.326-07:002011-03-24T15:42:21.326-07:00ok, you guys
don't laugh
i used a decimal le...ok, you guys<br /><br />don't laugh<br /><br />i used a decimal length from the get-go and then didn't see that if I'd used 1/2 rather than .5 I'd have a square root of 5/4 instead of a square root of 1.75<br /><br />oyCatherine Johnsonhttps://www.blogger.com/profile/03347093496361370174noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-41882651668705511562011-03-24T15:29:37.990-07:002011-03-24T15:29:37.990-07:00I've got to print these out & read!
Now I...I've got to print these out & read!<br /><br />Now I'm thinking...I wonder if the issue is that I was using decimals from the get-go ---<br /><br />hang on<br /><br />I'll look at my work<br /><br />I was doing it late last night, so cognitive flexibility wasn't at maximum strength---<br /><br />that reminds me of something else I want to postCatherine Johnsonhttps://www.blogger.com/profile/03347093496361370174noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-31003116259638356582011-03-24T14:49:27.231-07:002011-03-24T14:49:27.231-07:00My attempt, trying to put in as many little steps ...My attempt, trying to put in as many little steps as possible:<br /><br />*) given: AD = 1<br />*) given: AB and AD are sides of a square, and thus are equal<br />*) AD = 1 --> AB = 1<br />*) given: EB is the midpoint of AB<br />*) EB = ½ of AB --> EB = ½<br /><br />*) EBC is a right triangle, so we can use the pythagorean theorem: EB×EB + BC×BC = EC×EC<br />*) ½*½ + 1×1 = EC*EC<br />*) ¼ + 1 = EC×EC<br />*) EC = sqrt (1¼)<br /><br />*) given: FE = CE<br />*) EB = ½ from above, so<br />*) FB = sqrt(1¼) - ½<br /><br />*) sqrt(1¼) = sqrt(5/4)<br />*) sqrt(5/4) = sqrt(5)/sqrt(4)<br />*) sqrt(5)/sqrt(4) = sqrt(5)/2<br /><br />*) FB = sqrt(5)/2 = 1/2<br />*) FB = [sqrt(5) - 1]/2<br /><br />*) FB and BH are sides of a square, so BH = FB<br /><br />*) BH = [sqrt(5) - 1]/2<br /><br />-Mark RouloAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-2859727802043946532011-03-24T14:38:37.708-07:002011-03-24T14:38:37.708-07:00One of the hardest things to learn is the strength...One of the hardest things to learn is the strength to resist evaluating things numerically in the middle of a problem.Michael Weissnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-78118180868522103582011-03-24T14:28:20.894-07:002011-03-24T14:28:20.894-07:00Catherine, I'm curious what method you used to...Catherine, I'm curious what method you used to get your answer. Obviously your method is correct, because it leads to the right numerical result. At some point you must have taken a square root, right? What happens if you just retrace your steps, but resist the urge to turn that square root into a decimal, and leave it in its "unevaluated" form?<br /><br />I would attack the problem this way:<br /><br />Triangle BEC is a right triangle, with one leg twice as large as the other. In <i>any</i> such triangle, the hypotenuse is as long as the short leg times the square root of 5. Why? Because if the legs are x and 2x, then the Pythagorean Theorem gives you (hypotenuse)^2 = x^2 + (2x)^2. Simplifying the right-hand-side of the equation, you get 5x^2. (The 5 comes from squaring the 2 in the second term, to get 4 and then adding the 1 from the first term.) So the hypotenuse is (sqrt 5)*x.<br /><br />Going back to the problem, we now know that EC is as long as BE times sqrt(5), i.e. its length is sqrt(5) / 2.<br /><br />From the information in the problem, that is also the length of FE.<br /><br />So FB is that quantity, minus the length of BE... Or, in other words, FB = sqrt(5)/2 - 1/2. Over a common denominator, we have FB = (sqrt(5) - 1)/2.<br /><br />And, lastly, BH is just as long as FB.<br /><br />Extra credit writing assignment: What is the connection between this problem and the Golden Rectangle?Michael Weissnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-38335651760828348522011-03-24T14:12:29.947-07:002011-03-24T14:12:29.947-07:00BE is 1/2 (midpoint of square side, side=1)
BH is...BE is 1/2 (midpoint of square side, side=1)<br /><br />BH is side of square, equal to BF + BE, so BH = FE - 1/2 = CE - 1/2<br /><br />by Pythagoras, square(CE) = square(BE) + square(BC)<br /><br />so square(CE) = 1/4 + 4/4 = 5/4 and CE = square root(5/4) = (square root 5)/2<br /><br />(square root 5)/2 - 1/2 = (square root(5) - 1)/2<br /><br />(sorry for poor formatting and lazy use of =)<br /><br />- AndyAnonymousnoreply@blogger.com