HellBender Posted December 7, 2003 Share Posted December 7, 2003 This one was on today's SAT! See if any of you can figure it out, cause I sure cant. Sorry for the crude Paint drawing :) Okay. So here's the diagram. Given information: - all the shaded regions are equal - all the triangles are equal - all the triangles are right triangles - the area of the shaded regions is 20pi. So find out what x is. Crazy tough!! Link to comment Share on other sites More sharing options...
Cody Posted December 7, 2003 Share Posted December 7, 2003 are the parts of the triangles that cover the circle part of the 20 pi? Link to comment Share on other sites More sharing options...
bdevoes Posted December 7, 2003 Share Posted December 7, 2003 okay......if I am doing this right shaded areas are 20pi 3.14..*20=62.83 I assume degrees 4 shaded areas=251.33 ok......360-251.33=108.67 108.67/4=27.17......so if I am doing this right, the angle of the triangles in the circle is 27.17 degrees 180-(27.17+90)=62.83 x=62.83 degrees but how do you put that into an equation??? its been 7 years since the SAT for me......how bad did I do? Link to comment Share on other sites More sharing options...
Imperfect63 Posted December 7, 2003 Share Posted December 7, 2003 okay......if I am doing this rightshaded areas are 20pi 3.14..*20=62.83 I assume degrees 4 shaded areas=251.33 ok......360-251.33=108.67 108.67/4=27.17......so if I am doing this right, the angle of the triangles in the circle is 27.17 degrees 180-(27.17+90)=62.83 x=62.83 degrees but how do you put that into an equation??? its been 7 years since the SAT for me......how bad did I do? i think he meant the total area of the shaded regions is 20pi. overlooking that minor part, you can't subtracted area from 360. just by looking at units its not correct. 360 is in degrees and area is in some arbitrary unit squared. i dislike geometry :p mainly because i have an exam next week that involves quite a bit of it, hehe. i don't see how you can get it right now, guess my brain isn't working at 1am. Link to comment Share on other sites More sharing options...
CrzyCool Posted December 7, 2003 Share Posted December 7, 2003 It was a test section... wasn't on my copy. But ummm - that is damn hard. I had easy questions on all the sections (although I didn't simplify on one of em) Link to comment Share on other sites More sharing options...
Imperfect63 Posted December 7, 2003 Share Posted December 7, 2003 well heres some stuff that i can't really solve let a = angle of all the shaded regions 20pi = a / 360 * pi * r^2 20 *360 = a * r^2 if you can solve for a then its fairly easy. just 90 - (360 - a)/4 problem is i can't think of any way to get r. i think thats probably the approach, but i could be wrong. its probably something small i overlooked did they expect a numerical answer? if not, you can just substitute for a x = 90 - (360 - (20 * 360 / r^2)) / 4 Link to comment Share on other sites More sharing options...
KayMan2K Posted December 7, 2003 Share Posted December 7, 2003 Are the triangles drawn correctly? I mean, is the height of each triangle supposed to be equal to the radius? Not sure if because, obviously, the image isn't to scale. It would make a difference... Link to comment Share on other sites More sharing options...
lostspyder Posted December 7, 2003 Share Posted December 7, 2003 (edited) You need to know a side of the triangle to find anything. The old rule of thumb with triangles, 3 bits of indirecly related info and you got it all. Edited December 7, 2003 by lostspyder Link to comment Share on other sites More sharing options...
John Veteran Posted December 7, 2003 Veteran Share Posted December 7, 2003 that's what i was wondering too, how do you know how big the triangle is? :blink: you need the length of at least 1 side... Link to comment Share on other sites More sharing options...
alexander777 Posted December 7, 2003 Share Posted December 7, 2003 jeesh you guys must be products of public school, i graduated in 1983 and that problem is easy im gonna see how long it takes someone to get it if noone does then ill give you the answer. btw I learned triangle is equal on all sides :) Link to comment Share on other sites More sharing options...
lostspyder Posted December 7, 2003 Share Posted December 7, 2003 (edited) jeesh you guys must be products of public school, i graduated in 1983 and that problem is easy im gonna see how long it takes someone to get it if noone does then ill give you the answer. btw I learned triangle is equal on all sides :) umm no you didnt. There is no therom that will solve for the area of a triangle given two angles. The triangle could be any scale, theres nothing stoping the hypotitunuse from being infinity. How about you post the solution and we all laugh at you for cocking off your mouth and show you that you screwd up. Thats like saying the two triangles are ==. I suspect were not being told soemthing. Edited December 7, 2003 by lostspyder Link to comment Share on other sites More sharing options...
John Veteran Posted December 7, 2003 Veteran Share Posted December 7, 2003 lmfao, you can't solve that as an equilateral triangle :laugh: a triangle with 3-90 degree angles can't exist :wacko: at least not in my three dimensions... wherever you live, please tell me how it's possible :laugh: btw, i went to private school, so STFU :p Link to comment Share on other sites More sharing options...
soppychunk Posted December 7, 2003 Share Posted December 7, 2003 (e) Cannot be determined Link to comment Share on other sites More sharing options...
Evil Dragon Posted December 7, 2003 Share Posted December 7, 2003 Yeah, either you're not telling us something that we need to know, or there was an option that it cannot be solved, because that right there is impossible. ;) Link to comment Share on other sites More sharing options...
hifive Posted December 7, 2003 Share Posted December 7, 2003 Wow. Is this the SAT I or SAT II? I don't think they'll put a problem like this on the SAT I. Oh, and [iMO], the problem right now is impossible to solve. Link to comment Share on other sites More sharing options...
markus Posted December 7, 2003 Share Posted December 7, 2003 I think there is some detail you not telling us - or else i have forgetting all my math :s Link to comment Share on other sites More sharing options...
nomis_nehc Posted December 7, 2003 Share Posted December 7, 2003 (edited) actually, I have not give the problem too much thought, but if they are right triangle, it cannot be equilateral. also, it is possible to know the length of the sides... the two sides are equal... too tired to do the rest. Edited December 7, 2003 by nomis_nehc Link to comment Share on other sites More sharing options...
KayMan2K Posted December 7, 2003 Share Posted December 7, 2003 This quite tricky given only this little information. So far, we can deduce that 20pi = (r^2)pi*(x/90). So, we only need to express r as a function of x to solve, however that is the hard part that I do not have the time to solve right now. It may not even be possible. Link to comment Share on other sites More sharing options...
pink-hippo Posted December 7, 2003 Share Posted December 7, 2003 (edited) I could only find a formula for x in terms of r, the radius of the circle. This is the furthest I could go: First, cut the diagram in quarter, and deal with only one quarter of the diagram. We will get a right triangle and the shaded area of 5 pi, as follows: We know that the sector area of a circle is (1/2)(r^2)x = 5 pi where x is in radians. Solve for x, and we will get x = 10 pi / (r^2) By this equation, we can see that the value of x really depends on r. If r increases, you need to increase y (and thus decrease x) such that the shaded area is 5 pi. Due to the dependency between x and other non-given values such as the radius, more information is needed in order to find an actual value for x. Edited December 7, 2003 by Kiwi Boy Link to comment Share on other sites More sharing options...
KayMan2K Posted December 7, 2003 Share Posted December 7, 2003 (edited) We are actually getting closer then, we have two equations: 20pi = (r^2)pi*(x/90) x = 10 pi / (r^2) We just need to solve this... Errr, well almost because x is expressed in radians in the second equation but degrees in the first. So that needs to be converted. Furthermore, nevermind. Solving these two formulas (after correcting for degrees v radians) simply says that 1800 = 1800. Which is kinda useless in solving this problem. I am starting to agree, impossible without more information. Edited December 7, 2003 by KayMan2K Link to comment Share on other sites More sharing options...
nhut Posted December 7, 2003 Share Posted December 7, 2003 Not possible without some other bit of info. Link to comment Share on other sites More sharing options...
TranceSphere Posted December 7, 2003 Share Posted December 7, 2003 n/m :p Link to comment Share on other sites More sharing options...
caustiK Posted December 7, 2003 Share Posted December 7, 2003 i'm glad everyone is agreeing - i thought i had forgotten all my geometry for some reason, then i scrolled down ;) Link to comment Share on other sites More sharing options...
poind Posted December 7, 2003 Share Posted December 7, 2003 First off, I'd assume this was a noncounted experimental question. Secondly, knowing what the multiple choice answers were would help a ton. Stuff may or may not have been drawn to scale on the test, but I'd assume the triangle bases probably added up to the radius or the diameter, and with pi r squared for circle area, etc., one could probably figure out a good guesstimate based on answers provided. It's the kind of question where it's really helpful to see if answers are related to pi, r, etc., and perhaps even just do process of elimination. Getting a bit late for me to outright risk frying my brain on this one in-depth, especially as an outright solve. :) Link to comment Share on other sites More sharing options...
Gary_Player Posted December 7, 2003 Share Posted December 7, 2003 Apparently the SATs got harder since I took them 2 years ago ;) Link to comment Share on other sites More sharing options...
Recommended Posts