Tough SAT math problem


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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!!

post-7-1070773923.jpg

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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?

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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?

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.

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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

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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...

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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 by lostspyder
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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 :)

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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.

post-7-1070780901.jpg

Edited by lostspyder
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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

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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. ;)

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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 by nomis_nehc
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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.

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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:

stuff%20stuff.JPG

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 by Kiwi Boy
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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 by KayMan2K
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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. :)

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