12 Mar 09
Originally posted by geepamooglethis, in terms of the circle problem, is what i was trying to say. except that this explanation may be difficult for a grade-schooler to see if they get marks off on their paper. the problem, because of its assumption that there are two possible circles (but in fact one of the circles was IMpossible), was itself invalid.
The answer of x=2 is invalidated by the fact it would make the radius negative when the domain of radii permitted are non-negative numbers (Zero might be permitted for a circle which is in essence a point).
Since the "smaller" circle is invalid, the "difference" is also invalid.
Using the square root function isn't reversible in general, and it can ...[text shortened]... tly cover the triangle without gap or overlap, and their total area is 6 + 4 * sqrt(3)...
but that doesn't help you get the "right" answer on the test, if your teacher happened to miss this. in this case, one could quite accurately argue after the fact that this problem is majorly flawed, and would most likely be refunded whatever points were taken off for an "incorrect" answer. but it seems to me the better course of action might be to understand that your teacher made an error, to understand WHY that error invalidates the problem, but then to answer the question in such a way that follows what clearly was the intent of the teacher in constructing the problem.
from the standpoint of a problem constructor, this one is clearly "cooked."
12 Mar 09
2 edits
Originally posted by geepamoogleIf you draw it out, you can solve each triangle (there are 8 equal area triangles hidden in that drawing) the center equalateral triangle can be redrawn as two triangles where the vertical line just goes up from the spot where two of the circle tangents meet, straight up in the most obvious case, and you will find the area for each one of the 8 = (2*sqrt3)/4.
The answer of x=2 is invalidated by the fact it would make the radius negative when the domain of radii permitted are non-negative numbers (Zero might be permitted for a circle which is in essence a point).
Since the "smaller" circle is invalid, the "difference" is also invalid.
Using the square root function isn't reversible in general, and it can tly cover the triangle without gap or overlap, and their total area is 6 + 4 * sqrt(3)...
The apex of the little triangles has and angle of 30 degrees so it is a 30,60,90 degree triangle and the sin of 30 degrees is 0.5, so the one side is known, length of 1, divided by the sin of 30, or 0.5, 1/0.5=2 which is the hypotenuse, so the long bottom side by pythag (A^2 +b^2=C^2) C in this case is the hypotenuse, which is 2 so squared= 4. The one side you know is 1, so you have 4 - 1 =3 and you have to take the square root of 3 to get the answer, the square root of 3🙂. That is the length of the bottom side. For the area, draw two side by side as mirror images, with the height of 1 (the line in the center) and the two triangles side by side that way. So you now have a longer triangle with length of 2*sqrt3, right? and the height is 1. So the area of that double is 0.5 * 2Sqrt 3 which is the same as (2*sqrt3)/2. So half of that gives the area of one, which then would be (2*sqrt3)/4. Now you have exactly 8 of those beasts, right? so now you have ((2*sqrt3/4)*8), right? So that is the same as 2* (2*sqrt3) which is 4*Sqrt3. That is the total area of all 8 of those little triangles. Now we see we have those three rectangles, 1X2, so each one is 2 square, times three, which is 6. So the whole answer is 6+(4*sqrt3), which is the same answer as geep gave but I just showed the whole calculation, ok? I hope you can follow that.
12 Mar 09
Originally posted by sonhouseThank you, that was exactly what I was looking for. My problem, was that I was trying to solve it without actually drawing the triangle, which was making it difficult.
If you draw it out, you can solve each triangle (there are 8 equal area triangles hidden in that drawing) the center equalateral triangle can be redrawn as two triangles where the vertical line just goes up from the spot where two of the circle tangents meet, straight up in the most obvious case, and you will find the area for each one of the 8 = (2*sqrt3)/ ...[text shortened]... ame answer as geep gave but I just showed the whole calculation, ok? I hope you can follow that.
13 Mar 09
Originally posted by AetheraelIf I was to set such a question, if a student showed it to be invalid then it gets full marks. I would hope not to set such a question but accidents do happen.
this, in terms of the circle problem, is what i was trying to say. except that this explanation may be difficult for a grade-schooler to see if they get marks off on their paper. the problem, because of its assumption that there are two possible circles (but in fact one of the circles was IMpossible), was itself invalid.
but that doesn't help you ge ...[text shortened]... problem.
from the standpoint of a problem constructor, this one is clearly "cooked."
14 Mar 09
Originally posted by deriver69as a teacher, i fully agree and would always give full marks for an invalid question. from the standpoint of a student, though, i usually try to answer what i perceive to be the "intended question" of the teacher AND afterwards approach the teacher with an argument for invalidating the question. This covers all bases, including the possibility that the teacher is an unyielding egomaniac 🙂
If I was to set such a question, if a student showed it to be invalid then it gets full marks. I would hope not to set such a question but accidents do happen.
14 Mar 09
Originally posted by AetheraelWhat do you teach, Aethereal? (and at what level)
as a teacher, i fully agree and would always give full marks for an invalid question. from the standpoint of a student, though, i usually try to answer what i perceive to be the "intended question" of the teacher AND afterwards approach the teacher with an argument for invalidating the question. This covers all bases, including the possibility that the teacher is an unyielding egomaniac 🙂
15 Mar 09
1 edit
This sort of problem is not invalid. It's a standard sort of math problem. I've done zillions of problems with quadratics in which you had to dismiss one of the two answers for reasons like this.
I was an eighth grade Physical Science and sixth grade Earth Science teacher for one semester! Now I'm poor again 🙁
15 Mar 09
1 edit
Originally posted by AThousandYoungreread the question again carefully... there are definitely standard problems where you dismiss the negative answer as impossible, but this problem states that there are two circles in existence each with radius (x-5) and then asks you to compare their areas. but one of the circles can't exist, so how can we compare them?
This sort of problem is not invalid. It's a standard sort of math problem. I've done zillions of problems with quadratics in which you had to dismiss one of the two answers for reasons like this.
I was an eighth grade Physical Science and sixth grade Earth Science teacher for one semester! Now I'm poor again 🙁
the type of problem you're describing usually involves computing the area of a rectangle where each of the sides is expressed as a linear expression in x, and then using your answer for x to compute something else, like perimeter. in these cases when you solve you get two possible answers for the quadratic, but then must reject the negative answer because it doesn't make sense.
p.s. @ Palynka - to answer your question, i currently teach at a small extracurricular enrichment school. i have students ranging from 6th graders to high schoolers, and teach a number of different subjects including mathematics, essay writing, and SAT prep... but my main focus is math, as that was my major in college.
16 Mar 09
1 edit
The problem as stated includes the implicit assumption that the circles are different sizes.
However, given that one of the solutions to the quadratic is 2 (resulting in a radius of -3), we arrive at the following inconsistency.
A circle of radius -3 has a radius of 3, as I understand it, since anyone looking at the resulting circle would naturally take the radius to be positive. That makes x equal to 8 rather than 2, but x=8 results in a different answer for the area, a wrong answer.
17 Mar 09
Originally posted by geepamoogleIt also implies the circles have the same radius and area.
The problem as stated includes the implicit assumption that the circles are different sizes.
"The first question is as follows: There are two circles, each with a radius of (x-5), and an area of pi(4x+1). How much more area does the larger circle have than the smaller one?"
The first half of the question implies the circles have the same radius and same area.
The second half implies a difference in size.
For the rest, radii can't be negative, so x=2 is not a valid solution. Discarding it leaves only x=12, making the second half of the question trivial.
In this case I would give a student who put thought into this full credit. I usually don't give credit to student who don't write anything down, or less to those that write gibberish.
I reward them for their analasys of the problems. If a problem is flawed then they still should be able to make some form of analasys.
17 Mar 09
2 edits
Originally posted by TheMaster37I once got a nice comment from my first physics teacher. She had assigned a projectile motion problem. I was able to demonstrate that the basketball player made his basket while the ball was going UP, and therefore he threw it through the bottom of the basket. I felt good 🙂
It also implies the circles have the same radius and area.
"The first question is as follows: There are two circles, each with a radius of (x-5), and an area of pi(4x+1). How much more area does the larger circle have than the smaller one?"
The first half of the question implies the circles have the same radius and same area.
The second half imp blems. If a problem is flawed then they still should be able to make some form of analasys.
I see that the problem is indeed poorly written...
Well...actually...it's not.
The answer is clearly 0 by inspection, now that I see TheMaster37's post. pi(4x+1) - pi(4x+1) = 0.
No, wait...the problem refers to a "larger" and a "smaller" circle. Therefore the problem is poorly written.
17 Mar 09
2 edits
I missed the wording concerning both circles having a radius of x-5...
While I understood what was intended by that, the wording does unintentionally indicate that x is the same in both cases, making the circles identical.
This a vague and imprecise language at best, and it would probably have also been better to have stated explicitly the assumption implicitly spoken of in the actual question, that the circles are of two different sizes.
Good catch there.