Well, the easiest way to find the area of the shaded region is to subtract the area of the white rectangle from the total area. You can also divide the shaded region into smaller rectangles and compute the area of each and add them up. However, there isn't enough info on the bit you've shown to come up with a numerical answer; is there any further info given on the dimensions in some other part of the problem?
However, there isn't enough info on the bit you've shown to come up with a numerical answer; is there any further info given on the dimensions in some other part of the problem?
Let x be the height of the white rectangle and let y be width. Then:
Total area = 12in x 16in = 192in^2
White rect area = xy
Shaded area = 192in^2 - xy
Not the most useful answer, but AN answer.
There really was NO further info supplied than the height and width of the outside rectangle? Nothing about the relationship between the inner and outer boundaries, such as the shaded area is a uniform "x" inches deep or anything like that? I'm just having trouble comprehending that. Weird.
You could have some fun by measuring the sides, finding the ratios, deducing the unknown values, and then calculating the area. This would be a "hands-on" solution.
You need one side of the white rectangle showing a length. Then, you could set up a proportion to find the unknown side. Or, if the drawing were on a graph, showing the dimensions of the shapes used in the graph, you could figure it out.
This is typical of drawings in geometry problems--the leaving off of a number that is needed to solve the problem.
The new assistant superintendent of curriculum is so good (thus far) I've been contemplating pulling out all the grade disputes we should have had with Ms. K over the years...
Even forty-two's answer is over-specific. As written, you can't even determine the size of the figure without considering the cutout. None of the edges are specified as either parallel or perpendicular. Similarly, you cannot specify the size of the internal cutout, because you don't know that it's a rectangle (in addition to not knowing the length of any side).
Certainly, both figures look like rectangles, but this is a math test, not a carpentry test. Precision is required.
None of the edges are specified as either parallel or perpendicular. Similarly, you cannot specify the size of the internal cutout, because you don't know that it's a rectangle (in addition to not knowing the length of any side).
Certainly, both figures look like rectangles, but this is a math test, not a carpentry test. Precision is required.
I have to get you to take a look at one of the Singapore problems.
I can't tell whether the same issue pertains. (Will have to scan it.)
You have to look at the questions in context of what was studied in class. The second question says:
"In simplest form.."
This is a clue that tells you the answer is probably an expression. I would also assume that the class has studied a little bit about algebra and knows what "simplify" means.
The first problem looks like a labeling mistake, but after seeing the second problem, perhaps the course required students to define variables as needed. If so, then forty-two's response is correct.
I like Doug's response, but I don't think that would earn you any Brownie Points with the teacher. You could, however, say that you were applying your critical thinking skills.
That reminds me of the old right triangle problem that asks the student to "find" X.
Oh, if I were in the class, I'd probably reply something like:
"Assuming that the complete figure is intended to be a rectangle, and that the unshaded portion of the figure is also intended to be a rectangle, and assuming that the figure is intended to be drawn on a Euclidean plane, (forty-two's answer). (I wouldn't have included the Euclidean plane part in Jr. High. 8-)
Being a snarky SOB isn't a new thing for me, and that problem deserves snark.
Conventionally, a circle is either explicitly defined as such or shown with an explicit radius or diameter.
Certainly, both figures look like rectangles, but this is a math test, not a carpentry test. Precision is required.
This may be an unwarranted assumption! :-)
In any event, I was thinking about the snarky solution of copying the diagram to cardboard. Then cutting out the outer rectangle and weighing it. Then removing the inner rectangle and weighing it.
Now I have the ratio of the two areas. I can calculate the outer area ... so I can finish the problem.
Well, the easiest way to find the area of the shaded region is to subtract the area of the white rectangle from the total area. You can also divide the shaded region into smaller rectangles and compute the area of each and add them up. However, there isn't enough info on the bit you've shown to come up with a numerical answer; is there any further info given on the dimensions in some other part of the problem?
ReplyDeleteI would agree with your answer for the angle supplement - was it marked wrong or does the answer key have something different?
ReplyDeleteHowever, there isn't enough info on the bit you've shown to come up with a numerical answer; is there any further info given on the dimensions in some other part of the problem?
ReplyDeleteThat was my question.
This is all the info provided.
It can't be done, right?
I had a brain-freeze on the second problem - same one C. had.
ReplyDeleteI thought this was supposed to have a numerical answer; then this morning I realized it wasn't.
Just checking to make sure I did it right.
Well, technically you CAN do it:
ReplyDeleteLet x be the height of the white rectangle and let y be width. Then:
Total area = 12in x 16in = 192in^2
White rect area = xy
Shaded area = 192in^2 - xy
Not the most useful answer, but AN answer.
There really was NO further info supplied than the height and width of the outside rectangle? Nothing about the relationship between the inner and outer boundaries, such as the shaded area is a uniform "x" inches deep or anything like that? I'm just having trouble comprehending that. Weird.
No, it can't be done.
ReplyDeleteYou could have some fun by measuring the sides, finding the ratios, deducing the unknown values, and then calculating the area. This would be a "hands-on" solution.
ReplyDeleteIt cannot be done.
ReplyDeleteYou need one side of the white rectangle showing a length. Then, you could set up a proportion to find the unknown side. Or, if the drawing were on a graph, showing the dimensions of the shapes used in the graph, you could figure it out.
This is typical of drawings in geometry problems--the leaving off of a number that is needed to solve the problem.
Niki
Thank you!
ReplyDeleteIt's funny, because I'm.... not bad at algebra 1.... but I'm still not confident enough to assume that I'm not missing something.
I looked at this and thought: this can't be done.
But then I thought: I'm missing something.
This was it; this was all the information given ON A TEST.
Christopher got this one wrong.
The new assistant superintendent of curriculum is so good (thus far) I've been contemplating pulling out all the grade disputes we should have had with Ms. K over the years...
ReplyDeleteBut that would be wrong.
Talking to the new super was almost surreal.
ReplyDeleteIt was like being a person again.
Even forty-two's answer is over-specific. As written, you can't even determine the size of the figure without considering the cutout. None of the edges are specified as either parallel or perpendicular. Similarly, you cannot specify the size of the internal cutout, because you don't know that it's a rectangle (in addition to not knowing the length of any side).
ReplyDeleteCertainly, both figures look like rectangles, but this is a math test, not a carpentry test. Precision is required.
None of the edges are specified as either parallel or perpendicular. Similarly, you cannot specify the size of the internal cutout, because you don't know that it's a rectangle (in addition to not knowing the length of any side).
ReplyDeleteCertainly, both figures look like rectangles, but this is a math test, not a carpentry test. Precision is required.
I have to get you to take a look at one of the Singapore problems.
I can't tell whether the same issue pertains. (Will have to scan it.)
Do you need to specify that a figure is a circle?
I think I found another wrong answer on one of Christopher's old tests tonight.
ReplyDeleteI can certainly see why "Math Dad" was so exercised about the tests & the grading.
You have to look at the questions in context of what was studied in class. The second question says:
ReplyDelete"In simplest form.."
This is a clue that tells you the answer is probably an expression. I would also assume that the class has studied a little bit about algebra and knows what "simplify" means.
The first problem looks like a labeling mistake, but after seeing the second problem, perhaps the course required students to define variables as needed. If so, then forty-two's response is correct.
I like Doug's response, but I don't think that would earn you any Brownie Points with the teacher. You could, however, say that you were applying your critical thinking skills.
That reminds me of the old right triangle problem that asks the student to "find" X.
Oh, if I were in the class, I'd probably reply something like:
ReplyDelete"Assuming that the complete figure is intended to be a rectangle, and that the unshaded portion of the figure is also intended to be a rectangle, and assuming that the figure is intended to be drawn on a Euclidean plane, (forty-two's answer). (I wouldn't have included the Euclidean plane part in Jr. High. 8-)
Being a snarky SOB isn't a new thing for me, and that problem deserves snark.
Conventionally, a circle is either explicitly defined as such or shown with an explicit radius or diameter.
Certainly, both figures look like rectangles, but this is a math test, not a carpentry test. Precision is required.
ReplyDeleteThis may be an unwarranted assumption! :-)
In any event, I was thinking about the snarky solution of copying the diagram to cardboard. Then cutting out the outer rectangle and weighing it. Then removing the inner rectangle and weighing it.
Now I have the ratio of the two areas. I can calculate the outer area ... so I can finish the problem.
Whoo-hooo! Go me!
-Mark Roulo