# How to solve this Shaded Area Problem?

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#1
Hi all, can someone show the detailed solution for this question? Rather challenging. Thanks in advance!
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4 weeks ago
#2
(Original post by Darren888)
Hi all, can someone show the detailed solution for this question? Rather challenging. Thanks in advance!
Sorry solutions are not allowed.
What can you determine / what is difficult for you?
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4 weeks ago
#3
What have you done so far?
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4 weeks ago
#4
(Original post by mqb2766)
Sorry solutions are not allowed.
What can you determine / what is difficult for you?
Am I missing something? There doesn't seem any way to determine the actual position of the bottom right hand corner of the unshaded triangle, which would seem to make the problem impossible.
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#5
(Original post by DFranklin)
Am I missing something? There doesn't seem any way to determine the actual position of the bottom right hand corner of the unshaded triangle, which would seem to make the problem impossible.
Thanks, that thought crossed my mind too.
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4 weeks ago
#6
It is useful to notice that the triangle made from AB and the vertex closest to E is similar to the one created by the intersection of AE with DC the point C and also the vertex closest to E. Also the fact that the provided side lengths are said to be of squares.
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4 weeks ago
#7
(Original post by Darren888)
Thanks, that thought crossed my mind too.
Can you not set up the position as a variable, then equate areas to find it?
Or if for some reason, it's independent of the position, pick a value that makes the calc easy.
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#8
(Original post by mqb2766)
Can you not set up the position as a variable, then equate areas to find it?
Or if for some reason, it's independent of the position, pick a value that makes the calc easy.
What areas are you saying need to be equal?
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4 weeks ago
#9
(Original post by DFranklin)
What areas are you saying need to be equal?
If x was the position of the bottom unshaded triangle point
All could be worked out in terms of x (or independent of it).
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#10
(Original post by mqb2766)
Can you not set up the position as a variable, then equate areas to find it?
Or if for some reason, it's independent of the position, pick a value that makes the calc easy.
I tried to split the RHS triangle GFE, but to no avail in terms of coming up with the unknowns. My paper is a mess lol.

I just printed out a fresh copy. Will try again. 🙂
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4 weeks ago
#11
(Original post by mqb2766)
If x was the position of the bottom unshaded triangle point
All could be worked out in terms of x (or independent of it).
I think introducing a variable for an unmarked position, and giving an answer in terms of that variable is a bit of a stretch. particularly given the way the question is worded. Note that it not only shows you the lengths of ABCD and EFGH in the diagram, but expressly tells you in words as well when this isn't needed. It would seem very strange to jump from that "hand-holding" to "oh, but you need to decide for yourself to put in a variable for the position of the bottom RH corner and give an answer in terms of that variable".

[I did momentarily wonder if it was one of those "everything cancels" problems where changing x doesn't actually change the area, but that turns out not to be the case. I think they just messed up the diagram and the bottom right corner of the triangle is supposed to be at E to be honest].
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4 weeks ago
#12
(Original post by DFranklin)
I think introducing a variable for an unmarked position, and giving an answer in terms of that variable is a bit of a stretch. particularly given the way the question is worded. Note that it not only shows you the lengths of ABCD and EFGH in the diagram, but expressly tells you in words as well when this isn't needed. It would seem very strange to jump from that "hand-holding" to "oh, but you need to decide for yourself to put in a variable for the position of the bottom RH corner and give an answer in terms of that variable".

[I did momentarily wonder if it was one of those "everything cancels" problems where changing x doesn't actually change the area, but that turns out not to be the case. I think they just messed up the diagram and the bottom right corner of the triangle is supposed to be at E to be honest].
I'd agree, but the question was raised about whether it's solvable and that's one way to do it and I agree about not being able to choose a trivial position, but it's one thing for the OP to check.
So either
* Spot the magic bit of geometry
* Slog it out with areas
* Decide the problem is wrong

Edit - going down the variable position - area approach seems to give a quadratic in x. So not too bad.
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#13
(Original post by Darren888)
I tried to split the RHS triangle GFE, but to no avail in terms of coming up with the unknowns. My paper is a mess lol.

I just printed out a fresh copy. Will try again. 🙂
Useful to see roughly what you've tried. Also where does the problem come from?
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4 weeks ago
#14
(Original post by DFranklin)
Am I missing something? There doesn't seem any way to determine the actual position of the bottom right hand corner of the unshaded triangle, which would seem to make the problem impossible.
If you where to look in to my suggestion you can quite easily find the position of the bottom right vertex.
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4 weeks ago
#15
(Original post by mqb2766)
Edit - going down the variable position - area approach seems to give a quadratic in x. So not too bad.
Pretty sure the area should be a linear function of x FWIW.
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4 weeks ago
#16
(Original post by Fufuabc123)
It is useful to notice that the triangle made from AB and the vertex closest to E is similar to the one created by the intersection of AE with DC the point C and also the vertex closest to E. Also the fact that the provided side lengths are said to be of squares.
I don't see why. I can't see that we have *any* hard information about the horizontal position of the vertex in question other than it lying between C and E.
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4 weeks ago
#17
(Original post by mqb2766)
I'd agree, but the question was raised about whether it's solvable and that's one way to do it and I agree about not being able to choose a trivial position, but it's one thing for the OP to check.
So either
* Spot the magic bit of geometry
* Slog it out with areas
* Decide the problem is wrong

Edit - going down the variable position - area approach seems to give a quadratic in x. So not too bad.
I tend to agree with everything DFranklin said, but just for interest, a vector product approach yields a linear function in x.
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4 weeks ago
#18
(Original post by old_engineer)
I tend to agree with everything DFranklin said, but just for interest, a vector product approach yields a linear function in x.
Yes, that works (PRSOM). It's somewhat overkill here, but there's also a general formula for the area of a polygon given the vertices, and that formula is linear in every coordinate.

But I think the most straightforward approach is to calculate the 3 shaded areas - finding the area of a triangle and trapezium (where you know the base and heights) is pretty easy.
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4 weeks ago
#19
It would help to know some context of the question also.
Is it in a specific section that could be used to hint as to the working out process? -that would help identify whether there is information missing (ie, question is wrong)
or if, in context there are assumptions that can be made that we wouldn’t make out of context
or whether variables should be introduced as suggested.
2
4 weeks ago
#20
(Original post by GabiAbi84)
It would help to know some context of the question also.
Is it in a specific section that could be used to hint as to the working out process? -that would help identify whether there is information missing (ie, question is wrong)
or if, in context there are assumptions that can be made that we wouldn’t make out of context
or whether variables should be introduced as suggested.
Agreed. Even seeing what Q11/Q13 are would give an indication of what's expected.

[Although as I remarked recently - it does seem that there are an *awful* lot of badly written questions going around at the minute. I'm really glad I'm no longer of an age where I need to do aptitude tests, because the current ones seem pretty moronic].
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