This one has me stumped - entrance exam for junior high school in Japan

vancouverron

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“In the square above, the area of the shaded parts is 36 cm2. What is the length of one of the sides of the square?”

The above is an entrance-exam question from 6th grade to junior high school. I have not been able to solve it. Look forward to hearing your solution to this.
Vancouverron
 
View attachment 10675

“In the square above, the area of the shaded parts is 36 cm2. What is the length of one of the sides of the square?”

The above is an entrance-exam question from 6th grade to junior high school. I have not been able to solve it. Look forward to hearing your solution to this.
Vancouverron
Is it sum of the shaded parts or each of the shaded parts?
 
View attachment 10675

“In the square above, the area of the shaded parts is 36 cm2. What is the length of one of the sides of the square?”

The above is an entrance-exam question from 6th grade to junior high school. I have not been able to solve it. Look forward to hearing your solution to this.
Vancouverron

Presumably it's the total area of the shaded portion whose area is given.

I have a sort of trick solution that makes it very easy. We know that the area of a parallelogram is dependent only on the base and the height perpendicular to it, so if you slide the slanted strip down to the bottom, keeping the same vertical dimension but making it horizontal, the shaded area will be unchanged. Do the same with the vertical strip, and the picture is easier to work with.

If you know a little algebra, you can call the length of each side of the square x, and write an equation for the shaded area, then solve for x.

If you can't use any algebra, then find the area of the parallelogram at the intersection, then find the area of the remaining part of each "road", and from that find the "height" of those parts.

You will get essentially the same result algebraically if you work with the figure as given, though it will be a little more complicated.

Those are some hints. Please let us know what level of knowledge you have (in both algebra and geometry), what you have tried, and where you are stuck. This is certainly not a very easy problem at your level (or any level, really), but if we put together what you know, we should be able to help you finish your work.
 
Nice problem!

Actually, i started by printing it out and tried to rearrange the pieces into more familiar shapes ( just for fun).
I just wondered if there was a visual way of cutting through it.
Would like to see what the OP has tried out so far. We can probably build on it.
Tell us your ideas, however vague!
 
Presumably it's the total area of the shaded portion whose area is given.

I have a sort of trick solution that makes it very easy. We know that the area of a parallelogram is dependent only on the base and the height perpendicular to it, so if you slide the slanted strip down to the bottom, keeping the same vertical dimension but making it horizontal, the shaded area will be unchanged. Do the same with the vertical strip, and the picture is easier to work with.

If you know a little algebra, you can call the length of each side of the square x, and write an equation for the shaded area, then solve for x.

If you can't use any algebra, then find the area of the parallelogram at the intersection, then find the area of the remaining part of each "road", and from that find the "height" of those parts.

You will get essentially the same result algebraically if you work with the figure as given, though it will be a little more complicated.

Those are some hints. Please let us know what level of knowledge you have (in both algebra and geometry), what you have tried, and where you are stuck. This is certainly not a very easy problem at your level (or any level, really), but if we put together what you know, we should be able to help you finish your work.
If we need to rotate the shaded areas and make those horizontal and vertical - would the common shaded area remain same?

That is why I thought that the individual shaded areas were given (the common area would be counted twice then).
 
If we need to rotate the shaded areas and make those horizontal and vertical - would the common shaded area remain same?

That is why I thought that the individual shaded areas were given (the common area would be counted twice then).

Rotating would mess it up; but it's important that it is not the width, but the vertical distance across the slanted path, that is given as 2 inches. That's what suggested to me not rotating, but sliding (a shear transformation). That path retains the same area when the left end is slid down to the bottom, retaining the same height (actually base, in terms of the area formula), and so does the intersection. I probably didn't make my idea entirely clear; the result of the transformation I suggested is a figure with a 2-inch strip along the bottom and the right side that is shaded. The areas will be unchanged, but the problem is easier to think about.

However, I am not at all sure that any 6th grader could understand this. What they should be able to do is to see that the area of the shaded part is 2x + 2x - 2^2 (in one of several forms), which can be easily solved for x.
 
After reading some of the replies above I thought one way to solve this would be to calculate the overlapped area of the two sections. The parallelogram formed by the intersection turns into a square of 2cm by 2cm. Hence the overlap area becomes 2 x 2, i.e. 4cm2. The total area by the two paths = 36cm2 - 4cm2 = 32cm2. One side of the square would therefore be 5.66 cm. Any thoughts?

Just for grins: I am an ancient Electrical Engineer. As is evident by my question in the first place, I would have skipped this problem in a test environment...
 
This one has me stumped - entrance exam for junior high school in Japan - solved?

Rotating would mess it up; but it's important that it is not the width, but the vertical distance across the slanted path, that is given as 2 inches. That's what suggested to me not rotating, but sliding (a shear transformation). That path retains the same area when the left end is slid down to the bottom, retaining the same height (actually base, in terms of the area formula), and so does the intersection. I probably didn't make my idea entirely clear; the result of the transformation I suggested is a figure with a 2-inch strip along the bottom and the right side that is shaded. The areas will be unchanged, but the problem is easier to think about.

However, I am not at all sure that any 6th grader could understand this. What they should be able to do is to see that the area of the shaded part is 2x + 2x - 2^2 (in one of several forms), which can be easily solved for x.


OK, I think I finally got it figured out: Focus on the parallelogram created by the intersection of the two paths. This is the small parallelogram that includes the areas of the vertical path and the slanted path. As you can see in the attached drawing, the overlap-area is 4cm2. Given that, the area of the vertical path is 16cm2, which gives a length of 8cm for the sides of the square. Took me long enough to figure out...
Thank you all for hanging in there for this "fun" problem! Best wished for the season to all,
Vancouverron
 

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After reading some of the replies above I thought one way to solve this would be to calculate the overlapped area of the two sections. The parallelogram formed by the intersection turns into a square of 2cm by 2cm. Hence the overlap area becomes 2 x 2, i.e. 4cm2. The total area by the two paths = 36cm2 - 4cm2 = 32cm2. One side of the square would therefore be 5.66 cm. Any thoughts?

Just for grins: I am an ancient Electrical Engineer. As is evident by my question in the first place, I would have skipped this problem in a test environment...

I'm not sure what you mean by "the total area by the paths"; but there is no square here whose area is 32 cm2, so taking the square root of that is inappropriate. I think you just leaped a little too far in your thinking.

I did this: the total shaded area is 36 cm2; the overlap is 4cm2, so the remaining area is 32 cm2. Each path has half of this (16 cm2), so the remaining length of each path is 8 cm. That makes each side of the square 10 cm.

I'd love to know the curriculum in the schools from which students are expected to do this. It's a nice challenge.
 
View attachment 10675

“In the square above, the area of the shaded parts is 36 cm2. What is the length of one of the sides of the square?”

The above is an entrance-exam question from 6th grade to junior high school. I have not been able to solve it. Look forward to hearing your solution to this.
Vancouverron
OK, I think I finally got it figured out: Focus on the parallelogram created by the intersection of the two paths. This is the small parallelogram that includes the areas of the vertical path and the slanted path
@Vancouverron, I wish you had posted your result.
I think the the solution is really simple. Suppose that \(\displaystyle \bf{\mathcal{s}}\) is the length of the given square,
Now cut-out the shaded area from that square and reassemble the remaining four parts into a new square with sides of length \(\displaystyle {{\mathcal{s}}}-2\).

The area of the new square is \(\displaystyle (\mathcal{s} -2)^2\).
Thus \(\displaystyle (\mathcal{s} -2)^2+36=\mathcal{s}^2\) (area of the large square) which implies that \(\displaystyle \mathcal{s}=10\).
Do we agree?
 
@Vancouverron, I wish you had posted your result.
I think the the solution is really simple. Suppose that \(\displaystyle \bf{\mathcal{s}}\) is the length of the given square,
Now cut-out the shaded area from that square and reassemble the remaining four parts into a new square with sides of length \(\displaystyle {{\mathcal{s}}}-2\).

The area of the new square is \(\displaystyle (\mathcal{s} -2)^2\).
Thus \(\displaystyle (\mathcal{s} -2)^2+36=\mathcal{s}^2\) (area of the large square) which implies that \(\displaystyle \mathcal{s}=10\).
Do we agree?

I avoided this method, as it seems likely to be more algebra than the students taking the test would know; but the visual part of it is very nice. It's easy to see the sliding together of the four parts as long as you slide pairs together vertically first to make two rectangles with the same height, and then slide horizontally. I suppose then you could revert to my idea, since in effect you now have two 2-cm-wide borders on the small square, and can see that its area is 2x + 2x + 4 = 36, and so on (no quadratics needed, even briefly).
 
@Vancouverron, I wish you had posted your result.
I think the the solution is really simple. Suppose that \(\displaystyle \bf{\mathcal{s}}\) is the length of the given square,
Now cut-out the shaded area from that square and reassemble the remaining four parts into a new square with sides of length \(\displaystyle {{\mathcal{s}}}-2\).

The area of the new square is \(\displaystyle (\mathcal{s} -2)^2\).
Thus \(\displaystyle (\mathcal{s} -2)^2+36=\mathcal{s}^2\) (area of the large square) which implies that \(\displaystyle \mathcal{s}=10\).
Do we agree?

nice one pka!
 
In Japan by grade six, students have a strong base in simple algebra.
This is at least true for some kids. Back in the early sixties, when I was a senior in high school taking an AP calculus course, the strongest boy in the class, a freshman, was an exchange student from Japan. Admittedly, he was an unusually gifted boy, but he had been extremely well trained.
 
I got help to get into Purdue University because whites are one of the minorities in the Quantum program. Asians were the majority. And boy oh boy were they the curve breakers.

-Dan
 
alas - the light came on re This one has me stumped - entrance exam for junior high

pka: Elegant solution that I will need to study a bit to really get it.

However, and Dr. Peterson also mentioned it, at 6th grade the students did not yet have algebra, so the solution needed to be solved with simple geometry. They way I finally solved this: The slanted path (A1) has the same area as the vertical path (A2). The overlap area is 4cm2.
36cm2 = A1 + A2 - 4, each path = 20cm2, giving a length of 10cm per side.

Thank you all for your responses and your help.
 
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