vancouverron
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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
The total shaded area is 36 cm2, i.e. the area of the rectangular shape (2cm in width), plus the area of the parallelogram together is 36 cm2.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
If we need to rotate the shaded areas and make those horizontal and vertical - would the common shaded area remain same?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).
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...
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
@Vancouverron, I wish you had posted your result.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?
In Japan by grade six, students have a strong base in simple algebra.I avoided this method, as it seems likely to be more algebra than the students taking the test would know
@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?
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.In Japan by grade six, students have a strong base in simple algebra.