What is the area of the region in which squares ABCD and EFGH overlap?

ShaliniShalini

If we look at S2: it says distance of point C from point E and F is same and is equal to 2 root2

It means C is the center of the square EFGH. We are also given that side of square EFGH is 4

So, we can easily get the side of the small square asked; since we know the size of it now(which is 2)

Hope this helps!

Hi all,

Can someone please explain why B alone is correct?

IMO B means that the EFGH square is hinged in C, but it still can rotate around it (thus giving different areas while superposing with ABCD, right?).

If EFGH is both blocked in translation and rotation (i.e., 1+2) then we can draw conclusions on the overlapping region --> Ans C

nick1816 has the best explanation above.

Follow his picture underneath his explanation.


Statement 2:

Because we are given that EC = CF = 2 * sqrt(2)

And we know that EFGH is a square with side length of 4

Triangle ECF must be a right triangle because the sides are in the Ratio of: 1 : 1 : sqrt(2)

Or

2 * sqrt(2) : 2 * sqrt(2) : 4

Now, because we know that the diagonal of any square is = (side) * sqrt(2) ——-> we know that the full diagonal EG = (4) * sqrt(2)


Since EC and EF are (1/2) of the full diagonal of Square EFGH and we know they meet at a 90 degree angle, it must be the case that point C is the exact center of square EFGH —— because of the Rule that the Two diagonals of a square are perpendicular Bisectors of each other.

This is the only inference we can make. The next part is the tricky part.

The natural trap is to think that, in the overlap region, we have a Square around its Diagonal EC. This would mean E would have to be the center of Square ABCD. However, we can not infer that.

Since the diagrams are not drawn to scale for Data Sufficiency questions, it could be that the figure is actually more like case 2 in Nick’s diagram.


Nick proves how the Area will always equal 4.

The easiest way to think about it is to imagine 2 identical squares that have identical side lengths of 4. Square ABCD stays stationary at all times.

We attach square EFGH such that the center is thumbtacked at Vertex C. We can spin square EFGH around like a wheel of fortune (a square wheel of fortune)

If you start out with the squares perfectly perpendicular to each other:

where E is also the center of square ABCD

which means that the overlap region cuts through the midpoints of the Sides of square ABCD

And finally this overlap region is a square itself

——-> it must be the case that EC is the diagonal of the Square Overlap region.

You will then find that the side length of that square is 2 and the Area of that Square Overlap region = (2) * (2) = 4


No matter how you spin square EFGH around it’s center of Point C, the region covered by the square will never change.

You can try attaching 2 smaller pieces of square paper to see this fact visually. No matter how you spin square EFGH around it’s center C, the Overlap Region’s Area will never change.

nick1816 just goes through and proves this fact by using similar triangles.

I believe, other than following that approach, the next best way is to “visualize” why statement 2 is sufficient and the Area will always be 4.

However, you can always prove this by following his work above.




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