Find area of region that is common to both squares.

Question

A 3-meter square and a 4-meter square overlap as shown in the diagram.D is the center of the 3-meter square. Find the area of the region DGFE.

I 've tried to form right triangles in such region but the only lenght i know is the altitude dropped from corner D.

You should ask yourself why this question is soluble. Does it matter that the large square is 4-metre square? What about if it was 5, 6, whatever? Does the relative angle of the squares matter? Why, or why not? — AmpleMimic, Aug 14 '15 at 13:45
i've already tried ,but i couldnt find the angle measures of the angles which are not vertices of the squares. — Nameless, Aug 14 '15 at 13:54
What would the area be if the angles were right-angles? Does area vary as a function of angle? — AmpleMimic, Aug 14 '15 at 14:02
Call the midpoint of the right side of the square $R$, and the midpoint of the bottom side of the square $B$. Compare the triangles $DGR$ and $BDE$. — Akiva Weinberger, Aug 14 '15 at 14:25
I don't think there is an answer for your question at this stage, because you have not defined even one coordinate of the bigger square, (The other three points don't touch the squares in graph), nor have you given any angle of intersection. — Aditya Agarwal, Aug 14 '15 at 14:44
Try rotating the big square, keeping one corner at D. Does the overlap area change? Why not? — jbuddenh, Aug 14 '15 at 15:09
By rotating the big square it seems like the overlap area is 1/4 of the area of the small square.But can someone confirm ? — Nameless, Aug 14 '15 at 15:20
See columbus8myhw's comment, which shows you how to rigorously prove it. — user21820, Aug 14 '15 at 15:59

score 0 · Accepted Answer · edited Aug 18 '15 at 15:15

The angel does not have an impact. Watching your drawing, the half of the small square is split into 3 shapes in which the overlap is constructed exactly from the two non overlapping. Let me know if this is not clear, and I will add a drawing.

I decided to add the drawing.

Edit:

After more thought I conclude that this is actually a general property of symmetry that may apply to higher dimensions.

In the case of the square, you have two symmetry lines, vertical, which cut a square into four identical areas sections. Rotating around the intersection of the two vertical lines, the center of the square, will maintain four sections of equal area. In your case you look really at only one of this sections, which may be derived from using two vertical rays emitting from the center of the square.

This property of splitting a 2D symmetric shape into sections of equal area may be applied to any shape. What is really nice that it can be applied to higher dimensions. For example if you take a cube and split it with three plans through the centers of its sides, you get eight small cubes and an intersection point. Now, if you rotate the three plans around this intersection point (the center of the cube) or rotate the cube about the three intersecting plans, you get volumetrically sections that are each equal 1/8 of the cube volume, no matter how the rotation is made about the center.

This approach to the problem removes the need of proving congruency.

Thanks for your help,i finally solved the problem by showing that triangles EKJ and EIH (according to your drawing) are congruent.One last question :how could you tell (without showing the congruence of the triangles) that the overlap area is formed by the other two non overlapping? — Nameless, Aug 14 '15 at 18:19
You could try to calculate directly the areas, but you might still need the congruency proof. I will further think about. Do you approve the answer? — Moti, Aug 14 '15 at 19:20
the diagram does not make it clear why the areas shaded are equivalent, could you perhaps also add the congruency ideas? — General ASWalter, Sep 28 '23 at 14:40

Find area of region that is common to both squares.

1 Answers1