# Two Rectangles in a Rectangle

In the applet, ABCD, EFGH, and IFJH are all rectangles. Show that

the total of red areas equals the blue area.

the sum of the areas of EFGH and IFJH equals that of ABCD.

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Copyright © 1996-2018 Alexander BogomolnySince the diagonals of a rectangle (and even those of a parallelogram) meet at the common midpoint and since the two internal rectangles share one diagonal FH, their diagonals are equal and centers coincide. This point O is also the center of ABCD. Indeed, being the midpoint of FH it lies on the horizontal axis of ABCD. Being the midpoint of, say, EG, it lies on the vertical axis of ABCD.

We found that EG = IG and both pass through the center of ABCD. Hence, triangle GOJ is isosceles and

CG = BI,

such that BCGI is another rectangle and AIGD ie yet another one.

Obviously, Area(IFG) = Area(BCGI)/2 and Area(IGH) = Area(AIGD)/2. It follows that the area of the quadrilateral IFGH is half that of ABCD. The same of course is true for the quadrilateral EFJH:

Area(IFGH) = Area(EFJH) = Area(ABCD)/2.

We are in the framework of a variant of the Carpets Theorem. With _{1}_{1}

The second result is an immediate consequence of the first.

### References

- J. Konhauser, D. Velleman, S. Wagon,
*Which Way Did the Bicycle Go?*, MAA, 1996, #52

### Carpets Theorem

- The Carpets Theorem
- Carpets in a Parallelogram
- Carpets in a Quadrilateral
- Carpets in a Quadrilateral II
- Square Root of 2 is Irrational
- Carpets Theorem With Parallelograms
- Two Rectangles in a Rectangle
- Bisection of Yin and Yang
- Carpets in Hexagon
- Round Carpets
- A Property of Semicircles
- Carpets in Triangle
- Carpets in Triangle, II
- Carpets in Right Triangle
- Piecewise Carpets in Parallelogram

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Copyright © 1996-2018 Alexander Bogomolny69103240