Solution

We may make three observations:

1. Sides AD and of the big rectangle are equal.
2. BC also serves as the side of the small rectangle.
3. (Looking at AD,) Two small sides of a small rectangle fit exactly in the side of the big rectangle.

From #3, the small side of the big rectangle is the small side of the small rectangle, i.e., units. From #2 (and #1), the big side of the small rectangle equals 10 units. The area of a rectangle equals the product of its sides. Therefore, the area of a small rectangle equals ·10 = square units. small rectangles fit into the big one, making its area three times as large. It follows that the area of rectangle ABCD equals 3·50 = 150 unit².

Let's do this in a little more general way. Let x and y denote the small and the large dimensions of the small rectangle. This makes the area of the small rectangle and the area of the big rectangle 3xy. On the other hand, the small side of the big rectangle is whereas its big side measures . It follows that the area of rectangle ABCD is also given by 2x·(x + y). The two quantities are equal:

If x = 0, the problem degenerates into a case that requires no calculations. The big rectangles fills zero area as does a small rectangle. So assume x ≠ . This assumption allows us to divide both sides of the equation by x:

which shows that

For x = 5, y = and we are done.

References