I tired counting all the rectangles I could see, but that didn"t work. How do I approach this?
Go step by step.
First Picture: 1 rectangle
Second Picture: 2 additional rectangles. The small rectangle, which has been added and the big one, which contains the two small rectangles.
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Third picture: The big rectangle. Then two rectangles, which contains 2 small linked rectangles. And the small rectangle, which has been added
Fourth picture: Only one small rectangle.
Fifth Picture: The rectangle, which contains the two small rectangle and the small additional rectangle.
You go on like this. Then sum the amount of rectangles.
answered Jul 20 "14 at 15:34
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Each rectangle has two vertical lines and two horizontal lines.
There are five vertical lines in the picture, we can label them 1, 2, 3, 4, 5.
If the leftmost edge is 1: Then the top and bottom are uniquely determined, and it is easy to see that 3 or 4 must be the right edge. 2 options.
If the leftmost edge is 2: Then the rightmost edge is 3 or 4 (2 choices), and in either case there are 3 horizontal segments that can serve as the top/bottom($inom32 =3$ choices). So this gives $2 cdot 3 = 6$. 6 options.
If the leftmost edge is 3: If the rightmost edge is 5 there is only one rectangle. If the rightmost edge is 4, there are 5 horizontal segments for top and bottom, so $inom52 = 10$ choices. Hence 11 options.
If the leftmost edge is 4: Then the rightmost edge is 5, and there are four horizontal segments yielding $inom42 = 6$ possible rectanges. 6 options
The total is 2 + 6 + 11 + 6 = 25.
See more: Comparing Two Things Using Like Or As, What Is A Simile And Metaphor
answered Apr 8 "16 at 22:41
Alex ZornAlex Zorn
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