I exhausted counting every the rectangles I might see, however that didn"t work. How do I approach this?  Go action by step.

First Picture: 1 rectangle

Second Picture: 2 extr rectangles. The tiny rectangle, which has actually been added and the large one, which consists of the two small rectangles.

You are watching: How many four sided figures are in this diagram

Third picture: The big rectangle. Then 2 rectangles, which contains 2 tiny linked rectangles. And also the tiny rectangle, which has been added

Fourth picture: only one tiny rectangle.

Fifth Picture: The rectangle, which includes the two little rectangle and also the little additional rectangle.

You walk on like this. Then amount the quantity of rectangles. re-publishing
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answered Jul 20 "14 in ~ 15:34 callculus42callculus42
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Each rectangle has actually two vertical lines and also two horizontal lines.

There are five vertical lines in the picture, we have the right to label castle 1, 2, 3, 4, 5.

If the leftmost edge is 1: then the top and bottom room uniquely determined, and it is easy to check out that 3 or 4 have to be the best edge. 2 options.

If the outward edge is 2: then the rightmost edge is 3 or 4 (2 choices), and in either case there space 3 horizontal segments that can serve together the top/bottom($\binom32 =3$ choices). Therefore this offers $2 \cdot 3 = 6$. 6 options.

If the leftmost edge is 3: If the rightmost leaf is 5 there is only one rectangle. If the rightmost sheet is 4, there space 5 horizontal segments for top and bottom, therefore $\binom52 = 10$ choices. For this reason 11 options.

If the outward edge is 4: climate the rightmost edge is 5, and there are four horizontal segments yielding $\binom42 = 6$ possible rectanges. 6 options

The full is 2 + 6 + 11 + 6 = 25.

See more: Comparing Two Things Using Like Or As, What Is A Simile And Metaphor

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answer Apr 8 "16 at 22:41 Alex ZornAlex Zorn
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