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.

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

**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.

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Folshort

answered Jul 20 "14 at 15:34

callculus42callculus42

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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

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Folshort

answered Apr 8 "16 at 22:41

Alex ZornAlex Zorn

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