I worn down counting all the rectangles I can see, yet that didn"t work. How do I technique this?

Go action by step.

**First Picture**: 1 rectangle

**Second Picture**: 2 extr rectangles. The small rectangle, which has been included and the huge one, which includes the two small rectangles.

You are watching: How many four-sided figures

**Third picture**: The large rectangle. Then two rectangles, which includes 2 little linked rectangles. And the little rectangle, which has actually been added

**Fourth picture**: only one tiny rectangle.

**Fifth Picture**: The rectangle, which includes the two tiny rectangle and the tiny additional rectangle.

You go on favor this. Then sum the quantity of rectangles.

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answer Jul 20 "14 in ~ 15:34

callculus42callculus42

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Each rectangle has actually two upright lines and two horizontal lines.

There are five vertical present in the picture, we deserve to label lock 1, 2, 3, 4, 5.

**If the outward edge is 1:** then the top and also bottom room uniquely determined, and also it is simple to see that 3 or 4 need to be the best edge. **2 options**.

**If the leftmost edge is 2:** then the rightmost edge is 3 or 4 (2 choices), and also in either situation there are 3 horizontal segments that can serve as the top/bottom($inom32 =3$ choices). So this provides $2 cdot 3 = 6$. **6 options**.

**If the leftmost edge is 3:** If the rightmost leaf is 5 there is just one rectangle. If the rightmost sheet is 4, there are 5 horizontal segments for top and bottom, for this reason $inom52 = 10$ choices. Therefore **11 options**.

**If the leftmost edge is 4:** then the rightmost leaf is 5, and also there are four horizontal segments yielding $inom42 = 6$ feasible rectanges. **6 options**

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

See more: Are Rabbits Herbivores Carnivores Or Omnivores ? Are Rabbits Herbivores, Omnivores Or Carnivores

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reply Apr 8 "16 in ~ 22:41

Alex ZornAlex Zorn

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