I tired counting all the rectangles I can check out, however that didn"t work. How perform I technique this?

Go action by step.

**First Picture**: 1 rectangle

**2nd Picture**: 2 extra rectangles. The little rectangle, which has actually been included and also the big one, which includes the two small rectangles.

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

**4th picture**: Only one small rectangle.

**Fifth Picture**: The rectangle, which has the 2 little rectangle and the small extra rectangle.

You go on prefer this. Then amount the amount of rectangles.

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answered Jul 20 "14 at 15:34

callculuscallculus

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Each rectangle has two vertical lines and also 2 horizontal lines.

Tbelow are five vertical lines in the photo, we can label them 1, 2, 3, 4, 5.

**If the lefta lot of edge is 1:** Then the top and also bottom are uniquely established, and it is basic to check out that 3 or 4 must be the right edge. **2 options**.

**If the leftthe majority of edge is 2:** Then the rightmost edge is 3 or 4 (2 choices), and in either instance tright here are 3 horizontal segments that can serve as the top/bottom($inom32 =3$ choices). So this offers $2 cdot 3 = 6$. **6 options**.

**If the lefta lot of edge is 3:** If the rightthe majority of edge is 5 there is only one rectangle. If the rightthe majority of edge is 4, tbelow are 5 horizontal segments for peak and bottom, so $inom52 = 10$ choices. Hence **11 options**.

**If the leftmost edge is 4:** Then the rightmany edge is 5, and also tright here are four horizontal segments yielding $inom42 = 6$ possible rectanges. **6 options**

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

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answered Apr 8 "16 at 22:41

Alex ZornAlex Zorn

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