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In the figure, ABCD is a square, and all the dots are evenly spaced: each vertical or horizontal distance between two adjacent dots is 3 units. Find the area of the shaded region.

Re: In the figure, ABCD is a square, and all the dots are evenly spaced: [#permalink]

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30 Oct 2017, 09:46

1

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Bunuel wrote:

In the figure, ABCD is a square, and all the dots are evenly spaced: each vertical or horizontal distance between two adjacent dots is 3 units. Find the area of the shaded region.

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The dots are evenly spaced; distances between dots are equal. As long as the shapes' side lengths are made with the same number of dots, they are congruent.

"Move" some of the shaded parts, see diagram, left square.

Right hand square: Now the shaded area is half the square's area.

One side of the square = 4 dots * 3 units = 12 Square area = 12\(^2\) = 144

Re: In the figure, ABCD is a square, and all the dots are evenly spaced: [#permalink]

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31 Oct 2017, 12:32

I got 72, which is B. All I did was count the triangles. There are probably more efficient ways.
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I'd love to hear any feedback or ways to improve my problem solving. I make a lot of silly mistakes. If you've had luck improving on stupid mistakes, I'd love to hear how you did it.

Re: In the figure, ABCD is a square, and all the dots are evenly spaced: [#permalink]

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01 Nov 2017, 10:20

IMO B

Total area of the square is (3*4)^2= 144 If we counted the number of blue triangles on the left hand-side, we will realize they complete the blue ones on the left hand-side to make half of the total area of the square. Therefore, the shaded area is 144/2 = 72
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