Shaded area have already explained many of these issues. The trick is to calculate this figure as addition or subtraction of figures known.
Basically, if we look at the figure, it may be the sum of three squares least a triangle, or half of a rectangle least two small rectangles.
If we interpret as the sum of squares less than a triangle, we could calculate the area of \u200b\u200bsquares (16 + 9 + 1 = 26) and subtract the area of \u200b\u200bthe triangle (base of 4 + 3 + 1 = 8, height 4), which would match up base 2, ie, 26 - 8 * 4 / 2 = 26 - 16 = 10.
If we interpret it as a medium rectangle least two small rectangles, notes that the two small rectangles measure both 3 by 1, ie, the two have area 6, while the large rectangle measuring 8 by 4, ie, has area 32 . Its half is area 16, and if we take 6, is exactly 10, as before.
Another resource could be grid-paper and count squares, matching those that are cut with others who also are in the same way, but would be harder to see with precision.
If you know Pick's theorem, a graph can count the points that lie within and are on the edge of the figure and apply the formula (interior - edge / 2 - 1). Works where the vertices are on the grid.
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