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GMAT Prep - Coordinate System

Expert replies
by Blast » Sun Nov 23, 2008 6:12 pm
It probably looks harder than it actually is
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Source: — Problem Solving |

by lunarpower » Wed Nov 26, 2008 3:57 am
ok, i'm going to assume you don't need help with region Q.

with the other regions, there are two approaches that i think are easier than most others. here they are:

* use the formula for the area of a triangle, (1/2)(base)(height).
just be aware that "base" may not necessarily be horizontal, and that "height" may not necessarily be vertical.
for region P, the base is 4 (extending from (0, 1) to (4, 1)), and the height is 1 (easiest to visualize when drawn from (2, 2) downward, hitting perpendicular at (2, 1)). so area = (1/2)(4)(1) = 2.
for region S, the base is 3 (extending from (5, 1) to (5, 4)), and the height is 1 (easiest to visualize when drawn from (4, 3) rightward, hitting perpendicular at (4, 4)). so area = (1/2)(3)(1) = 1.5.
you can break R up into 2 triangles, each of which has base 2 and height 1.

* break P, R, and S up into "half rectangles".
for instance, split P** up by drawing a vertical line from (2, 2) to (2, 1).
the left half is exactly half of a 2x1 rectangle, so its area is 1.
the right half is exactly half of another 2x1 rectangle, so its area is 1.
the total area is 2.
you can do likewise with regions R and S.

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** "split P"! ha!
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Another way to solve for R

by prasadjoglekar » Sun Nov 30, 2008 7:13 pm
Another way to solve for R:

R is a square and therefore a rhombus. Area of a rhombus is (d1 * d2) / 2 [d1, d2 being the diagonals].

Each diagonal is 2, hence area is 2.
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