2014 AMC 10A Problems/Problem 16
In rectangle , , , and points , , and are midpoints of , , and , respectively. Point is the midpoint of . What is the area of the shaded region?
Note that the region is a kite; hence its diagonals are perpendicular and it has area for diagonals of length and . Since as both and are midpoints of parallel sides of rectangle and , we let . Now all we need to do is to find .
Let the other two vertices of the kite be and with closer to than . This gives us . Now let . We thus find that the equation of is and that of is . Solving this system gives us , so the -coordinate of is ; in other words, is from . By symmetry, is also the same distance from , so as we have . Hence the area of the kite is .
Let the area of the shaded region be . Let the other two vertices of the kite be and with closer to than . Note that . The area of is and the area of is . We will solve for the areas of and in terms of x by noting that the area of each triangle is the length of the perpendicular from to and to respectively. Because the area of = based on the area of a kite formula, for diagonals of length and , . So each perpendicular is length . So taking our numbers and plugging them into gives us Solving this equation for gives us
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