2014 AMC 10A Problems/Problem 16
Contents
Problem
In rectangle , , , and points , , and are midpoints of , , and , respectively. Point is the midpoint of . What is the area of the shaded region?
Solution 1
Denote . Then . Let the intersection of and be , and the intersection of and be . Then we want to find the coordinates of so we can find . From our points, the slope of is , and its -intercept is just . Thus the equation for is . We can also quickly find that the equation of is . Setting the equations equal, we have . Because of symmetry, we can see that the distance from to is also , so . Now the area of the kite is simply the product of the two diagonals over . Since the length , our answer is .
[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1);
// do not look pair X = (1/3,2/3); pair Y = (2/3,2/3);
draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed);
label("",A,NW); label("",B,NE); label("",C,SE); label("",D,SW); label("",E,E); label("",F,S); label("",G,W); label("",H,N); label("",Y,E); label("",X,W);
label("",(0.25,0),S);
label("",(0.75,0),S);
label("",(1,0.5),E);
label("",(1,1.5),E);
[/asy]
Solution 2
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
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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