2013 AMC 10B Problems/Problem 16

Problem

In triangle $ABC$ , medians $AD$ and $CE$ intersect at $P$ , $PE=1.5$ , $PD=2$ , and $DE=2.5$ . What is the area of $AEDC$ ?

\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)} $==Solution== Let us use mass points: Assign$ B $mass$ 1 $. Thus, because$ E $is the midpoint of$ AB $,$ A $also has a mass of$ 1 $. Similarly,$ C $has a mass of$ 1 $.$ D $and$ E $each have a mass of$ 2 $because they are between$ B $and$ C $and$ A $and$ B $respectively. Note that the mass of$ D $is twice the mass of$ A $, so AP must be twice as long as$ PD $. PD has length$ 2 $, so$ AP $has length$ 4 $and$ AD $has length$ 6 $. Similarly,$ CP $is twice$ PE $and$ PE=1.5 $, so$ CP=3 $and$ CE=4.5 $. Now note that triangle$ PED $is a$ 3-4-5 $right triangle with the right angle$ DPE $. This means that the quadrilateral$ AEDC $is a kite. The area of a kite is half the product of the diagonals,$ AD $and$ CE $. Recall that they are$ 6 $and$ 4.5 $respectively, so the area of$ AEDC $is$ 6*4.5/2=\boxed{\textbf{(B)} 13.5} $==Solution 2== Note that triangle$ DPE $is a right triangle, and that the four angles that have point$ P $are all right angles. Using the fact that the centroid ($ P $) divides each median in a$ 2:1 $ratio,$ AP=4 $and$ CP=3 $. Quadrilateral$ AEDC $is now just four right triangles. The area is$ \frac{4\cdot 2+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}$

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2013 AMC 10B Problems/Problem 16

Problem