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 $AEDC$ ?

$\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}$ $[asy] pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("P",P,SSE); [/asy]$

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 $\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}$

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2013 AMC 10B Problems/Problem 16

Problem

Solution

Solution 2