Difference between revisions of "2013 AMC 10B Problems/Problem 16"

Revision as of 18:03, 1 February 2015

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 $AEDC$ ?

$\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}15$ $[asy] pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("P",P,SSE); [/asy]$

Solution

Solution 1

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

Solution 3

From the solution above, we can find the length of the diagonals to be 6 and 4.5. Now, since AEDC is a trapezoid, we use the area formula to find that the total area is $\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{B}$

@@ Line 38: / Line 38: @@
 ===Solution 2===
 Note that triangle <math>DPE</math> is a right triangle, and that the four angles that have point <math>P</math> are all right angles. Using the fact that the centroid (<math>P</math>) divides each median in a <math>2:1</math> ratio, <math>AP=4</math> and <math>CP=3</math>. Quadrilateral <math>AEDC</math> is now just four right triangles. The area is <math>\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}</math>
+===Solution 3===
+From the solution above, we can find the length of the diagonals to be 6 and 4.5. Now, since AEDC is a trapezoid, we use the area formula to find that the total area is <math>\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{B}</math>
 == See also ==

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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