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

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==Problem==
 
==Problem==
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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?
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In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>.  What is the area of <math>AEDC</math>?
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(A) 13 (B) 13.5 (C) 14 (D) 14.5 (E) 15
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\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}<math>
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==Solution==
 
==Solution==
 
Let us use mass points:
 
Let us use mass points:
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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=(B) 13.5.
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Assign </math>B<math> mass </math>1<math>.  Thus, because </math>E<math> is the midpoint of </math>AB<math>, </math>A<math> also has a mass of </math>1<math>.  Similarly, </math>C<math> has a mass of </math>1<math>.  </math>D<math> and </math>E<math> each have a mass of </math>2<math> because they are between </math>B<math> and </math>C<math> and </math>A<math> and </math>B<math> respectively.  Note that the mass of </math>D<math> is twice the mass of </math>A<math>, so AP must be twice as long as </math>PD<math>.  PD has length </math>2<math>, so </math>AP<math> has length </math>4<math> and </math>AD<math> has length </math>6<math>.  Similarly, </math>CP<math> is twice </math>PE<math> and </math>PE=1.5<math>, so </math>CP=3<math> and </math>CE=4.5<math>.  Now note that triangle </math>PED<math> is a </math>3-4-5<math> right triangle with the right angle </math>DPE<math>.  This means that the quadrilateral </math>AEDC<math> is a kite.  The area of a kite is half the product of the diagonals, </math>AD<math> and </math>CE<math>.  Recall that they are </math>6<math> and </math>4.5<math> respectively, so the area of </math>AEDC<math> is </math>6*4.5/2=\boxed{\textbf{(B)} 13.5}<math>
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==Solution 2==
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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 2+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}$

Revision as of 23:16, 21 February 2013

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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