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

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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>?
 
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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<math>\qquad\textbf{(A) }13\qquad\textbf{(B) }13.5\qquad\textbf{(C) }14\qquad\textbf{(D) }14.5\qquad\textbf{(E) }15</math>
 
 
<asy>
 
<asy>
 +
unitsize(0.2cm);
 
pair A,B,C,D,E,P;
 
pair A,B,C,D,E,P;
 
A=(0,0);
 
A=(0,0);
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dot(E);
 
dot(E);
 
dot(P);
 
dot(P);
−
label("A",A,NNW);
+
label("A",A,SW);
−
label("B",B,NNE);
+
label("B",B,SE);
−
label("C",C,ENE);
+
label("C",C,N);
−
label("D",D,ESE);
+
label("D",D,NE);
 
label("E",E,SSE);
 
label("E",E,SSE);
−
label("P",P,SSE);
+
label("P",P,SSW);
 
</asy>
 
</asy>
  
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<math>\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15</math>
  
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==Solution 1==
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==Solution 1 ( mass points) ==
 
Let us use mass points:
 
Let us use mass points:
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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>. Since the diagonals of quadrilaterals <math>AEDC</math>, <math>AD</math> and <math>CE</math>, are perpendicular, the area of <math>AEDC</math> is <math>\frac{6 \times 4.5}{2}=\boxed{\textbf{(B)} 13.5}</math>
+
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 <math>AP</math> 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>. Since the diagonals of quadrilaterals <math>AEDC</math>, <math>AD</math> and <math>CE</math>, are perpendicular, the area of <math>AEDC</math> is <math>\frac{6 \times 4.5}{2}=\boxed{\textbf{(B)} 13.5}</math>
  
 
==Solution 2==
 
==Solution 2==
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From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.
 
From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.
 +
 +
==Solution 5==
 +
We know that <math>[AEDC]=\frac{3}{4}[ABC]</math>, and <math>[ABC]=3[PAC]</math> using median properties. So Now we try to find <math>[PAC]</math>. Since <math>\triangle PAC\sim \triangle PDE</math>, then the side lengths of <math>\triangle PAC</math> are twice as long as <math>\triangle PDE</math> since <math>D</math> and <math>E</math> are midpoints. Therefore, <math>\frac{[PAC]}{[PDE]}=2^2=4</math>. It suffices to compute <math>[PDE]</math>. Notice that <math>(1.5, 2, 2.5)</math> is a Pythagorean Triple, so <math>[PDE]=\frac{1.5\times 2}{2}=1.5</math>. This implies <math>[PAC]=1.5\cdot 4=6</math>, and then <math>[ABC]=3\cdot 6=18</math>. Finally, <math>[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}</math>.
 +
 +
~CoolJupiter
 +
 +
==Solution 6==
 +
As from Solution 4, we find the area of <math>\triangle DPE</math> to be <math>\frac{3}{2}</math>. Because <math>DE = \frac{5}{2}</math>, the altitude perpendicular to <math>DE = \frac{6}{5}</math>. Also, because <math>DE || AC</math>, <math>\triangle ABC</math> is similar to <math>\triangle{DBE}</math> with side length ratio <math>2:1</math>, so <math>AC=5</math> and the altitude perpendicular to <math>AC = \frac{12}{5}</math>. The altitude of trapezoid <math>ACDE</math> is then <math>\frac{18}{5}</math> and the bases are <math>\frac{5}{2}</math> and <math>5</math>. So, we use the formula for the area of a trapezoid to find the area of <math>ACDE = \boxed{13.5}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 15:28, 30 October 2024

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 $AEDC$?

[asy] unitsize(0.2cm); 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,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,SSE); label("P",P,SSW); [/asy]

$\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15$

Solution 1 ( mass points)

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 $DPE$. Since the diagonals of quadrilaterals $AEDC$, $AD$ and $CE$, are perpendicular, the area of $AEDC$ is $\frac{6 \times 4.5}{2}=\boxed{\textbf{(B)} 13.5}$

Solution 2

Note that triangle $DPE$ is a right triangle, and that the four angles (angles $APC, CPD, DPE,$ and $EPA$) 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 that the lengths of the diagonals are $6$ and $4.5$. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is $\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{\textbf{(B)} 13.5}$

Solution 4

From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.

Solution 5

We know that $[AEDC]=\frac{3}{4}[ABC]$, and $[ABC]=3[PAC]$ using median properties. So Now we try to find $[PAC]$. Since $\triangle PAC\sim \triangle PDE$, then the side lengths of $\triangle PAC$ are twice as long as $\triangle PDE$ since $D$ and $E$ are midpoints. Therefore, $\frac{[PAC]}{[PDE]}=2^2=4$. It suffices to compute $[PDE]$. Notice that $(1.5, 2, 2.5)$ is a Pythagorean Triple, so $[PDE]=\frac{1.5\times 2}{2}=1.5$. This implies $[PAC]=1.5\cdot 4=6$, and then $[ABC]=3\cdot 6=18$. Finally, $[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}$.

~CoolJupiter

Solution 6

As from Solution 4, we find the area of $\triangle DPE$ to be $\frac{3}{2}$. Because $DE = \frac{5}{2}$, the altitude perpendicular to $DE = \frac{6}{5}$. Also, because $DE || AC$, $\triangle ABC$ is similar to $\triangle{DBE}$ with side length ratio $2:1$, so $AC=5$ and the altitude perpendicular to $AC = \frac{12}{5}$. The altitude of trapezoid $ACDE$ is then $\frac{18}{5}$ and the bases are $\frac{5}{2}$ and $5$. So, we use the formula for the area of a trapezoid to find the area of $ACDE = \boxed{13.5}$

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
All AMC 10 Problems and Solutions

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