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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>?
  
−
<math>\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}</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);
 
B=(80,0);
 
B=(80,0);
 
C=(20,40);
 
C=(20,40);
−
D=(50,30);
+
D=(50,20);
 
E=(40,0);
 
E=(40,0);
 
P=(33.3,13.3);
 
P=(33.3,13.3);
Line 23: Line 23:
 
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>
  
−
==Solution==
+
<math>\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15</math>
 +
 
 +
==Solution 1==
 
Let us use mass points:
 
Let us use mass points:
−
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>
+
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==
−
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>
+
Note that triangle <math>DPE</math> is a right triangle, and that the four angles (angles <math>APC, CPD, DPE,</math> and <math>EPA</math>) 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 that the lengths of the diagonals are <math>6</math> and <math>4.5</math>. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is <math>\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{\textbf{(B)} 13.5}</math>
 +
 
 +
==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 <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 ==
 +
{{AMC10 box|year=2013|ab=B|num-b=15|num-a=17}}
 +
 
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Problems]]
 +
{{MAA Notice}}

Revision as of 18:33, 11 January 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

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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