Difference between revisions of "2007 AMC 8 Problems/Problem 8"

m (Problem)
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In trapezoid <math>ABCD</math>, <math>\overline{AD}</math> is perpendicular to <math>\overline{DC}</math>,
 
In trapezoid <math>ABCD</math>, <math>\overline{AD}</math> is perpendicular to <math>\overline{DC}</math>,
<math>AD = AB = 3</math>, and <math>DC = 6</math>. In addition, <math>E</math> is on
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<math>AD = AB = 3</math>, and <math>DC = 6</math>. In addition, <math>E</math> is on <math>\overline{DC}</math>, and <math>\overline{BE}</math> is parallel to <math>\overline{AD}</math>. Find the area of <math>\triangle BEC</math>.
<math>\overline{DC}</math>, and <math>\overline{BE}</math> is parallel to <math>\overline{AD}</math>. Find the area of
 
<math>\triangle BEC</math>.
 
 
 
<center>
 
 
<asy>
 
<asy>
 
defaultpen(linewidth(0.7));
 
defaultpen(linewidth(0.7));
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label("$6$", E, S);
 
label("$6$", E, S);
 
</asy>
 
</asy>
</center>
 
  
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18</math>
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18</math>
  
== Solution ==
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== Solution 1 (Area Formula for Triangles) ==
 
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Clearly, <math>ABED</math> is a square with side-length <math>3.</math> By segment subtraction, we have <math>EC = DC - DE = 6 - 3 = 3.</math>
We know that <math>ABED</math> is a square with side length <math>3</math>. We subtract <math>DC</math> and <math>DE</math> to get the length of <math>EC</math>.
 
  
<math>EC = DC - DE = 6 - 3 = 3</math>
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The area of <math>\triangle BEC</math> is <cmath>\frac12\cdot EC\cdot BE = \frac12\cdot3\cdot3 = \boxed{\textbf{(B)}\ 4.5}.</cmath>
  
We are trying to find the area of <math>\triangle BEC</math>.
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~Aplus95 (Solution)
  
So, <math>\frac{1}{2} \cdot 3 \cdot 3 = \boxed{\textbf{(B)}\ 4.5}</math>
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~MRENTHUSIASM (Revision)
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=7|num-a=9}}
 
{{AMC8 box|year=2007|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:03, 23 July 2021

Problem

In trapezoid $ABCD$, $\overline{AD}$ is perpendicular to $\overline{DC}$, $AD = AB = 3$, and $DC = 6$. In addition, $E$ is on $\overline{DC}$, and $\overline{BE}$ is parallel to $\overline{AD}$. Find the area of $\triangle BEC$. [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, NW); label("$3$", A--D, W); label("$3$", A--B, N); label("$6$", E, S); [/asy]

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$

Solution 1 (Area Formula for Triangles)

Clearly, $ABED$ is a square with side-length $3.$ By segment subtraction, we have $EC = DC - DE = 6 - 3 = 3.$

The area of $\triangle BEC$ is \[\frac12\cdot EC\cdot BE = \frac12\cdot3\cdot3 = \boxed{\textbf{(B)}\ 4.5}.\]

~Aplus95 (Solution)

~MRENTHUSIASM (Revision)

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 AJHSME/AMC 8 Problems and Solutions

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