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

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== Problem ==
 
== Problem ==
  
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In trapezoid <math>ABCD</math>, <math>AD</math> is perpendicular to <math>DC</math>,
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In trapezoid <math>ABCD</math>, <math>\overline{AD}</math> is perpendicular to <math>\overline{DC}</math>,
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<math>AD</math> = <math>AB</math> = <math>3</math>, and <math>DC</math> = <math>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>.
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<math>DC</math>, and <math>BE</math> is parallel to <math>AD</math>. Find the area of
 
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<math>\triangle BEC</math>.
 
−
 
 
 
<asy>
 
<asy>
 
defaultpen(linewidth(0.7));
 
defaultpen(linewidth(0.7));
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label("$3$", A--D, W);
 
label("$3$", A--D, W);
 
label("$3$", A--B, N);
 
label("$3$", A--B, N);
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label("$6$", E, S);</asy>
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label("$6$", E, S);
 +
</asy>
 +
 
 +
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18</math>
  
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<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4.5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 18</math>
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== Solution 1 (Area Formula for Triangles) ==
 +
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>
  
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== Solution ==
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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>
 +
~Aplus95 (Solution)
  
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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>.
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~MRENTHUSIASM (Revision)
  
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<math>EC = DC - DE = 6 - 3 = 3</math>
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== Solution 2 (Area Subtraction) ==
 +
Clearly, <math>ABED</math> is a square with side-length <math>3.</math>
  
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We are trying to find the area of <math>\triangle BEC</math>.
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Let the brackets denote areas. We apply area subtraction to find the area of <math>\triangle BEC:</math>
 +
<cmath>\begin{align*}
 +
[BEC]&=[ABCD]-[ABED] \\
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&=\frac{AB+CD}{2}\cdot AD - AB^2 \\
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&=\frac{3+6}{2}\cdot 3 - 3^2 \\
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&=\boxed{\textbf{(B)}\ 4.5}.
 +
\end{align*}</cmath>
 +
~MRENTHUSIASM
  
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So, <math>\frac{1}{2} * 3 * 3 = 4.5</math>, <math>\boxed{B}</math>
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==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=omFpSGMWhFc
  
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<center>[[Image:AMC8_2007_8S.png]]</center>
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==Video Solution by WhyMath==
 +
https://youtu.be/Qdbpdc-Khg4
  
 
==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}}

Latest revision as of 16:37, 28 October 2024

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)

Solution 2 (Area Subtraction)

Clearly, $ABED$ is a square with side-length $3.$

Let the brackets denote areas. We apply area subtraction to find the area of $\triangle BEC:$ \begin{align*} [BEC]&=[ABCD]-[ABED] \\ &=\frac{AB+CD}{2}\cdot AD - AB^2 \\ &=\frac{3+6}{2}\cdot 3 - 3^2 \\ &=\boxed{\textbf{(B)}\ 4.5}. \end{align*} ~MRENTHUSIASM

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=omFpSGMWhFc

Video Solution by WhyMath

https://youtu.be/Qdbpdc-Khg4

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
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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