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Difference between revisions of "2007 AMC 8 Problems/Problem 8"
m (→‎Problem: Fixed LaTeX.)
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In trapezoid <math>ABCD</math>, <math>AD</math> is perpendicular to <math>DC</math>,
 
In trapezoid <math>ABCD</math>, <math>AD</math> is perpendicular to <math>DC</math>,
−
<math>AD</math> = <math>AB</math> = <math>3</math>, and <math>DC</math> = <math>6</math>. In addition, <math>E</math> is on
+
<math>AD = AB = 3</math>, and <math>DC = 6</math>. In addition, <math>E</math> is on
 
<math>DC</math>, and <math>BE</math> is parallel to <math>AD</math>. Find the area of
 
<math>DC</math>, and <math>BE</math> is parallel to <math>AD</math>. Find the area of
 
<math>\triangle BEC</math>.
 
<math>\triangle BEC</math>.
  
 +
<center>
 
<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);
−
label("$6$", E, S);</asy>
+
label("$6$", E, S);
 +
</asy>
 +
</center>
  
−
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4.5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(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 ==
 
== Solution ==

Revision as of 05:17, 22 July 2021

Problem

In trapezoid $ABCD$, $AD$ is perpendicular to $DC$, $AD = AB = 3$, and $DC = 6$. In addition, $E$ is on $DC$, and $BE$ is parallel to $AD$. 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

We know that $ABED$ is a square with side length $3$. We subtract $DC$ and $DE$ to get the length of $EC$.

$EC = DC - DE = 6 - 3 = 3$

We are trying to find the area of $\triangle BEC$.

So, $\frac{1}{2} \cdot 3 \cdot 3 = \boxed{\textbf{(B)}\ 4.5}$

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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