2007 AMC 8 Problems/Problem 8: Difference between revisions
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We are trying to find the area of <math>\triangle BEC</math>. | We are trying to find the area of <math>\triangle BEC</math>. | ||
So, <math>\frac{1}{2} \cdot 3 \cdot 3 = \boxed{\textbf{(B)}\ | So, <math>\frac{1}{2} \cdot 3 \cdot 3 = \boxed{\textbf{(B)}\ 4.5}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=7|num-a=9}} | {{AMC8 box|year=2007|num-b=7|num-a=9}} | ||
Revision as of 20:02, 24 December 2012
Problem
In trapezoid 
, 
is perpendicular to 
,
= 
= 
, and = 
. In addition, 
is on
, and 
is parallel to . Find the area of
.
![[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]](http://latex.artofproblemsolving.com/b/4/e/b4ebd938ad5c620ff045bf0311f56ffd33605c3b.png)
Solution
We know that 
is a square with side length . We subtract and 
to get the length of 
.
We are trying to find the area of .
So, 
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 | ||
