Difference between revisions of "2007 AMC 8 Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | In trapezoid <math>ABCD</math>, <math>AD</math> is perpendicular to <math>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 | + | <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>DC</math>, and <math>BE</math> is parallel to <math>AD</math>. Find the area of | ||
| − | <math>\triangle BEC</math>. | ||
| − | |||
| − | |||
<asy> | <asy> | ||
defaultpen(linewidth(0.7)); | defaultpen(linewidth(0.7)); | ||
| Line 21: | Line 17: | ||
label("$6$", E, S); | label("$6$", E, S); | ||
</asy> | </asy> | ||
| − | |||
<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 == | + | == 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> | ||
| + | |||
| + | 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) | ||
| + | |||
| + | ~MRENTHUSIASM (Revision) | ||
| − | + | == Solution 2 (Area Subtraction) == | |
| + | Clearly, <math>ABED</math> is a square with side-length <math>3.</math> | ||
| − | <math> | + | Let the brackets denote areas. We apply area subtraction to find the area of <math>\triangle BEC:</math> |
| + | <cmath>\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*}</cmath> | ||
| + | ~MRENTHUSIASM | ||
| − | + | ==Video Solution by SpreadTheMathLove== | |
| + | https://www.youtube.com/watch?v=omFpSGMWhFc | ||
| − | + | ==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}} | {{MAA Notice}} | ||
Latest revision as of 16:37, 28 October 2024
Contents
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/d/c/c/dccbcc1d762d975a6d3a54f56433f5da50a2502d.png)
Solution 1 (Area Formula for Triangles)
Clearly, 
is a square with side-length 
By segment subtraction, we have 
The area of is ![\[\frac12\cdot EC\cdot BE = \frac12\cdot3\cdot3 = \boxed{\textbf{(B)}\ 4.5}.\]](http://latex.artofproblemsolving.com/1/5/e/15e2685d98ebf59ef4397dfc6c5ea113770772fa.png)
~Aplus95 (Solution)
~MRENTHUSIASM (Revision)
Solution 2 (Area Subtraction)
Clearly, is a square with side-length
Let the brackets denote areas. We apply area subtraction to find the area of 
![\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*}](http://latex.artofproblemsolving.com/9/8/c/98c91fef6cf3f20a4dd54765f0ea2b5c168f135d.png)
~MRENTHUSIASM
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
Video Solution by WhyMath
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
