Difference between revisions of "2005 AMC 8 Problems/Problem 13"
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| − | + | == Problem == | |
| + | The area of polygon <math> ABCDEF</math> is 52 with <math> AB=8</math>, <math>BC=9</math> and <math>FA=5</math>. What is <math> DE+EF</math>? | ||
| + | |||
| + | <asy> | ||
| + | pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); | ||
| + | draw(a--b--c--d--e--f--cycle); | ||
| + | draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); | ||
| + | draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); | ||
| + | draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); | ||
| + | draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); | ||
| + | draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); | ||
| + | label("$A$", a, NW); | ||
| + | label("$B$", b, NE); | ||
| + | label("$C$", c, SE); | ||
| + | label("$D$", d, SW); | ||
| + | label("$E$", e, SW); | ||
| + | label("$F$", f, SW); | ||
| + | label("5", (0,6.5), W); | ||
| + | label("8", (4,9), N); | ||
| + | label("9", (8, 4.5), E); | ||
| + | </asy> | ||
| + | |||
| + | <math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math> | ||
| + | |||
| + | == Solution == | ||
| + | Notice that <math>AF + DE = BC</math>, so <math>DE=4</math>. Let <math>O</math> be the intersection of the extensions of <math>AF</math> and <math>DC</math>, which makes rectangle <math>ABCO</math>. The area of the polygon is the area of <math>FEDO</math> subtracted from the area of <math>ABCO</math>. | ||
| + | |||
| + | <cmath>\text{Area} = 52 = 8 \cdot 9- EF \cdot 4</cmath> | ||
| + | |||
| + | Solving for the unknown, <math>EF=5</math>, therefore <math>DE+EF=4+5=\boxed{\textbf{(C)}\ 9}</math>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | Similar to solution 1, <math>AF + DE = BC</math>, so <math>DE=4</math>. Split the polygon into two rectangles by extending the <math>DE</math> so it intersects <math>AB</math>. Let's say the length of <math>FE</math> is equal to <math>x</math>. We can form the equation: <cmath>5x + 9(8-x) = 52</cmath>. We see that <math>x = 5</math>, so we can add <math>5 + 4 = \boxed{\textbf{(C)}\ 9}</math>. | ||
| + | - ArjunBhumula | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/abSgjn4Qs34?t=1908 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | == See Also == | ||
| + | {{AMC8 box|year=2005|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:37, 5 May 2025
Problem
The area of polygon 
is 52 with 
, 
and 
. What is 
?
![[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label("$A$", a, NW); label("$B$", b, NE); label("$C$", c, SE); label("$D$", d, SW); label("$E$", e, SW); label("$F$", f, SW); label("5", (0,6.5), W); label("8", (4,9), N); label("9", (8, 4.5), E); [/asy]](http://latex.artofproblemsolving.com/1/a/9/1a9b3504d94fa699730b91e95cc71d2a3c84db6b.png)
Solution
Notice that 
, so 
. Let 
be the intersection of the extensions of 
and 
, which makes rectangle 
. The area of the polygon is the area of 
subtracted from the area of .
![\[\text{Area} = 52 = 8 \cdot 9- EF \cdot 4\]](http://latex.artofproblemsolving.com/c/f/c/cfc8abb5efa3c6ac223b72d3961eea22b85e9e92.png)
Solving for the unknown, 
, therefore 
.
Solution 2
Similar to solution 1, , so . Split the polygon into two rectangles by extending the 
so it intersects 
. Let's say the length of 
is equal to 
. We can form the equation: ![\[5x + 9(8-x) = 52\]](http://latex.artofproblemsolving.com/5/0/d/50d03bf56e6a486d4d80ba2901f2e31a77516f56.png)
. We see that 
, so we can add 
.
- ArjunBhumula
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=1908
~ pi_is_3.14
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
