Difference between revisions of "2011 AMC 8 Problems/Problem 13"
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==Solution== | ==Solution== | ||
The overlap length is <math>2(15)-25=5</math>, so the shaded area is <math>5 \cdot 15 =75</math>. The area of the whole shape is <math>25 \cdot 15 = 375</math>. The fraction <math>\dfrac{75}{375}</math> reduces to <math>\dfrac{1}{5}</math> or 20%. Therefore, the answer is <math>\boxed{ \textbf{(C)}\ \text{20} }</math> | The overlap length is <math>2(15)-25=5</math>, so the shaded area is <math>5 \cdot 15 =75</math>. The area of the whole shape is <math>25 \cdot 15 = 375</math>. The fraction <math>\dfrac{75}{375}</math> reduces to <math>\dfrac{1}{5}</math> or 20%. Therefore, the answer is <math>\boxed{ \textbf{(C)}\ \text{20} }</math> | ||
| + | |||
==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=mYn6tNxrWBU | https://www.youtube.com/watch?v=mYn6tNxrWBU | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=12|num-a=14}} | {{AMC8 box|year=2011|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:10, 17 December 2023
Problem
Two congruent squares, 
and 
, have side length 
. They overlap to form the by 
rectangle 
shown. What percent of the area of rectangle is shaded?
![[asy] filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); label("D",(0,0),S); label("R",(25,0),S); label("Q",(25,15),N); label("A",(0,15),N); filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black); label("S",(10,0),S); label("C",(15,0),S); label("B",(15,15),N); label("P",(10,15),N);[/asy]](http://latex.artofproblemsolving.com/2/3/6/236d17f1bb6b7c18318444d3cce64f632976b385.png)
Solution
The overlap length is 
, so the shaded area is 
. The area of the whole shape is 
. The fraction 
reduces to 
or 20%. Therefore, the answer is 
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=mYn6tNxrWBU
See Also
| 2011 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
