2011 AMC 8 Problems/Problem 13: Difference between revisions
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==Problem== | ==Problem== | ||
Two congruent squares, <math>ABCD</math> and <math>PQRS</math>, have side length <math>15</math>. They overlap to form the <math>15</math> by <math>25</math> rectangle <math>AQRD</math> shown. What percent of the area of rectangle <math>AQRD</math> is shaded? | Two congruent squares, <math>ABCD</math> and <math>PQRS</math>, have side length <math>15</math>. They overlap to form the <math>15</math> by <math>25</math> rectangle <math>AQRD</math> shown. What percent of the area of rectangle <math>AQRD</math> is shaded? | ||
<asy> | <asy> | ||
filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); | filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); | ||
| Line 14: | Line 15: | ||
<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25 </math> | <math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25 </math> | ||
==Solution 2== | |||
The length of BP is 5. the ratio of the areas is <math>\dfrac{5\cdot 15}{25\cdot 15}=\dfrac{5}{25}=20\% </math> | |||
-Megacleverstarfish15 | |||
==Solution 3(similar to Solution 1)== | |||
To find the overlap length, we do the total length of the squares and subtract <math>25</math>(side length of figure). <math>(15 + 15) - 25 = 5</math>, so the overlap length is <math>5</math>. To find what percentage of <math>AQRD</math> is shaded, we divide the shaded part by the area of the <math>AQRD</math>, so the percentage is <math>\dfrac{15 \cdot 5}{15 \cdot 25}</math> = <math>\dfrac{5}{25}</math> = <math>\dfrac{1}{5}</math> = <math>\dfrac{20}{100}</math> = <math>20</math>%, so the answer is <math>\boxed{ \textbf{(C)}\ \text{20} }</math>. | |||
~NXC | |||
==Video Solution== | |||
https://www.youtube.com/watch?v=mYn6tNxrWBU | |||
~==SpreadTheMathLove== | |||
==Video Solution by WhyMath== | |||
https://youtu.be/VLS29yiMHSw | |||
==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}} | |||
Latest revision as of 13:50, 12 January 2025
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 2
The length of BP is 5. the ratio of the areas is 
-Megacleverstarfish15
Solution 3(similar to Solution 1)
To find the overlap length, we do the total length of the squares and subtract (side length of figure). 
, so the overlap length is 
. To find what percentage of is shaded, we divide the shaded part by the area of the , so the percentage is 
= 
= 
= 
= 
%, so the answer is 
.
~NXC
Video Solution
https://www.youtube.com/watch?v=mYn6tNxrWBU
~==SpreadTheMathLove==
Video Solution by WhyMath
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
