Difference between revisions of "2011 AMC 8 Problems/Problem 13"
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+ | ==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); | ||
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<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== | ||
+ | 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> | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=mYn6tNxrWBU | ||
+ | |||
+ | ~==SpreadTheMathLove== | ||
==See Also== | ==See Also== | ||
− | {{AMC8 box|year=2011|num-b= | + | {{AMC8 box|year=2011|num-b=12|num-a=14}} |
+ | {{MAA Notice}} |
Latest revision as of 10:00, 13 January 2024
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?
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
Solution 2
The length of BP is 5. the ratio of the areas is
-Megacleverstarfish15
Video Solution
https://www.youtube.com/watch?v=mYn6tNxrWBU
~==SpreadTheMathLove==
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 |
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